LIMIT OF FUNCTIONS
Intuitive Motivation for Limits
Formal Definition (Epsilon – Delta)
Theorems on Limits
One – Sided Limits
Infinite Limits
Limits at Infinity
EDUC 612 - CALCULUS FOR MATHEMATICS TEACHERS
Saturday (2:00 – 5:30)
SUBMITTED BY:
GROUP 1 (SECTION 1)
Abarquez, Fredelyn A.
Abendan, Nina Fatima V.
Algar, Ervie L.
Anunciado, Donna Kristina G.
Armenta, Ruth B.
Dionaldo, Arnel B.
INTUITIVE MOTIVATION FOR LIMITS
I. Introduction
The limit wonders, “If you can see everything except a single value, what do you think is
there?”.
When our prediction is consistent and improves the closer we look, we feel confident in
it. And if the function behaves smoothly, like most real-world functions do, the limit is
where the missing point must be.
The limit was a certain value that quantities approach closer and closer
to but never coincide with or exceed. The expression "approaching" implied "not to be
equal" and "not to go beyond.” Limits are a strategy for making confident predictions.
II. Learning outcomes:
By the end of the lesson, students are able to:
•
•
•
Define limits
Understand the concepts of limits
Solve simple limits
III. Pre – test
Complete the table and use the result to estimate the limit.
1. lim
𝑥−2
𝑥→2 𝑥 2 −4
x
f(x)
1.9
2. Evaluate.
lim (2𝑥 2 − 3𝑥 + 4)
𝑛→5
3. lim
𝑥
𝑥→10 2
𝑥−2
4. lim 𝑥+1
𝑥→1
1.99
1.999
2.001
2.01
2.1
IV. Content Inputs and Discussions
Limits is one of the most fundamental concepts of calculus. The foundation of calculus
was not entirely solid during the time of Leibniz and Newton, but later developments on
the concept, particularly the
definition by Cauchy, Weierstrass and other
mathematicians established its firm foundation. In the discussion below, I shall introduce
the concept of limits intuitively as it appears in common problems.
Circumference and Limits
If we are going to approximate the circumference of a circle using the perimeter of an
inscribed polygon, even without computation, we can observe that as the number of
sides of the polygon increases, the better the approximation. In fact, we can make the
perimeter of the polygon as close as we please to the circumference of the circle by
choosing a sufficiently large number of sides. Notice that no matter how large the
number of sides our polygon has, its perimeter will never exceed or equal the
circumference of the circle.
Figure 1 – As the number of side of the polygons increases, its perimeter gets closer to
the circumference of the circle.
In a more technical term, we say that the limit of the perimeter of the inscribed polygon
as the number of its sides increases without bound (or as the number of sides of the
inscribed polygon approaches infinity) is equal to the circumference of the circle. In
symbol, if we let n be the number of sides of the inscribed polygon,
be the perimeter
of a polygon with n sides, and C be the circumference of the circle, we can say that the
limit of
as
n is equal to C. Compactly, we can write
.
Functions and Limits
Consider the function
where x is a natural number. Calculating the values of
the function using the first 20 natural numbers and plotting the points in the -plane,
we arrive at the table and the graph in Figure 2.
Figure 2.
First, we see that as the value of x increases, the value of f(x) decreases and
approaches 0. Furthermore, we can make the value of f(x) as close to 0 as we please
by choosing a sufficiently large x. We also notice that no matter how large the value of x
is, the value of f(x) will never reach 0.
Hence, we say that the limit of
as the value of x increases without bound is
equal to 0, or equivalently the limit of
symbol, we write the limit of
as x approaches infinity is equal to 0. In
as
or more compactly the
Tangent line and Limits
Recall that the slope of a line is its “rise” over its “run”. The formula of slope m of a line
is
, given two points with coordinates (x1,y1) and (x2,y2). One of the
famous ancient problems in mathematics was the tangent problem, which is getting the
slope of a line tangent to a function at a point. In the Figure 3, line n is tangent to the
function f at point P.
Figure 3 – Line n is tangent to the function f at point P.
.
If we are going to compute for the slope of the line tangent line, we have a big problem
because we only have one point, and the slope formula requires two points. To deal
with this problem, we select a point Q on the graph of f, draw the secant line PQ and
move Q along the graph of f towards P. Notice that as Q approaches P (shown as Q'
and Q''), the secant line gets closer and closer to the tangent line. This is the same as
saying that the slope the secant line is getting closer and closer to the slope of the
tangent line. Similarly, we can say that as the distance between the x-coordinates of P
and Q is getting closer and closer to 0, the slope of the secant line is getting closer and
closer to the slope of the tangent line.
Figure 4 – As point Q approaches P, the slope of the secant line is getting closer and
closer to the slope of the tangent line.
If we let h be the distance between the x-coordinates of and ,
P and Q, be the
slope of the secant line PQ and mt be the slope of the tangent line, we can say that the
limit of the slope of secant line as h approaches 0 is equal to the slope of the tangent
line. Concisely, we can write
Area and Limits
Another ancient problem is about finding the area under a curve as shown in the
leftmost graph in Figure 5. During the ancient time, finding the area of a curved plane
was impossible.
Figure 5 – As the number of rectangles increases, the sum of the area of the rectangles
is getting closer and closer to the area of the bounded plane under the curve.
We can approximate the area above in the first graph in Figure 5 by constructing
rectangles under the curve such that one of the corners of the rectangle touches the
graph as shown in the second and third graph in Figure 5. We can see that as we
increase the number of rectangles, the better is our approximation of the area under the
curve. We can also see that no matter how large the number of rectangles is, the sum
its areas will never exceed (or equal) the area of the plane under the curve. Hence, we
say that as the number of rectangles increases without bound, the sum of the areas of
the rectangles is equal to the area under the curve; or the limit of the sum of the areas
of the rectangles as the number of rectangles approaches infinity is equal to the area of
the plane under the curve.
If we let A be the area under the curve, Sn be the sum of the areas of n rectangles, then
we can say that the limit of Sn as n approaches infinity is equal to A. Concisely, we can
write
.
Numbers and Limits
We end with a more familiar example usually found in books. What if we want to find the
limit of 2x + 1 as x approaches 3?
To answer the question, we must find the value 2x + 1 where x is very close to 3. Those
values would be numbers that are very close to 3 – some slightly greater than 3 and
some slightly less than 3. Place the values in a table we have
From the table, we can clearly see that as the value of x approaches 3, the value of 2x +
1 approaches 7. Concisely, we can write the
Intuitive Definition of a Limit
To make it simple, the limit of a function is what the function "approaches" when
the input (the variable "x" in most cases) approaches a specific value.
Let's analyze this graph:
As the variable x approaches a, the function f(x) approaches L The limit of an
actual function may look like this:
This is the function f(x) equals x squared. As "x" approaches 1, f(x) approaches 1.
To express this we write:
Solving Simple Limits
Many limits are very easy to solve. Let's start with some:
Let's think. What will happen to the function when x approaches 1 more and more?
Let's take our calculator and make a table:
The function clearly approaches 3, right? Let's see what happens if x approaches
1, but takes values greater than one.
It also approaches 3...So, when this happens, we write:
Limit of the Sum of Functions
Now, let's suppose we have two functions :
g(x) doesn't have an "x", so it is constant. This means its value is six, no matter
what the "x" is. What will happen if we add these functions and try to find the limit?
This problem is the same as the previous one. We don't need to make a table to
know that when x approaches 2, x squared will approach 4. Six al ways will be six.
So:
Here we can note two important properties of the limit of a function:
The first one is that the limit of the sum of two or more functions equals the sum of
the limits of each function.
The second one is that the limit of a constant equals the same constant. By a
"constant" we mean any number.
Limit of a Product
In our first example:
We used another important property of the limit of a function. Can you see which
one?
This is similar to the property about sums, but with products:
The limit of the product of two or more functions equals the product of the limits of
each function.
This also means that whenever you have a function multiplied by any number you
can do this:
That is, you can take the number out of the limit sign. Another example:
Limit of a Quotient
As you probably expect by now, the limit of the quotient of two functions equals the
quotient of the limits. For example:
In this example we used all the properties we learned. Using these you can solve
many simple limits.
At first, you should think what properties you are using to solve your limits. But as
you practice more, you'll see that you can simply replace "x" for the value it is
approaching.
Let's see another example:
V. Post test
Evaluate.
1. lim (8 − 3𝑥 + 12𝑥 2 )
𝑛→2
6+4𝑡
𝑡→−3 𝑡 2 + 1
2. lim
3. lim 4𝑥
𝑛→3
4. Find the limit using table.
lim
𝑥−2
𝑥→2 𝑥 2 −𝑥−2
Reinforcement Activity
Give 4 real life scenario wherein the study of limits is applied.
References
http://www.intuitive-calculus.com/limit-of-a-function.html
https://www.onlinemathlearning.com/limits-calculus.html
https://tutorial.math.lamar.edu/problems/calci/computinglimits.aspx
http://mathandmultimedia.com/2009/12/22/intro-to
https://betterexplained.com/articles/an-intuitive-introduction-to-limits/
https://www.ck12.org/book/ck-12-precalculus-concepts/section/14.3/
“Don’t limit your challenges, challenge your limits.”
-
Anonymous
Answer key
Pre- test
1.
x
1.9
1.99
1.999
f(x)
0.25641
0.25063
0.25006
1
The evidence suggests that the limit is 4.
2.001
0.24994
2.01
0.24938
2.1
0.2439
2.001
0.33322
2.01
0.33223
2.1
0.32258
2. 39
3. 5
4. -
1
2
Post test
1. 50
3
2. - 5
3. 12
4.
x
1.9
1.99
1.999
f(x)
0.34483
0.33445
0.33344
1
The evidence suggests that the limit is 3.
Reinforcement Activity
Answers may vary.
(FORMAL DEFINITION (EPSILON-DELTA)
I. INTRODUCTION:
People think that mathematics is so complicated because they do not realize
how complicated life it is. This segment will focus on understanding the formal definition
of a limit while also exploring its different applications in mathematics. The Epsilon- Delta
𝜀 − 𝛿 definition of a limit was 1st used by Augustin-Louis Cauchy, formally defined by
Bernard Bolzano and its modern definition was provided by Karl Weirestrass (Grabiner,
1983).Many refer to this as the “EPSILON-DELTA” referring to the letters “𝜀” (Epsilon)
and “𝜹” (Delta) of Greek alphabet that defines a limit at a finite point that has a finite
value.
In mathematics, a limit is the value that a function or sequence “approaches” as
the input or index “approaches” of some value. At the same time, limits are essential to
calculus and mathematical analysis in general that are used to define continuity,
derivatives and integrals. Also, limit is more about how much you are allowed to do
something and it gives us reasonable estimate that can be applied in real life situation.
Whatever or whenever we are going to do everything has a limit in this world we live in
but some of us are just not aware of it. Exploring and playing with limits is the key that
opens the door for the learners and set them up for a successful career in advance
mathematics. The Formal definition of a limits is quite possibly one of the most challenging
you will first encounter in studying calculus.
According to Edward Kasner that
mathematics is man’s own handiwork, subject only to the limitations imposed by the laws
of thought.
Thus, this materials shall guide you step by step as you discover and understand
the lesson prepared for you. Discussions and exercises are carefully stated in order for
you to engage of how the beauty of formal definition of limits (Epsilon-Delta) it is.
II. Learning Outcomes:
At the end of the lesson, the learners are able to:
•
•
•
Define the formal definition of a limit
Discuss and graph the epsilon delta definition of a limit
Use the Epsilon-Delta definition to prove the limit laws.
General Instructions: To successfully use this materials, be sure to follow instructions
carefully. There is no need to hurry as long as you learn the lessons.
Read the content carefully.
Perform all the drills and exercises in the lesson.
Test yourself.
Record all the information in your notebook.
Summarize what you have learn.
If there is anything you do not understand in this lesson, see your teacher
or somebody who can help you but never ask someone to do the activity
and do it by yourself honestly.
III. PRETEST
Direction: Read each of the following statements and identify of what is being asked.
_______________1. What is epsilon-delta definition means?
_______________2. Who is the mathematician who 1st used the formal definition
(epsilon-delta)?
_______________3. In what century that the formal definition of a limit
(epsilon-delta) formally define?
_______________4. Who is the mathematician formally defined the formal definition
of
a limit (epsilon-delta)?
________________5. Who is provided the modern definition of formal definition of a
limit (epsilon-delta)?
IV. CONTENT INPUTS AND DISCUSSION
The formal definition of a limit Epsilon-Delta (𝜺 − 𝜹) is one of the most
elegant and creative definitions in math. At the same time, that was 1st used by AugustinLouis Cauchy in his equations while trying to solve formulas in his limit arguments which
became the basis for the rigorously defining continuity. However, he never gave a formal
definition in terms of his variables. Although, his contemporaries who more formally
defined the limit was by Bernard Bolzano and its modern definition was provided by Karl
Weierstrass (Grabiner, 1983) who continued to use the variables that was established.
The Epsilon (𝜀) and Delta (𝛿) are the Greek letters and their lowercase
version is used as the variables in the definition. Although, the "𝜀" is used as the number
of that is added or subtracted from the limit while the "𝛿" is used as the number of that is
added or subtracted from x to get the horizontal range used in the limit. On the other
hand, let’s see together and try to understand the definition by formalizing the notion or
phrases and words. Weather you say it in words or write it symbolically it’s supposed to
mean the same thing.
Here is the definition that can be written symbolically in a couple of ways. One
is to write “approaches” as an arrow: as 𝑥 → 𝑎, 𝑓 (𝑥 ) → 𝐿 and we are considering what
it means to say as x approaches a, 𝑓(𝑥) approaches to"𝐿". At this moment, "𝒙" is a
variable and "𝒂 “is a specific number, "𝑓" is the function under consideration and "𝐿" is
another specific number called the limit. So, the other one is to use the limit of notation
that can be written in lim 𝑓(𝑥) = 𝐿 which can read the expressions as the “limit of f(x) as
𝑥→𝑎
x approaches to 𝑎 is equals to L. Which means that for each positive number 𝜀, there is
a positive number 𝛿 which may defend on 𝜀, So, whatever x is a number not equal to 𝒂
but within 𝛿 𝑜𝑓 𝑎, it is the case that 𝑓(𝑥 ) is within 𝜀 𝑜𝑓 𝐿.
So, here is the given definition of limits (epsilon-delta) below.
DEFINITION OF A LIMITS (EPSILON-DELTA)
Let 𝑓(𝑥) be a function define for all 𝑥 ≠ 𝑎 over an open interval containing 𝑎.
Let 𝐿 be a real number. Then lim 𝑓(𝑥 ) = 𝐿 if for every number 𝜀 > 0 there exists some
𝑥→𝑎
real number 𝛿 > 0 so that if 0 < |𝑥 − 𝑎| < 𝛿 then |𝑓(𝑥 ) − 𝐿| < 𝜀.
For the statement above, we can easily understand if we break it down phrase by phrase.
So, the statement involves something called universal quantifier (𝑓𝑜𝑟 𝑒𝑣𝑒𝑟𝑦 𝜀 > 0), an
existential quantifier (𝑡ℎ𝑒𝑟𝑒 𝑒𝑥𝑖𝑠𝑡𝑠 𝑎 𝛿 > 0) and especially the conditional statement
(𝑖𝑓 0 < |𝑥 − 𝑎| < 𝛿 𝑡ℎ𝑒𝑛 |𝑓 (𝑥 ) − 𝐿| < 𝜀). Otherwise, Let’s take a look at the figure below
which breaks down the definition and translate each part.
Translation of the Epsilon-Delta of the limit
DEFINITION
𝟏. 𝐅𝐨𝐫 𝐞𝐯𝐞𝐫𝐲 𝛆 > 𝟎
𝟐. 𝐭𝐡𝐞𝐫𝐞 𝐞𝐱𝐢𝐬𝐭𝐬 𝐚 𝛅 > 𝟎
TRANSLATION
𝟏. 𝑭𝒐𝒓 𝒆𝒗𝒆𝒓𝒚 𝒑𝒐𝒔𝒊𝒕𝒊𝒗𝒆 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆
𝜺 𝒇𝒓𝒐𝒎 𝑳
𝟐. 𝑻𝒉𝒆𝒓𝒆 𝒊𝒔 𝒂 𝒑𝒐𝒔𝒊𝒕𝒊𝒗𝒆 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝜹 𝒇𝒓𝒐𝒎 𝒂
𝟑. 𝐬𝐮𝐜𝐡 𝐭𝐡𝐚𝐭
𝟑. 𝒔𝒖𝒄𝒉 𝒕𝒉𝒂𝒕
𝟒. 𝐢𝐟 𝟎 < |𝐱 − 𝐚| < 𝛅,
𝐭𝐡𝐞𝐧 |𝐟(𝐱) − 𝐋| < 𝛆
𝟒. 𝒊𝒇 𝒙 𝒊𝒔 𝒄𝒍𝒐𝒔𝒆𝒓 𝒕𝒉𝒂𝒏 𝜹 𝒕𝒐 𝒂 𝒂𝒏𝒅
𝒙 ≠ 𝒂, 𝒕𝒉𝒆𝒏 𝒇(𝒙)𝒊𝒔 𝒄𝒍𝒐𝒔𝒆𝒓 𝒕𝒉𝒂𝒏 𝜺 𝒕𝒐 𝑳.
So, let’s proceed to the figure example of a definition of limits (𝜀 − 𝛿) below graphically.
𝑹𝒆𝒈𝒊𝒐𝒏 𝑰
y
𝑳+𝜺
𝑹𝒆𝒈𝒊𝒐𝒏 𝑰𝑰
𝑳
𝑹𝒆𝒈𝒊𝒐𝒏 𝐼𝐼
𝑳−𝜺
x
𝒂
𝒂−𝜹
𝒂+𝜹
The figure above shows the values of a function 𝑓 (𝑥 ) at three different points
which are close to each other. 𝜹 Represents the change in the value of x and the 𝜺 shows
the change in the values of the limit of the function at these points. These parameters
formalize the notion of the points being really close to each other and the meaning of the
phrases like x approaching the 𝑥 = 𝑎. Otherwise, using the formal definition of the limits
can be written above that is called the “epsilon-delta” definition and what the definition is
trying to say can be explained with the figure above. According to the definition, there is
a certain number 𝜀 > 0, that we pick. The two horizontal lines in the figure represent
𝐿 + 𝜀 𝑎𝑛𝑑 𝐿 − 𝜀. After that, the definition says there is another number 𝛿 > 0 out there
that we need to determine. It allows us to add those vertical lines in the figure above
representing 𝑎 + 𝛿, 𝑎 − 𝛿. For any point that is there between region I,
0 < |𝑥 − 𝑎| < 𝛿. If we now identify the point on the graph will lie in the intersection of the
green and gray region. This means that this function value 𝑓 (𝑥 ) will be closer to 𝐿 than
either of 𝐿 + 𝜀 𝑎𝑛𝑑 𝐿 − 𝜀 or |𝑓(𝑥 ) − 𝐿| < 𝜀.
Using the definition to calculate the limits
As an example, let’s consider a function𝑓 (𝑥 ) = 𝑥². Using the limit definition that
mention above prove that lim 𝑥 2 = 0.
𝑥→0
For this case, 𝐿 = 0 𝑎𝑛𝑑 𝑎 = 0. consider any arbitrary number 𝜀 > 0.
|𝑥 2 − 0| < 𝜀
⇒ |𝑥 2 | < 𝜀
The goal is to find a number 𝛿, such that,
|𝑥 2 | < 𝜀 𝑤ℎ𝑒𝑟𝑒 0 < |𝑥 | < 𝛿
|𝑥 2 | < 𝜀
⇒ |𝑥| < √𝜀
We can choose our 𝛿 = √𝜀
For verification, one should check that for given condition with 𝛿, the value of expression
should not change more than 𝜀. In this particular case, it means that we need to make
sure that |𝑥 2 | < 𝜀. So, let’s use the previously derived result |𝑥 | < √𝜀. Squaring it,
|𝑥 2 | < 𝜀 and that is the result we were looking for. Hence, verified.
On the other hand, let’s see and understand some example problems with these
concepts.
Example Problem 1: For the given function f(x), find lim 𝑓(𝑥),
𝑥→4
𝑓 (𝑥 ) =
Solution:
Using the substitution rule, 𝑓(𝑥 ) =
lim 𝑓(𝑥)
𝑥→4
𝑥+5
𝑥→4 𝑥
⇒ lim
⇒lim
4+5
𝑥→4 4
9
⇒4
𝑥+5
𝑥
𝑥+5
𝑥
Example Problem 2: For the given function f(x), prove using the epsilon delta
definition of the limit,lim 𝑓 (𝑥 ) = 6.
𝑥→2
𝑓(𝑥 ) = 5𝑥 − 4
Solution:
For this case, L=6 and a=2. Consider any arbitrary number 𝜀 > 0.
|5𝑥 − 4 − 6| < 𝜀
⇒|5𝑥 − 4 − 6| < 𝜀
The goal is to find a number 𝛿 , such that,
|5𝑥 − 4 − 6| < 𝜀 𝑤ℎ𝑒𝑟𝑒 0 < |𝑥 − 2| < 𝛿
|5𝑥 − 4 − 6| < 𝜀
⇒ 5|𝑥 − 2| <
𝜀
5
𝜀
We can choose our 𝛿 = 5
For verification, one should check that for given condition with 𝛿, the value of the
expression should not change more than 𝜀. In this particular case, it means that we need
to make sure that 5𝑥 − 4 − 6| < 𝜀.
Let’s take the expression and use the derived result.
⇒ 5|𝑥 − 2| < 5
𝜀
5
=|5𝑥 − 4 − 6| < 𝜀
So, after trying to understand those problem examples with a concepts. Let’s proceed by
proving limit laws.
PROVING LIMIT LAWS
We now demonstrate how to use the epsilon-delta definition of a limit to
construct a rigorous proof one of the limit laws. The triangle inequality is used at a key
point of the proof, so we review first this key property of absolute value.
DEFINITION
The triangle inequality states that if 𝑎 and 𝑏 are any real numbers
Then, |𝑎 + 𝑏| ≤ |𝑎| + |𝑏|.
Proof: We prove the following limit law: if lim f(x) = L and lim g(x) = M,
x→a
x→a
Then lim(f(x) + g(x)) = L + M.
x→a
𝐿𝑒𝑡 𝜀 > 0.
Choose 𝛿ı > 0 𝑠𝑜 𝑡ℎ𝑎𝑡 𝑖𝑓 0 < |𝑥 − 𝑎| < 𝛿𝚤, 𝑡ℎ𝑒𝑛 |𝑓 (𝑥 ) − 𝐿| < 𝜀/2.
Choose 𝛿₂ > 0 𝑠𝑜 𝑡ℎ𝑎𝑡 𝑖𝑓 0 < | 𝑥 − 𝑎| < 𝛿₂, 𝑡ℎ𝑒𝑛 | 𝑔(𝑥 ) − 𝑀| < 𝜀/2.
Choose 𝛿 = 𝑚𝑖𝑛{𝛿ı, 𝛿₂}.
Assume 0 < |𝑥 − 𝑎| < 𝛿.
Thus, 0 < |𝑥 − 𝑎| < 𝛿ı 𝑎𝑛𝑑 0 < |𝑥 − 𝑎| < 𝛿₂.
Hence,|(𝑓 (𝑥 ) + 𝑔(𝑥 )) − (𝐿 + 𝑀)| = |(𝑓(𝑥 ) − 𝐿) + (𝑔(𝑥 ) − 𝑀)| ≤ |𝑓 (𝑥 ) − 𝐿| + |𝑔(𝑥 ) −
𝜀
𝜀
𝑀| < 2 + 2 = 𝜀
We explore now what it means for a limit not to exist. The limit lim 𝑓(𝑥) does
𝑥→𝑎
not exist if there is no real number 𝐿 𝑓𝑜𝑟 𝑤ℎ𝑖𝑐ℎ lim 𝑓 (𝑥 ) = 𝐿. Thus, for all real numbers L,
𝑥→𝑎
lim 𝑓 (𝑥 ) ≠ 𝐿. To understand what this means, we look at each part of the definition of
𝑥→𝑎
lim 𝑓 (𝑥 ) = 𝐿 together with its opposite. A translation of the definition is given in the figure
𝑥→𝑎
below.
Translation of the definition of 𝐥𝐢𝐦 𝒇(𝒙) = 𝑳 and its Opposite
𝒙→𝒂
DEFINITION
OPPOSITE
𝟏. 𝑭𝒐𝒓 𝒆𝒗𝒆𝒓𝒚 𝜺 > 𝟎,
𝟏. 𝑻𝒉𝒆𝒓𝒆 𝒆𝒙𝒊𝒔𝒕 𝜺 > 𝟎 𝒔𝒐 𝒕𝒉𝒂𝒕
𝟐. 𝒕𝒉𝒆𝒓𝒆 𝒊𝒔 𝒂 𝜹 > 𝟎 𝒔𝒐 𝒕𝒉𝒂𝒕
2. 𝒇𝒐𝒓 𝒆𝒗𝒆𝒓𝒚 𝜹 > 𝟎
𝟑. 𝒊𝒇 𝟎 < |𝒙 − 𝒂| < 𝜹,
𝒕𝒉𝒆𝒏 |𝒇(𝒙) − 𝑳| < 𝜺.
𝟑. 𝑻𝒉𝒆𝒓𝒆 𝒊𝒔 𝒂𝒏 𝒙 𝒔𝒂𝒕𝒊𝒔𝒇𝒚𝒊𝒏𝒈
𝟎 < |𝒙 − 𝒂| < 𝜹 𝒔𝒐 𝒕𝒉𝒂𝒕
|𝒇(𝒙) ≥ 𝜺.
Finally, we may state what it means for a limit not to exist. The limit lim (𝑓(𝑥) does not
𝑥→𝑎
exist if for every real number L, there exists real number 𝜀 > 0 𝑠𝑜 𝑡ℎ𝑎𝑡 𝑓𝑜𝑟 𝑎𝑙𝑙 𝛿 >
0, 𝑡ℎ𝑒𝑟𝑒 𝑖𝑠 𝑎𝑛 𝑥 𝑠𝑎𝑡𝑖𝑠𝑓𝑦𝑖𝑛𝑔 0 < |𝑥 − 𝑎| < 𝛿, 𝑠𝑜 𝑡ℎ𝑎𝑡|𝑓 (𝑥 ) − 𝐿| ≥ 𝜀. At the same, time lets
apply the figure below to show that limit does not exist.
SHOWING THAT A LIMIT DOES NOT EXIST
Show that lim
|𝑥|
𝑥→0 𝑥
does not exist. The graph of 𝑓 (𝑥 ) =
1
0
𝑓 (𝑥 ) =
1
|𝑥|
𝑥
|𝑥|
𝑥
is shown here:
SOLUTION:
1
Suppose that L is a candidate for a limit. Choose 𝜀 = 2 .
Let 𝛿 > 0. 𝐸𝑖𝑡ℎ𝑒𝑟 𝐿 ≥ 0 𝑜𝑟 𝐿 < 0. 𝐼𝑓 𝐿 ≥ 0, 𝑡ℎ𝑒𝑛 𝑙𝑒𝑡 𝑥 = −𝛿.
𝛿
𝛿
Thus,|𝑥 − 0| = │ − 2 − 0│ = 2 < 𝛿
And │
𝛿
2
𝛿
−
2
|− |
1
− 𝐿│ = |−1 − 𝐿| = 𝐿 + 1 ≥ 1 > 2 = 𝜀.
𝛿
On the other hand, If L< 0, then let 𝑥 = 2 .
𝛿
Thus, |𝑥 − 0| = │ 2 − 0│ =
And│
𝛿
| |
2
𝛿
2
𝛿
2
<𝛿
1
− 𝐿│ = |1 − 𝐿| = |𝐿|+≥ 1 > 2 = 𝜀.
Thus, for any value of 𝐿, lim
|𝑥|
𝑥→0 𝑥
≠ 𝐿.
Throughout This chapter, we are finally come to an end of various types of limits
to rigorous formal definitions. Since formal definition of epsilon delta is a famous topic in
the world of calculus that might be the first abstract topic that has a puzzled math for
student’s centuries but in order to understand the language of mathematics, the only
perquisites are curiosity and imagination. For instance, mathematical thinking is
necessary for understanding, exploring and using the ideas.
“Teachers can open the door but you must inter it by yourself”
V. POST TEST
PROBLEM SOLVING
Instructions: In the following exercises, write the appropriate 𝜀 − 𝛿 definition for each of
the given statements.
1. lim 𝑔(𝑡) = 𝑀
𝑡→𝑏
2. lim ∅(𝑥 ) = 𝐿
𝑥→𝑎
3. lim 𝑓( 𝑥) = 𝑁
𝑥→𝑎
4. lim ℎ(𝑥 ) = 𝐿
𝑥→𝑐
5. lim 𝑓 (𝑥 ) = 𝑀
𝑥→𝑎
VI. REINFORCEMENT ACTIVITY:
PROVING A STATEMENT
Complete the proof that lim (4𝑥 + 1) = −3 by filling in the blanks.
𝑥→−1
Let_______.
Choose 𝛿 = ____________.
Assume 0 < |𝑥 − ______| < 𝛿.
Thus,
|_____-__|=|_______|=|__| |_______|=__|______|<______=_____=𝜀
Therefore, lim (4𝑥 + 1) = −3
𝑥→−1
References:
Grabiner,J.(1983)Who Gave you the epsilon? Cauchy and the origins of Rigorous
Calulus.The American Mathematical Monthly, 90(3):185194,doi:10.2307/2975545,JSTORN2975545, archived from the original on May,4 th
2009,retrieved june 7th,2020.
Pelias (2018).Basic Calculus-First Edition.Rex Book Store.No.856 Nicanor Reyes,Sr.
Street,Manila Philippines.ISBN:978-9712378140.
PatrickjMT.(2015).Precise Definition of a limit-Understanding the
definition(video).RetrievedJune 7, 2020,from
https://www.youtube.com/watch?v=Goax2x2_Emo.
MAT137-2016/17L5201Resources
https://www.khanacademy(2013).org?math?ap-calculus-ab/ab-limits-new/ab-limitsoptional/v/proving-a-limit-using-epsilon-delta-definition
ANSWER KEY:
PRETEST
1. The Epsilon (𝜀) and Delta (𝛿) are the Greek letters and their lowercase version is
used as the variables in the definition. The "𝜀" is used as the number of that is added or
subtracted from the limit while the "𝛿" is used as the number of that is added or
subtracted from x to get the horizontal range used in the limit.
2. Augustin-Louis Cauchy
3. 19th centuryS
4. Bernard Bolzano
5. Karl Weirestrass
POST TEST
1 . 𝐹𝑜𝑟 𝑒𝑣𝑒𝑟𝑦 𝜀 > 0, 𝑡ℎ𝑒𝑟𝑒 𝑒𝑥𝑖𝑠𝑡𝑠 𝑎 𝛿 > 0 𝑠𝑜 𝑡ℎ𝑎𝑡 𝑖𝑓 0 < |𝑡 − 𝑏| < 𝛿, 𝑡ℎ𝑒𝑛 |𝑔(𝑡) − 𝑀| < 𝜀
2. 𝐹𝑜𝑟 𝑒𝑣𝑒𝑟𝑦 𝜀 > 0, 𝑡ℎ𝑒𝑟𝑒 𝑒𝑥𝑖𝑠𝑡 𝑎 𝛿 > 0 𝑠𝑜 𝑡ℎ𝑎𝑡 𝑖𝑓 0 < |𝑥 − 𝑎| < 𝛿, 𝑡ℎ𝑒𝑛|∅(𝑥 ) − 𝐴| < 𝜀
3. 𝐹𝑜𝑟 𝑒𝑣𝑒𝑟𝑦 𝜀 > 0, 𝑡ℎ𝑒𝑟𝑒 𝑒𝑥𝑖𝑠𝑡 𝑎 𝛿 > 0 𝑠𝑜 𝑡ℎ𝑎𝑡 𝑖𝑓 0 < |𝑥 − 𝑎| < 𝛿, 𝑡ℎ𝑒𝑛|𝑓(𝑥) − 𝑁| < 𝜀
4. 𝐹𝑜𝑟 𝑒𝑣𝑒𝑟𝑦 𝜀 > 0, 𝑡ℎ𝑒𝑟𝑒 𝑒𝑥𝑖𝑠𝑡 𝑎 𝛿 > 0 𝑠𝑜 𝑡ℎ𝑎𝑡 𝑖𝑓 0 < |𝑥 − 𝑐 | < 𝛿, 𝑡ℎ𝑒𝑛|ℎ(𝑥 ) − 𝐿| < 𝜀
5. 𝐹𝑜𝑟 𝑒𝑣𝑒𝑟𝑦 𝜀 > 0, 𝑡ℎ𝑒𝑟𝑒 𝑒𝑥𝑖𝑠𝑡 𝑎 𝛿 > 0 𝑠𝑜 𝑡ℎ𝑎𝑡 𝑖𝑓 0 < |𝑥 − 𝑎| < 𝛿, 𝑡ℎ𝑒𝑛|𝑓(𝑥 ) − 𝑀| < 𝜀
REINFORCEMENT ACTIVITY
Complete the proof that lim (4𝑥 + 1) = −3 by filling in the blanks.
𝑥→−1
Let 𝜀 < 0.
𝜀
Choose 𝛿 = 4.
Assume 0 < |𝑥 − (−1)| < 𝛿.
𝜀
Thus, |(4𝑥 + 1) − (−3)| = |4𝑥 + 4| = |4||𝑥 + 1| < 4𝛿 = 4 (4) = 𝜀
Therefore, lim (4𝑥 + 1) = −3
𝑥→−1
THEOREM ON LIMITS
I. INTRODUCTION
Calculus is the broad area of mathematics dealing with such topics as
instantaneous rates of change, areas under curves, and sequences and series.
Underlying all of these topics is the concept of a limit, which consists of analyzing the
behavior of a function at points ever closer to a particular point, but without ever actually
reaching that point. Calculus has two basic applications: differential calculus and
integral calculus.
One of the topics in Calculus is Theorem on Limits. In this note you will be able
to learn when a limit does or does not exist. However, it is neither desirable nor
practical, in every instance, to reach a conclusion about the existence of a limit based
on graph or on a table of numerical values. We must be able to evaluate a limit, or
discern its non-existence, in a somewhat mechanical fashion. The Theorem that we
shall consider in this note is a means.
II. LEARNING OUTCOMES:
•
•
to evaluate limits
to manipulate certain examples so that the theorems may be used
III. PRETEST
Evaluate the following limits, if they exist.
1.) lim f(x)
x→−6
2.) lim f(x)
x→1
3.) lim(8 − 3x + 12x 2 )
x→2
4.) lim
6+4𝑡
𝑡→−3 𝑡 2+1
5.) lim
𝑥→−5
𝑥 2 −25
𝑥 2 +2𝑥−15
IV. CONTENT INPUTS AND DISCUSSION
Definition:
LIMITS in mathematics, a limit is the value that
a function (or sequence) approaches as the input (or index) approaches
some value.[1] Limits are essential to calculus and mathematical analysis, and
are used to define continuity, derivatives, and integrals.
Theorem – a theoretical proposition, statement, or formula embodying
something to be proved from other propositions or formulas.
We will now prove that a certain limit exists, namely the limit
of f (x) = x as x approaches any value c. (That f(x) also approaches c should be
obvious.)
THEOREM. If f (x) = x, then for any value c that we might name:
For, if a sequence of values of the variable x approaches c as a limit (The limit of a
variable. We say that a sequence of values of a variable v approaches a
number l as a limit (a number not a term in the sequence), if, beginning with a certain
term vn, and for any subsequent term we might name, the absolute value of v n − l is
less than any positive number we name, however small.
When that condition is satisfied, we write v
l.), then a sequence of values
of the function f(x) = x will also approach c as a limit
(The limit of a function of a variable.
We say that a function f(x) approaches a limit L as x approaches c if the
sequence of values of x, both from the left and from the right, causes the sequence
of values of f(x) to satisfy the definition of "approaches a limit":
If that is the case, then we write:
"The limit of f(x) as x approaches c is L.")
For example,
Theorems on limits
To help us calculate limits, it is possible to prove the following.
Let f and g functions of a variable x. Then, if the following limits exist:
lim 𝑓 = 𝐴, 𝑎𝑛𝑑 lim 𝑔 = 𝐵
𝑥→𝑙
𝑥→𝑙
1.) lim(𝑓 + 𝑔) = 𝐴 + 𝐵.
𝑥→𝑙
2.) lim(𝑓𝑔) = 𝐴𝐵.
𝑥→𝑙
𝑓
𝐴
3.) lim 𝑔 = 𝐵 , 𝑖𝑓 𝐵 𝑖𝑠 𝑛𝑜𝑡 0.
𝑥→𝑙
In other words:
1) The limit of a sum is equal to the sum of the limits.
2) The limit of a product is equal to the product of the limits.
3) The limit of a quotient is equal to the quotient of the limits,
3) provided the limit of the denominator is not 0.
Also, if c does not depend on x -- if c is a constant – then
4.) lim 𝑐 = 𝑐.
𝑥→𝑙
For example, lim 5 = 5.
𝑥→4
To see that, let x approach 4: e.g.,
4
1 1 1
1
1
4 4 4
4
. . .
2 4 8 16 32
Then the value of 5 -- or any constant -- does not change. It is constant!
When c is a constant factor, but f depends on x, then
5.) lim 𝑐 𝑓 = 𝑐 lim 𝑓,
𝑥→𝑙
𝑥→𝑙
A constant factor may pass through the limit sign.
For example,
Example 1.
Problem 1. prove the following:
Solution. x2 = x · x. And we have proved that lim 𝑥 exists, and is equal to
𝑥→4
4. Therefore, lim 𝑥 ∙ 𝑥 = 4 ∙ 4
𝑥→4
That is,
It should be clear from this example that to evaluate the limit of any power
of x as x approaches any value, simply evaluate the power at that value.
Problem 2. Evaluate
lim 𝑥 5
𝑥→2
Problem 3. Evaluate the following limits, and justify your answers.
a.) lim (𝑥 3 + 𝑥)
𝑥→4
b.) lim (𝑥 2 + 1)
𝑥→4
Limits of polynomials
We might think that to evaluate a limit as x approaches a value, all we do is evaluate
the function at that value. And for the most part that is true One of the most
important classes of functions for which that is true are the polynomials. A polynomial
in x has this general form:
where n is a whole number, and an
0.
Therefore, according to the Theorems on limits, to name the limit of a
polynomial as x approaches any value c, simply evaluate the polynomial at that
value.
If P(x) is a polynomial, then
lim 𝑃(𝑥) = 𝑃(𝑐)
𝑥→𝑐
(In the following Topic we will see that is equivalent to saying that polynomials
are continuous functions. )
It is important to state again that when we write
lim 𝑃(𝑥 ) = 𝑃 (𝑐 ),
𝑥↔𝑐
the variable x is never equal to c, and therefore P(x) is
never equal to P(c) Both c and P(c) are approached as limits. The point is, we
can name the limit simply by evaluating the function at c.
Problem 4. Evaluate lim (5𝑥 4 − 4𝑥 3 + 3𝑥 2 − 2𝑥 + 1)
𝑥→−1
Problem 5. Evaluate lim(𝑥 + 𝑐)
𝑥→𝑐
Problem 6. Evaluate
lim 3𝑡 2 − 5𝑡 + 1
𝑡→−1
Problem 7. Evaluate lim 4𝑥 3 + 6𝑥 2 ℎ + 4𝑥ℎ2
ℎ→0
Example 2. Consider the function g(x) = x + 2, whose graph is a simple straight
line. And just to be perverse let the following function f(x) not be defined for
x = 2. That is, let
In other words, the point (2, 4) does not belong to the function; it is not on the
graph.
Yet the limit as x approaches 2 -- whether from the left or from the right -- is 4
For, every sequence of values of x that approaches 2, can come as close to 2
as we please. (The limit of a variable is never a member of the sequence, in any
case.) Hence the corresponding values of f(x) will come closer and closer to 4.
“The only way to discover the limits of the possible is to go beyond them into the
impossible”
- Arthur C. Clarke
V. POSTTEST
Evaluate the following limits, if they exist.
1.) lim f(x)
x→−6
2.) lim f(x)
x→1
3.) lim(8 − 3x + 12x 2 )
x→2
6+4𝑡
𝑡→−3 𝑡 2+1
𝑥 2 −25
4.) lim
5.) lim
𝑥→−5 𝑥 2 +2𝑥−15
VI. REINFORCEMENT ACTIVITY
Evaluate the following limit.
1.)
2.)
3.)
4.)
5.)
lim 𝑥 5
𝑥→3
lim (4𝑥 2 − 2𝑥 + 1)
𝑥→2
lim 3𝑥 2 − 4𝑥 3 + 7𝑥 − 5
𝑥→−1
lim 5𝑥 2
𝑥→4
lim
𝑥 2 −2
𝑥→4 𝑥+1
References
-
https://en.wikipedia.org/wiki/Limit_(mathematics)
-
http://archives.math.utk.edu/visual.calculus/1/limits.18/index.html
-
https://www.dictionary.com/browse/theorem
-
https://www.math.utah.edu/lectures/math1210/3PostNotes.pdf
ANSWER KEY
PRETEST
1.) lim f(x) = lim (7 − 4x)
x→−6
x→−6
= (7 − 4(−6))
= (7 − (−24))
= 31
2.) lim f(x) = lim(7 − 4x)
x→1
x→1
= (7 − 4(1))
= (7 − 4)
=3
3.) lim (8 − 3𝑥 + 12𝑥 2 ) = (8 − 3(2) + 12(22 ))
𝑥→2
= (8 − 6 + 12(4))
= (8 − 6 + 48)
= −6 + 56
= 50
4.) lim
6+4𝑡
𝑡→−3 𝑡 2 +1
=
=
=
6+4(−3)
−32 +1
6+(−12)
9+1
−6
10
=
5.) lim
𝑥→−5
−3
5
𝑥 2 −25
𝑥 2 +2𝑥−15
= lim
(𝑥−5)(𝑥+5)
𝑥→−5 (𝑥−3)(𝑥+5)
(𝑥−5)
= (𝑥−3)
−5−5
= −5−3
=
−10
−8
−5
5
= −4 or 4
PROBLEM 3.
a.) lim (𝑥 3 + 𝑥) = (43 + 4)
𝑥→4
= 64 + 4
= 68
b.) lim (𝑥 2 + 1) = (42 + 1)
𝑥→4
= 16 + 1
= 17
REINFORCEMENT ACTIVITY
1.) Lim 𝑥 5 = 35
𝑥→3
= 243
2.) lim 4𝑥 2 − 2𝑥 + 1 = 4(2)2 − 2(2) + 1
𝑥→2
= 4(4) − 4 + 1
= 16 − 4 + 1
= 13
3𝑥 2 −4𝑥 3 +7𝑥−5
3.) lim
𝑥→−1
2𝑥 2 +3𝑥+4
=
=
=
3(−1)2−4(−1)3 +7(−1)−5
2(−1)2 +3(−1)+4
3+4−7−5
2−3+4
−5
3
4.) lim 5𝑥 2 = 5(42 )
𝑋→4
= 5(16)
= 80
5.) lim
𝑥 2 −2
𝑋→4 𝑋+1
42 −2
=
4+1
=
=
16−2
5
14
5
PROTTEST
1.) lim f(x) = lim (7 − 4x)
x→−6
x→−6
= (7 − 4(−6))
= (7 − (−24))
= 31
2.) lim f(x) = lim(7 − 4x)
x→1
x→1
= (7 − 4(1))
= (7 − 4)
=3
3. ) lim (8 − 3𝑥 + 12𝑥 2 ) = (8 − 3(2) + 12(22 ))
𝑥→2
= (8 − 6 + 12(4))
= (8 − 6 + 48)
= −6 + 56
= 50
6 + 4𝑡 6 + 4(−3)
=
𝑡→−3 𝑡 2 + 1
−32 + 1
4. ) lim
=
=
=
5.) lim
𝑥→−5
6+(−12)
9+1
−6
10
−3
5
𝑥 2 −25
𝑥 2 +2𝑥−15
= lim
𝑥→−5 (𝑥−3)(𝑥+5)
(𝑥−5)
= (𝑥−3)
−5−5
= −5−3
=
(𝑥−5)(𝑥+5)
−10
−8
−5
5
= −4 or 4
ONE-SIDED LIMITS
I. INTRODUCTION
One of the most important topics in mathematics is calculus. Calculus was first
studied formally in the 17th century by well-known scientists and mathematicians like
Isaac Newton and Gottfried Leibniz, while it is probable that it was used as early as the
Greek era. It is a branch of mathematics that deals with functions, limits, derivatives, and
integrals. Throughout the history of mathematics, this discipline has left an indelible mark.
It has been able to create a new mathematical system overtime and was used in a variety
of applications.
One of the main focus of Calculus is Limit. The concept of limit of a function is
crucial to understanding calculus. It is used to define some of the more essential notions
in calculus, such as continuity, a function's derivative, and a function's definite integral.
Limits are used as real-life approximations to calculating derivatives. It is very difficult to
calculate a derivative of complicated motions in real-life situations. So, to make
calculations, engineers will approximate a function using small differences in the a
function and then try and calculate the derivative of the function by having smaller and
smaller spacing in the function sample intervals. For example, when designing the engine
of a new car, an engineer may model the gasoline through the car's engine with small
intervals called a mesh, since the geometry of the engine is too complicated to get exactly
with simply functions such as polynomials. These approximations always use limits.
In this discussion, we will be focusing on one-sided limits. One-sided limits are
differentiated as right-hand limits (when the limit approaches from the right) and left-hand
limits (when the limit approaches from the left) whereas ordinary limits are sometimes
referred to as two-sided limits. Right-hand limits approach the specified point from positive
infinity. Lefthand limits approach this point from negative infinity. Under this discussion,
we are expected to understand that when we wish to find the limit of a function f(x) as it
approaches a point a and we cannot evaluate f(x) at a because it is undefined at that
point, we can compute the function's one-sided limits in order to find the desired limit. If
its one-sided limits are the same, then the desired limit exists and is the value of the onesided limits. If its one-sided limits are not the same, then the desired limit does not exist.
II. LEARNING OUTCOMES
1. State and define the term One-Sided Limit.
2. Compute the value of the one-sided limits
III. PRETEST
For nos. 1-4, refer your answer to the graph of f(x) given below. Choose the
letter of the correct answer.
1. What is the value of 𝑓(−4)
a. 0
b.1
2. What is the value of lim − 𝑓(𝑥)?
c. 3
d. does not exist
a. 1
b. 3
3. What is the value of lim + 𝑓(𝑥)?
c. 4
d. does not exist
a. -1
b. -2
4. What is the value of lim 𝑓(𝑥)?
c.1
d. 2
c. 4
d. does not exist
𝑥→−4
𝑥→−4
𝑥→−4
a. 0
b.2
IV. CONTENT INPUTS AND DISCUSSIONS
Definition:
A one-sided limit is the value the function approaches as the x-values
approach the limit from *one side only*. For example, f(x)=|x|/x returns -1 for
negative numbers, 1 for positive numbers, and isn't defined for 0. The onesided *right* limit of f at x=0 is 1, and the one-sided *left* limit at x=0 is -1.
As the name implies, with one-sided limits, we will only be looking at
one side of the point in question. Here are the definitions for the two one sided
limits.
Right-handed limit
We say
𝐥𝐢𝐦 𝒇(𝒙) = 𝑳
𝒙→𝒂+
Provided we can make 𝑓(𝑥 ) as close to L as we want for all x
sufficiently close to a with 𝑥 > 𝑎 without actually letting 𝑥 be 𝑎.
Left-handed limit
We say
𝐥𝐢𝐦 𝒇(𝒙) = 𝑳
𝒙→𝒂−
Provided we can make 𝑓(𝑥 ) as close to L as we want for all x
sufficiently close to a with 𝑥 < 𝑎 without actually letting 𝑥 be 𝑎.
Note that the change in notation is very minor and in fact might be missed
if you aren’t paying attention. The only difference is the bit that is under the “lim”
part of the limit. For the right-handed limit we now have x→a (note the “+”) which
means that we know will only look at x>a. Likewise, for the left-handed limit we
have x→a−(note the “-”) which means that we will only be looking at x<a.
Also, note that as with the “normal” limit (i.e. the limits from the previous
section) we still need the function to settle down to a single number in order for the
limit to exist. The only difference this time is that the function only needs to settle
down to a single number on either the right side of x=a or the left side
of x=a depending on the one-sided limit we’re dealing with.
So, when we are looking at limits it’s now important to pay very close
attention to see whether we are doing a normal limit or one of the one-sided
limits. Let’ now take a look at some problems below and look at one-sided
limits instead of the normal limit.
Example
Example 1: Estimate the value of the following limits
lim+ 𝐻(𝑡 ) 𝑎𝑛𝑑
𝑡→0
lim−𝐻(𝑡 )
𝑡→0
𝑤ℎ𝑒𝑟𝑒 𝐻(𝑡 ) = {
0 𝑖𝑓 𝑡 < 0
1 𝑖𝑓 𝑡 ≥ 0
Answer:
Let us observe first the graph so we would know what this
function looks like.
So we can see that if we stay to the right of 𝑡 = 0 (𝑖. 𝑒. 𝑡 > 0)
then the function is moving in towards a value of 1 as we get closer
and closer to 𝑡 = 0, but staying to the right. We can therefore say that
the right-handed limit is,
lim 𝐻(𝑡 ) = 1
𝑡→0+
Likewise, if we stay to the left of 𝑡 = 0 (𝑖. 𝑒. 𝑡 < 0) the function
is moving in towards a value of 0 as we get closer and closer to 𝑡 =
0, but staying to the left. Therefore, the left-handed limit is,
lim 𝐻(𝑡 ) = 0
𝑡→0−
In this example, we do get one-sided limits even though the
normal limit itself doesn’t exist.
Example 2: Estimate the value of the following limits.
𝜋
lim+ cos ( )
𝑡→0
Graph:
𝑡
𝜋
lim− cos ( )
𝑡→0
𝑡
We can see that both of the one-sided limits suffer the same
problem that the normal limit did in the previous section. The function
does not settle down to a single number on either side of t=0.
Therefore, neither the left-handed nor the right-handed limit will exist
in this case.
Take note: one-sided limits don’t have to exist just as normal limits
aren’t guaranteed to exist.
Example 3: Estimate the value of the following limits.
𝑥 2 +4𝑥−12
lim+ 𝑔(𝑥) 𝑎𝑛𝑑
𝑥→2
Answer:
Graph
lim− 𝑔(𝑥)
𝑥→2
𝑤ℎ𝑒𝑟𝑒 𝑔(𝑥) = {
𝑥 2 −2𝑥
6
𝑖𝑓 𝑥 ≠ 2
𝑖𝑓 𝑥 = 2
In this case regardless of which side of x=2 we are on the function is
always approaching a value of 4 and so we get,
lim 𝑔(𝑥) = 4
𝑥→2+
lim 𝑔(𝑥) = 4
𝑥→2−
Note that one-sided limits do not care about what’s happening at the point
any more than normal limits do. They are still only concerned with what is going
on around the point. The only real difference between one-sided limits and normal
limits is the range of x’s that we look at when determining the value of the limit.
Now let’s take a look at the first and last example in this section to get a
very nice fact about the relationship between one-sided limits and normal limits. In
the last example the one-sided limits as well as the normal limit existed and all
three had a value of 4. In the first example the two one-sided limits both existed,
but did not have the same value and the normal limit did not exist.
The relationship between one-sided limits and normal limits can be
summarized by the following fact.
Fact:
Given a function f(x) if,
𝐥𝐢𝐦 𝒇(𝒙) = 𝐥𝐢𝐦− 𝒇(𝒙) = 𝑳 ,
𝒙→𝒂+
𝒙→𝒂
then the normal limit will exist and
𝐥𝐢𝐦 𝒇(𝒙) = 𝑳
𝒙→𝒂
Likewise, if
𝐥𝐢𝐦 𝒇(𝒙) = 𝑳 ,
𝒙→𝒂
Then
𝐥𝐢𝐦 𝒇(𝒙) = 𝐥𝐢𝐦− 𝒇(𝒙) = 𝑳
𝒙→𝒂+
𝒙→𝒂
This fact can be turned around to also say that if two one-sided limits
have different values, i.e.,
𝐥𝐢𝐦 𝒇(𝒙) ≠ 𝐥𝐢𝐦− 𝒇(𝒙)
𝒙→𝒂+
𝒙→𝒂
then the normal limit will not exist.
This should make some sense. If the normal limit did exist then by the
fact the two one-sided limits would have to exist and have the same value by the
above fact. So, if the two one-sided limits have different values (or don’t even exist)
then the normal limit simply can’t exist.
Let’s take a look at one more example
Example 4: Given the graph below, compute the following given:
a. 𝑓 (−4)
b. lim − 𝑓(𝑥)
𝑥→−4
c.
lim 𝑓(𝑥)
𝑥→−4+
d. lim 𝑓(𝑥)
𝑥→−4
e. 𝑓 (1)
f. lim− 𝑓(𝑥)
𝑥→1
g. lim+ 𝑓(𝑥)
𝑥→1
h. lim 𝑓(𝑥)
𝑥→1
Graph
Answers:
a. f(-4) Doesn’t exist. There is no closed dot for this value
of x and so the function doesn’t exist at this point.
b. lim − 𝑓(𝑥) = 2. The function is approaching a value of 2
𝑥→−4
as x moves in towards -4 from the left.
c. lim + 𝑓(𝑥) = 2. The function is approaching a value of 2
𝑥→−4
as x moves in towards -4 from the right.
d. lim 𝑓(𝑥)=2. We can do this one of two ways. Either we
𝑥→−4
can use the fact here and notice that the two one-sided
limits are the same and so the normal limit must exist and
have the same value as the one-sided limits or just get the
answer from the graph.
e. 𝑓 (1)=4. The function will take on the y value where the
closed dot is.
f. lim− 𝑓(𝑥) = 4. The function is approaching a value of 4
𝑥→1
as x moves in towards 1 from the left.
g. lim+ 𝑓(𝑥) = −2. The function is approaching a value of -2
𝑥→1
as x moves in towards 1 from the right. Remember that the
limit does NOT care about what the function is actually
doing at the point, it only cares about what the function is
doing around the point. In this case, always staying to the
right of x=1, the function is approaching a value of -2 and
so the limit is -2. The limit is not 4, as that is value of the
function at the point.
h. lim 𝑓(𝑥) does not exist. The two one-sided limits both
𝑥→1
exist, however they are different and so the normal limit
doesn’t exist.
V. POSTTEST
Below is the graph of f(x). Find the value of the following:
1. 𝑓 (4)
2. lim− 𝑓(𝑥)
𝑥→4
3. lim+ 𝑓(𝑥)
𝑥→6
4. lim 𝑓(𝑥)
𝑥→−4
Graph
VI. REINFORCEMENT ACTIVITY
Homework:
Study further about the graphs of f(x) with one sided-limits then do the
activity blow
Sketch a graph of a function that satisfies each of the following conditions.
𝐥𝐢𝐦 𝒇(𝒙) = 𝟏
𝒙→𝟐−
𝐥𝐢𝐦 𝒇(𝒙) = −𝟒
𝒙→𝟐+
𝒇(𝟐) = 𝟏
REFERENCES
•
•
© 2003 - 2021 Paul Dawkins. https://tutorial.math.lamar.edu/.
Page Last Modiefied:01/21/2018.
https://tutorial.math.lamar.edu/Problems/CalcI/OneSidedLimits.a
spx
Answer key
Pretest
1.
2.
3.
4.
C. 3
B. 3
B. -2
D. does not exist
Posttest
1. 𝑓 (−4) = 3
2. lim− 𝑓(𝑥) = 2
𝑥→4
3. lim+ 𝑓(𝑥) = 5
𝑥→2
4. lim 𝑓(𝑥) = 4
𝑥→−1
Reinforcement Activity
There are literally an infinite number of possible graphs that we could
give here for an answer. However, all of them must have a closed dot on the graph
at the point (2,1), the graph must be approaching a value of 1 as it
approaches x=2 from the left (as indicated by the left-hand limit) and it must be
approaching a value of -4 as it approaches x=2 from the right (as indicated by the
right-hand limit).
Here is a
sketch
of
one
possible graph that
meets
these
conditions.
INFINITE LIMITS
I. INTRODUCTION
We will see in this note that infinity symbols,
−∞ (minus infinity)𝑎𝑛𝑑 ∞ (infinity), are notational devices used to indicate, in turn, that
a quantity becomes unbounded in the negative direction (in the Cartesian plane this
means to the left for x and download for y) and in the positive direction (to the right for x
and upward for y).
II. LEARNING OUTCOMES
▪ To evaluate the limit of a function at a point or to evaluate the limit of a function
from the right and left at a point.
▪ To describe the behavior of functions that do not have finite limits.
▪ To recognize an infinite limit.
III. PRETEST
Compute the following limits:
1
1.) lim 𝑥2
𝑥→0
2.) lim (𝑥 3 − 𝑥)
𝑥→∞
3.) lim
6
𝑥→0+ 𝑥 2
−4
4.) lim + 𝑥+2
𝑥→−2
IV. CONTENT INPUTS AND DISCUSSIONS
Definitions: Infinite Limits
The words Infinite Limit always refer to a limit that does not exist because the
function 𝑓 exhibits unbounded behavior: 𝑓 (𝑥 ) → −∞ 𝑜𝑟 𝑓(𝑥) → ∞.
We define three types of infinite limits.
1. Infinite limits from the left: Let f(x) be a function define at all values in an open
interval of the form (b,a).
i. If the values of f(x) increase without bound as the values of x (𝑤ℎ𝑒𝑟𝑒 𝑥 < 𝑎),
approach the number 𝑎, then we say that the limit as x approaches a from the left is
positive infinity and we write:
lim 𝑓(𝑥) = +∞
𝑥→𝑎−
ii. If the values of f(x) decrease without bound as the values of x (𝑤ℎ𝑒𝑟𝑒 𝑥 < 𝑎)
approach the number 𝑎, then we say that the limit as x approaches a from the left is
negative infinity and we write
lim 𝑓 (𝑥 ) = −∞
𝑥→𝑎−
2. Infinite limits from the right: Let f(x) be a function defined at all values in an open
interval of the form (a,c).
i. If the values of f(x) increase without bound as the values of x (where 𝑥 > 𝑎)
approach the number a, then we say that the limit as x approaches a from the left is
positive infinity and we write
lim 𝑓 (𝑥 ) = +∞
𝑥→𝑎+
ii. If the values of f(x) decrease without bound as the values of x (where 𝑥 > 𝑎)
approach the number a, then we say that the limit as x approaches a from the left is
negative infinity and we write
lim 𝑓 (𝑥 ) = −∞
𝑥→𝑎+
3. Two – sided infinite limit: Let f(x) be defined for all 𝑥 ≠ 𝑎 in an open interval
containing a
i. If the values of f (x) increase without bound as the values of x (where 𝑥 ≠ 𝑎)
approach the number a, then we say that the limit as x approaches a is appositive
infinity and we write
lim 𝑓(𝑥 ) = +∞
𝑥→𝑎
ii. If the values of f(x) decrease without bound as the values of x (where 𝑥 ≠ 𝑎)
approach the number a, then we say that the limit as x approaches a is negative infinity
and we write
lim 𝑓(𝑥 ) = −∞
𝑥→𝑎
It is important to understand that when we write the statements such as
lim 𝑓 (𝑥 ) = +∞ 𝑜𝑟 lim 𝑓(𝑥) = −∞ we are describing the behavior of the function, as we
𝑥→𝑎
𝑥→𝑎
have just defined it. We are not asserting that a limit exists. For the limit of a function
f(x) to exist at a, it must approach a real number L as x approaches a. That said, if, for
example, lim 𝑓(𝑥) = +∞, we always write lim 𝑓 (𝑥 ) = +∞ rather than lim 𝑓 (𝑥 )𝐷𝑁𝐸.
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
Example. Recognizing an Infinite Limit
Evaluate each of the following limits, if possible. Use table of functional values and
graph f(x) = 1/x to confirm your conclusion.
a. lim
1
𝑥→0− 𝑥
b. lim
1
𝑥→0+ 𝑥
1
c. lim 𝑥
𝑥→0
Solution
Begin by constructing a table of functional values.
X
-0.1
-0.01
-0.001
-0.0001
-0.00001
-0.000001
𝟏
𝒙
-10
-100
-1000
-10,000
-100,000
-1,000,000
x
0.1
0.01
0.001
0.0001
0.00001
0.000001
𝟏
𝒙
10
100
1000
10,000
100,000
1,000,000
a. The values of 1/x decrease without bound as x approaches 0 from the left. We
conclude that
1
lim = −∞.
𝑥→0− 𝑥
b. The values of 1/x increase without bound as x approaches 0 from the right. We
conclude that
1
lim = +∞.
𝑥→0+ 𝑥
c. Since lim
1
𝑥→0− 𝑥
= −∞ 𝑎𝑛𝑑 lim
1
𝑥→0+ 𝑥
= +∞ have different values, we conclude that
1
lim 𝐷 𝑁 𝐸.
𝑥→0 𝑥
The graph of f(x) = 1/x below confirms these conclusions.
The graph of f(x) = 1/x confirms that the limit as x approaches 0 does not exist.
“Once we accept our limits, we go beyond them.”
-
Albert Einstein
-
V. POSTTEST
1
1.) lim 𝑥2
𝑥→0
2.) lim (𝑥 3 − 𝑥)
𝑥→∞
3.) lim
6
𝑥→0+ 𝑥 2
−4
4.) lim + 𝑥+2
𝑥→−2
VI. REINFORCEMENT ACTIVITY
Evaluate each of the following limits, if possible. Use a table of functional values
and graph 𝑓(𝑥) = 1⁄𝑥 2 to confirm your conclusion.
a. lim
1
𝑥→0− 𝑥 2
1
b. lim
𝑥→0+ 𝑥 2
1
c. lim 𝑥2
𝑥→0
It is useful to point out that functions of the form 𝑓(𝑥) = 1⁄(𝑥 − 𝑎)𝑛 , where n is a positive
integer, have infinite limits as x approaches a from either the left or right. These limits
are summarized in the above definitions.
The function 𝑓(𝑥) = 1⁄(𝑥 − 𝑎)𝑛 has infinite limits at a.
REFERENCES
-
-
https://math.libretexts.org/Courses/Monroe_Community_College/MTH_210_
Calculus_I_(Professor_Dean)/Chapter_2_Limits/2.4%3A_Infinite_Limits#:~:t
ext=infinite%20limit%20A%20function%20has,one%2Dsided%20limit%20of
%20a
https://www.sfu.ca/mathcoursenotes/Math%20157%20Course%20Notes/sec_InfLimits.html
https://tutorial.math.lamar.edu/classes/calci/infinitelimits.aspx
ANSWER KEY
PRETEST/POSTTEST
Compute the following limits:
1
1.) lim 𝑥2
𝑥→0
We refer to the graph below. Let’s first look at the limit as 𝑥 → 0+, and notice that
increases without bound.
Therefore,
lim+
𝑥→0
As 𝑥 → 0− , we again see that
1
𝑥2
1
= +∞
𝑥2
increase without bound:
lim−
𝑥→0
1
= +∞
𝑥2
We conclude that
1
=∞
𝑥→0 𝑥 2
lim
2.) lim (𝑥 3 − 𝑥)
𝑥→∞
One might be tempted to write:
lim 𝑥 3 − lim 𝑥 = ∞ − ∞,
𝑥→∞
𝑥→∞
1
𝑥2
However, we do not know what ∞ − ∞ is, as ∞ is not a real number and so cannot be
treated like one.
Incidentally, the expression ∞ − ∞ is another indeterminate form.
We instead write:
lim (𝑥 3 − 𝑥) = lim 𝑥(𝑥 2 − 1)
𝑥→∞
𝑥→∞
As x becomes arbitrarily large, then both x and 𝑥 2 − 1 become arbitrarily large, and
hence their product 𝑥(𝑥 2 − 1) will also become arbitrarily large. Thus we see that
lim (𝑥 3 − 𝑥) = ∞
𝑥→∞
3.) lim
6
𝑥→0+ 𝑥 2
=∞
−4
4.) lim + 𝑥+2 = −∞
𝑥→−2
REINFORCEMENT ACTIVITY
1
a. lim
𝑥→0− 𝑥 2
1
b. lim
𝑥→0+ 𝑥 2
1
= +∞
= +∞
c. lim 𝑥2 = +∞
𝑥→0
LIMITS AT INFINITY
I. LEARNING OUTCOMES
a. Define limits at infinity
b. Evaluate the limits at infinity
II. PRETEST
1. Evaluate the following limits and identify if it gives positive infinity (∞)
2. or negative infinity (−∞)
𝟒𝒛𝟐 +𝒛𝟔
1.
𝐥𝐢𝐦 𝟏−𝟓𝒛𝟑
2.
𝐥𝐢𝐦
𝒛→∞
𝟒𝒛𝟐 +𝒛𝟔
𝒛→−∞ 𝟏−𝟓𝒛𝟑
3. 𝐥𝐢𝐦 𝟒𝒙𝟐 − 𝟏𝟖𝒙𝟑 + 𝟗
𝒙→−∞
4. 𝐥𝐢𝐦 𝟒𝒙𝟐 − 𝟏𝟖𝒙𝟑 + 𝟗
𝒙→∞
III. CONTENT INPUTS AND DISCUSSIONS
In the previous discussion, we saw limits that were infinity. In this section, we
will take a look at limits at infinity. By limits at infinity, we mean one of the following
two limits.
lim 𝑓(𝑥)
𝑥→∞
lim 𝑓(𝑥)
𝑥→−∞
We are going to be looking at what happens to a function if we let 𝑥 get a
very large in either the positive or negative sense. Also, we will also see that these
limits may also have infinity as a value.
Now, we will take note first the two facts about infinite limits that we will need
in this section.
Fact 1
1. If 𝑟 is a positive rational number and 𝑐 is any real number then,
𝑐
=0
𝑥→∞ 𝑥 𝑟
lim
2. If 𝑟 is a positive rational number, 𝑐 is any real number and 𝑥 𝑟 is define
for 𝑥 < 0 then,
𝑐
lim 𝑟 = 0
𝑥→∞ 𝑥
The first part of this fact should make sense if you think about it. Because
we are requiring 𝑟 > 0 we know that 𝑥 𝑟 will stay in the denominator. Next as we
increase 𝑥 then 𝑥 𝑟 will also increase. So, we have a constant divided by an
increasingly large number and so the result will be increasingly small. Or, in the
limit we will get zero.
The second part is nearly identical except we need to worry about 𝑥 𝑟 being
1
defined for negative 𝑥. This condition is here to avoid cases such as 𝑟 = 2. If
this 𝑟 were allowed we’d be taking the square root of negative numbers which
would be complex and we want to avoid that at this level.
Note as well that the sign of 𝑐 will not affect the answer. Regardless of the
sign of 𝑐 we’ll still have a constant divided by a very large number which will result
in a very small number and the larger 𝑥 get the smaller the fraction gets. The sign
of 𝑐 will affect which direction the fraction approaches zero (i.e. from the positive
or negative side) but it still approaches zero.
Examples:
Example 1: Evaluate
a. 𝐥𝐢𝐦 (𝟐𝒙𝟒 − 𝒙𝟐 − 𝟖𝒙)
𝒙→∞
Solution:
What we’ll do here is factor the largest power of x out of the whole
polynomial as follows,
lim (2𝑥 4 − 𝑥 2 − 8𝑥) = lim [𝑥 4 (2 −
𝑥→∞
𝑥→∞
1
8
)]
−
𝑥2 𝑥3
Now for each of the terms we have,
lim 𝑥 4 = ∞
𝑥→∞
1
8
lim (2 − 𝑥2 − 𝑥3 ) = 2
𝑥→∞
The first limit is clearly infinity and for the second limit we’ll use the fact
above on the last two terms.
𝐥𝐢𝐦 (𝟐𝒙𝟒 − 𝒙𝟐 − 𝟖𝒙) = ∞
𝒙→∞
𝟏
b. 𝐥𝐢𝐦 ( 𝒕𝟓 + 𝟐𝒕𝟑 − 𝒕𝟐 + 𝟖)
𝒙→∞ 𝟑
Solution:
1
1 2
1
8
lim ( 𝑡 5 + 2𝑡 3 − 𝑡 2 + 8) = lim [𝑡 5 ( + 2 − 3 + 5 )]
𝑥→∞ 3
𝑥→∞
3 𝑡
𝑡
𝑡
Now all we need to do is take the limit of the two terms. In the first
don’t forget that since we’re going out towards −∞ and we’re raising t to the
5th power that the limit will be negative (negative number raised to an odd
power is still negative). In the second term we’ll again make heavy use of
the fact above to see that is a finite number.
Therefore, the value of the limits is,
𝟏
𝐥𝐢𝐦 (𝟑 𝒕𝟓 + 𝟐𝒕𝟑 − 𝒕𝟐 + 𝟖) = −∞
𝒙→∞
Fact 2
If 𝑝(𝑥 ) = 𝑎𝑛 𝑥 𝑛 + 𝑎𝑛−1 𝑥 𝑛−1 + ⋯ + 𝑎𝑛 𝑥 + 𝑎0 is a polynomial of degree
𝑛 (𝑖. 𝑒. 𝑎𝑛 ≠ 0) then,
lim 𝑝(𝑥 ) = lim 𝑎𝑛 𝑥 𝑛
𝑥→∞
lim 𝑝(𝑥 ) = lim 𝑎𝑛 𝑥 𝑛
𝑥→∞
𝑥→−∞
𝑥→−∞
What this fact is really saying is that when we take a limit at infinity
for a polynomial all we need to really do is look at the term with the largest
power and ask what that term is doing in the limit since the polynomial will
have the same behavior.
Example:
Example 2: Evaluate the given limits.
a. 𝐥𝐢𝐦
𝒙→∞
𝟐𝒙𝟒 −𝒙𝟐 +𝟖𝒙
−𝟓𝒙𝟒 +𝟕
𝐥𝐢𝐦
𝒙→−∞
𝟐𝒙𝟒 −𝒙𝟐 +𝟖𝒙
−𝟓𝒙𝟒 +𝟕
Solution:
We first identify the largest power of 𝑥 in the denominator (and
yes, we only look at the denominator for this) and we then factor this out
of both the numerator and denominator. Doing this for the first limit gives,
lim
𝑥→∞
2𝑥 4 −𝑥 2 +8𝑥
−5𝑥 4 +7
= lim
𝑥→∞
1
8
+ )
𝑥2 𝑥3
7
𝑥 4 (−5+ 4 )
𝑥
𝑥 4 (2−
Once we’ve done this we can cancel the 𝑥 4 from both the
numerator and the denominator and then use the Fact 1 above to take
the limit of all the remaining terms. This gives,
lim
𝑥→∞
2𝑥 4 −𝑥 2 +8𝑥
−5𝑥 4 +7
= lim
𝑥→∞
=
1
8
+ )
𝑥2 𝑥3
7
(−5+ 4 )
𝑥
(2−
2+0+0
−5+0
2
= −5
In this case the indeterminate form was neither of the
“obvious” choices of infinity, zero, or -1 so be careful with make these kinds
of assumptions with this kind of indeterminate forms.
The second limit is done in a similar fashion. Notice however, that
nowhere in the work for the first limit did we actually use the fact that the
limit was going to plus infinity. In this case it doesn’t matter which infinity we
are going towards we will get the same value for the limit.
𝐥𝐢𝐦
𝒙→−∞
𝟐𝒙𝟒 −𝒙𝟐 +𝟖𝒙
−𝟓𝒙𝟒 +𝟕
= −
𝟐
𝟓
Example 3: Evaluate
a. 𝐥𝐢𝐦
√𝟑𝐱 𝟐 +𝟔
𝐱→∞ 𝟓−𝟐𝐱
Solution:
𝐥𝐢𝐦
√𝟑𝐱 𝟐 +𝟔
𝐱→−∞ 𝟓−𝟐𝐱
√𝑥 2 (3 + 62 )
√3𝑥 2 + 6
𝑥
lim
= lim
5
𝑥→∞ 5 − 2𝑥
𝑥→∞
𝑥(𝑥 − 2)
√𝑥 2 √3+
= lim
𝑥→∞
6
𝑥2
5
𝑥(𝑥−2)
This is where we need to be really careful with the square root in
the problem. Don’t forget that
√ 𝑥 2 = |𝑥 |
Square roots are ALWAYS positive and so we need the
absolute value bars on the 𝑥 to make sure that it will give a positive
answer. This is not something that most people ever remember
seeing in an Algebra class and in fact it’s not always given in an
Algebra class. However, at this point it becomes absolutely vital that
we know and use this fact. Using this fact the limit becomes,
6
|𝑥 | √(3 + 2 )
√3𝑥 2 + 6
𝑥
lim
= lim
5
𝑥→∞ 5 − 2𝑥
𝑥→∞
𝑥 (𝑥 − 2)
Now, we can’t just cancel the 𝑥. We first will need to get rid of
the absolute value bars. To do this let’s recall the definition of
absolute value.
𝑥 𝑖𝑓 𝑥 ≥ 0
|𝑥 | = {
−𝑥 𝑖𝑓 𝑥 < 0
In this case we are going out to plus infinity so we can safely
assume that the 𝑥 will be positive and so we can just drop the
absolute value bars. The limit is then,
6
𝑥 √(3 + 2 )
√3𝑥 2 + 6
𝑥
lim
= lim
5
𝑥→∞ 5 − 2𝑥
𝑥→∞
𝑥 (𝑥 − 2)
= 𝐥𝐢𝐦
𝒙→∞
𝟔
𝒙
√(𝟑+ 𝟐 )
𝟓
(𝒙−𝟐)
=
√𝟑+𝟎
𝟎−𝟐
=−
√𝟑
𝟐
Let’s now take a look at the second limit (the one with negative
infinity). In this case we will need to pay attention to the limit that we are
using. The initial work will be the same up until we reach the following
step.
6
|𝑥 | √(3 + 2 )
√3𝑥 2 + 6
𝑥
lim
= lim
5
𝑥→−∞ 5 − 2𝑥
𝑥→−∞
𝑥 (𝑥 − 2)
In this limit we are going to minus infinity so in this case we
can assume that 𝑥 is negative. So, in order to drop the absolute value
bars in this case we will need to tack on a minus sign as well. The limit
is then,
6
−𝑥 √(3 + 2 )
√3𝑥 2 + 6
𝑥
lim
= lim
5
𝑥→−∞ 5 − 2𝑥
𝑥→−∞
𝑥 (𝑥 − 2)
6
− √(3 + 2 )
𝑥
= lim
5
𝑥→−∞
(𝑥 − 2)
=
√𝟑
𝟐
So, as we saw in the last two examples sometimes the infinity in the
limit will affect the answer and other times it won’t. Note as well that it doesn’t
always just change the sign of the number. It can on occasion completely change
the value.
Now, let us review our knowledge about asymptotes (see previous
topic). Just as we can have vertical asymptotes defined in terms of limits, we can
also have horizontal; asymptotes defined in terms of limits.
Horizontal Asymptote
Definition:
The function 𝑓 (𝑥 ) will have a horizontal asymptote at 𝑦 = 𝐿 if
either of the following are true.
𝐥𝐢𝐦 𝒇(𝒙) = 𝑳
𝒙→∞
𝐥𝐢𝐦 𝒇(𝒙) = 𝑳
𝒙→−∞
Example:
Given
𝑓 (𝑥 ) =
8 − 4𝑥 2
9𝑥 2 + 5𝑥
Check if 𝑥 → ∞ 𝑖𝑓 lim 𝑓(𝑥) exists and is a finite number.
𝑥→−∞
Same goes with 𝑥 → −∞ 𝑖𝑓 lim 𝑓(𝑥)
𝑥→−∞
8
8
𝑥 2 ( 2 − 4)
2−4
8 − 4𝑥 2
−4
𝑥
𝑥
lim
=
lim
=
lim
=
2
5
5
𝑥→−∞ 9𝑥 + 5𝑥
𝑥→−∞ 2
𝑥→−∞
9
𝑥 (9 + 𝑥 )
9+𝑥
8
8
𝑥 2 ( 2 − 4)
8 − 4𝑥 2
2−4
−4
𝑥
𝑥
lim 2
= lim
= lim
=
5
5
𝑥→∞ 9𝑥 + 5𝑥
𝑥→∞ 2
𝑥→∞
9
𝑥 (9 + 𝑥)
9+𝑥
We know that there will be a horizontal asymptote for 𝑥 →
−∞ 𝑖𝑓 lim 𝑓(𝑥) exists and is a finite number. Likewise, We
𝑥→−∞
know that there will be a horizontal asymptote for 𝑥 →
∞ 𝑖𝑓 lim 𝑓(𝑥) exists and is a finite.
𝑥→∞
Therefore, from the first two parts, we can see that we will
get the horizontal asymptote.
𝟒
𝒚 = −𝟗
For both 𝑥 → −∞ and 𝑥 → ∞.
It’s an easy fact to give and we can use the previous example to illustrate
all the asymptote ideas we’ve seen in the both this section and the previous
section. The function in the last example will have two horizontal asymptotes. It will
also have a vertical asymptote. Here is a graph of the function showing these.
Limits at Infinity of other types of functions
Example:
1. Evaluate each of the following limits.
𝐥𝐢𝐦 𝒆𝒙
𝒙→∞
𝐥𝐢𝐦 𝒆𝒙
𝐥𝐢𝐦 𝒆−𝒙
𝒙→−∞
𝒙→∞
𝐥𝐢𝐦 𝒆−𝒙
𝒙→−∞
Answer:
𝐥𝐢𝐦 𝒆𝒙 = ∞
𝒙→∞
𝐥𝐢𝐦 𝒆−𝒙 = 𝟎
𝒙→∞
𝐥𝐢𝐦 𝒆𝒙 = 𝟎
𝒙→−∞
𝐥𝐢𝐦 𝒆−𝒙 = ∞
𝒙→−∞
𝟐
2. Evaluate 𝐥𝐢𝐦 𝒆𝟐−𝟒𝒙−𝟖𝒙
𝒙→∞
Answer:
In this part what we need to note is that in the limit the
exponent of the exponential does the following,
𝐥𝐢𝐦 𝟐 − 𝟒𝒙 − 𝟖𝒙𝟐 = −∞
𝒙→∞
So, the exponent goes to minus infinity in the limit and
so the exponential must go to zero in the limit using the ideas from
the previous set of examples. So, the answer here is,
𝟐
𝐥𝐢𝐦 𝒆𝟐−𝟒𝒙−𝟖𝒙 = 𝟎
𝒙→∞
𝟒
𝒕
𝒆
𝒙→−∞
3. Evaluate 𝐥𝐢𝐦
−𝟓𝒕𝟐 +𝟏
Answer:
Here, let’s fist note that,
𝐥𝐢𝐦 (𝒕𝟒 − 𝟓𝒕𝟐 + 𝟏) = ∞
𝒙→−∞
The exponent goes to infinity in the limit and so the
exponential will also need to fo to infinity in the limit. Or,
𝟒
𝒆𝒕
𝒙→−∞
𝐥𝐢𝐦
−𝟓𝒕𝟐+𝟏 = ∞
4. Evaluate 𝐥𝐢𝐦 𝐥𝐧(𝟕𝒙𝟑 − 𝒙𝟐 + 𝟏)
𝒙→∞
Answer:
So, let’s first look to see what the argument of the log is doing,
𝐥𝐢𝐦 𝟕𝒙𝟑 − 𝒙𝟐 + 𝟏 = ∞
𝒙→∞
The argument of the log is going to infinity and so the log must
also be going to infinity in the limit. The answer to this part is then,
𝐥𝐢𝐦 𝐥𝐧(𝟕𝒙𝟑 − 𝒙𝟐 + 𝟏) = ∞
𝒙→∞
𝟏
5. 𝐥𝐢𝐦 𝐥𝐧 𝒕𝟐−𝟓𝒕
𝒙→−∞
Answer:
First, note that the limit going to negative infinity here isn’t a
violation (necessarily) of the fact that we can’t plug negative numbers
into the logarithm. The real issue is whether or not the argument of
the log will be negative or not.
𝐥𝐢𝐦
𝒙→−∞
𝟏
𝒕𝟐 −𝟓𝒕
=𝟎
and let’s also note that for negative numbers (which we can
assume we’ve got since we’re going to minus infinity in the limit) the
denominator will always be positive and so the quotient will also
always be positive. Therefore, not only does the argument go to
zero, it goes to zero from the right. This is exactly what we need to
do this limit.
So, the answer here is,
𝟏
𝐥𝐢𝐦 𝐥𝐧 𝒕𝟐 −𝟓𝒕 = −∞
𝒙→−∞
6. Evaluate 𝐥𝐢𝐦 𝐭𝐚𝐧−𝟏 𝒙
and
𝒙→∞
𝐥𝐢𝐦 𝐭𝐚𝐧−𝟏 𝒙
𝒙→−∞
Answer:
All we really need to do here is look at the graph of
the inverse tangent. Doing this shows us that we have the
following value of the limit.
𝝅
a. 𝐥𝐢𝐦 𝐭𝐚𝐧−𝟏 𝒙 = 𝟐
𝒙→∞
b.
𝝅
𝐥𝐢𝐦 𝐭𝐚𝐧−𝟏 𝒙 = − 𝟐
𝒙→−∞
IV. POSTTEST
Evaluate each of the following limits.
1. 𝐥𝐢𝐦
𝟒𝒛𝟐 +𝒛𝟔
𝒛→∞ 𝟏−𝟓𝒛𝟑
𝟒𝒛𝟐 +𝒛𝟔
2. 𝐥𝐢𝐦
𝒛→−∞ 𝟏−𝟓𝒛𝟑
𝟑
3. 𝐥𝐢𝐦 √𝒕 + 𝟏𝟐𝒕 − 𝟐𝒕𝟐
𝒛→∞
𝟑
4. 𝐥𝐢𝐦 √𝒕 + 𝟏𝟐𝒕 − 𝟐𝒕𝟐
𝒛→−∞
V. REINFORCEMENT ACTIVITY
Homework:
Research more about writing down equations of any horizontal
asymptotes. Then, answer the problem below.
Given the f(x) below,
a. evaluate 𝐥𝐢𝐦 𝒇(𝒙)
𝒙→−∞
b. evaluate 𝐥𝐢𝐦 𝒇(𝒙)
𝒙→∞
c. Write down the equations(s) of any horizontal asymptotes for the
function.
1. 𝑓(𝑥 ) =
3𝑥 7−4𝑥2+1
5−10𝑥 2
VI. REFERENCES:
•
Paul Dawkins. 2003 – 2021. tutorial.math.lamar.edu.
https://tutorial.math.lamar.edu/classes/calci/limitsatinfinityi.aspx .
ANSWER KEY
1. −∞
2. ∞
3. −∞
4. ∞
Posttest
𝟏. 𝐥𝐢𝐦
𝟒𝒛𝟐 +𝒛𝟔
𝒛→∞ 𝟏−𝟓𝒛𝟑
= −∞
𝟒𝒛𝟐 +𝒛𝟔
𝟐. 𝐥𝐢𝐦
𝒛→−∞ 𝟏−𝟓𝒛𝟑
=∞
𝟑. 𝐥𝐢𝐦 𝟑√𝒕 + 𝟏𝟐𝒕 − 𝟐𝒕𝟐 = −∞
𝒛→∞
𝟑
𝟒. 𝐥𝐢𝐦 √𝒕 + 𝟏𝟐𝒕 − 𝟐𝒕𝟐 = −∞
𝒛→−∞
Reinforcement Activity:
Research more about writing down equations of any horizontal asymptotes.
Then, answer the problem below.
Given the f(x) below,
a. evaluate 𝐥𝐢𝐦 𝒇(𝒙)
𝒙→−∞
b. evaluate 𝐥𝐢𝐦 𝒇(𝒙)
𝒙→∞
c. Write down the equations(s) of any horizontal asymptotes for the
function.
𝑓 (𝑥 ) =
a. 𝐥𝐢𝐦
𝒙→−∞
3𝑥 7 −4𝑥 2 +1
5−10𝑥 2
3𝑥 7 − 4𝑥 2 + 1
5 − 10𝑥 2
1
= 𝐥𝐢𝐦
𝒙→−∞
𝑥 2 (3𝑥 5−4+ 2)
𝑥
5
𝑥 2 ( 2 −10)
𝑥
1
= 𝐥𝐢𝐦
𝒙→−∞
3𝑥 5 −4+ 2
𝑥
5
−10
𝑥2
−∞
= −𝟏𝟎 = ∞
b. 𝐥𝐢𝐦
𝒙→−∞
3𝑥 7 −4𝑥 2 +1
5−10𝑥 2
1
= 𝐥𝐢𝐦
𝒙→−∞
𝑥 2 (3𝑥 5−4+ 2)
𝑥
5
𝑥 2 ( 2 −10)
𝑥
1
= 𝐥𝐢𝐦
𝒙→−∞
3𝑥 5 −4+ 2
𝑥
5
−10
𝑥2
∞
= −𝟏𝟎 =
−∞
c. We know that there will be a horizontal asymptote for 𝑥 →
−∞ 𝑖𝑓 lim 𝑓(𝑥) exists and is a finite number. Likewise, We
𝑥→−∞
know that there will be a horizontal asymptote for 𝑥 →
∞ 𝑖𝑓 lim 𝑓(𝑥) exists and is a finite.
𝑥→∞
Therefore, from the first two parts, we can see that this
function will have no horizontal asymptotes since neither of
the two limits are finite.