Limits and Their Properties
A Preview of Calculus
Calculus exists because of limits. We'll talk more about limits later; for now, let's look at how
limits can take simple ideas and make them into…well, Calculus!
The Tangent Problem
We know all about lines—most importantly, we know all about the slope of a line. But why
should the idea of slope be restricted to lines? Can't we say the graph below is steep…
…while the next graph is rather shallow?
The first graph was steep; it had a large (far from zero) slope. The second graph was shallow; it
had a small (close to zero) slope.
The problem should be fairly evident—how do you describe the slope of this graph?
In some places, it is steep—but in others, it's shallow!
For lines, slope is constant—it doesn't matter where you measure; you'll get the same slope.
For curves, the slope changes as you look at different points along the curve. There's the rub: our
formula for slope uses any two points—we can't do that for curves.
We can approximate the slope of the curve (at a particular value of x) with a secant line.
Observe: (the curve is blue; the secant line is red)
HOLLOMAN’S COLLEGE ALGEBRA &
INTRODUCTION TO CALCULUS
CAICNOTES03, PAGE #1
First, the two points are (fairly) far apart—the red line (the secant line) doesn't show the slope of
the curve. However, as the points get closer together…
The secant line looks more and more like a tangent line, which shows the slope of the curve. The
trick is to use limits to determine what happens to the secant lines as those points get closer
together.
The Area Problem
The area under a curve is an important problem. We've got formulas for finding areas of
rectangles, circles, trapezoids…but not general curves. We'd want to do this in areas such as
probability.
Anyway—how are you going to find the area under a curve? Really, you can only estimate.
Here's an estimate of the area under a curve using four rectangles.
Perhaps you can see that the estimate isn't very good—some rectangles have too much area;
others, too little. The shaded area is 2.75.
Here's another estimate of the same area, using eight rectangles.
HOLLOMAN’S COLLEGE ALGEBRA &
INTRODUCTION TO CALCULUS
CAICNOTES03, PAGE #2
This is a little better—although some are too big and others are too small, the smaller width of
each rectangle allows it to conform to the shape of the curve better. The shaded area is 2.6875.
To get a really good estimate (or the actual answer of 2.6667), we need to use limits to see what
happens to the total area as the width of the rectangles approaches zero.
Finding Limits Graphically and Numerically
The Idea
It is of interest to mathematicians to know what happens to the value of a function as the input
(x) gets really close (arbitrarily close) to a particular value. This is usually because we cannot
plug in the specified x-value directly—perhaps because it is not in the domain. This process of
determining the behavior of the function near certain x-values is the process of limits.
The Idea, Graphically
Let's look at the function y =
1
around the x-value of 1. Note that 1 is not in the domain of
x −1
this function.
What happens to the value of the function (the y-value) as x gets really close to 1? Well, it
depends on which side you're looking at. On the left side of 1, the function is headed down. To
the right of 1, the function is going up. We call these the left- and right-hand limits, respectively.
Try this function: y =
x−2
. Note that neither 2, nor -2, are in the domain.
x2 − 4
What happens as x gets really close to 2? Perhaps we should zoom in on that area.
HOLLOMAN’S COLLEGE ALGEBRA &
INTRODUCTION TO CALCULUS
CAICNOTES03, PAGE #3
As x gets closer to 2 (from the left), the y-values appear to be headed towards 0.25. The same
occurs on the right.
The Idea, Numerically
Let's look at that last one, but from a numeric perspective. Here's a table of values for the
function, with x-values very close to (but slightly lower than) 2.
1.9
1.99
1.999
1.9999
1.99999
x
y 0.25641 0.250627 0.250063 0.250006 0.250001
The values of x are getting closer and closer to 2, and the y-values appear to be getting closer and
closer to 0.25. The same is true on the right:
2.00001
2.0001
2.001
2.01
2.1
x
y 0.249999 0.249994 0.249938 0.249377 0.243902
The Limit of a Function
First, notation: to talk about the limit of the function f(x) as x approaches the value a from the
left, we write lim− f ( x ) . The same limit, from the right, is written lim+ f ( x ) .
x→ a
x→ a
Now we can talk about the limit at a value. lim f ( x ) exists if lim− f ( x ) = lim+ f ( x ) . In other
x→a
x→ a
x→ a
words, if the left- and right-hand limits agree, then we can just talk about the limit, without any
qualifying words.
There is a technical definition (the epsilon-delta definition) for limits, and if you know that you
you're going to have to take more Calculus in college, then you need to know this definition.
Otherwise…
Evaluating Limits Analytically
Basic Limits
The limit of a constant is that constant. If b is a real number, then the limit as x approaches any
value a is lim b = b .
x→a
If a function is defined for x = a, then (usually) you can simply plug into the function to find the
limit. lim x = a , lim x n = a n , lim x = a , etc. The main exceptions are piece-wise defined
x→a
x→a
x→a
functions. Also, you can't find the limit of a radical equation at the endpoint (there aren't two
sides!).
Properties
For these definitions, assume that lim f ( x ) = L , lim g ( x ) = K , and b is a real number.
x→a
x→a
Constants can be factored out: lim b ⋅ f ( x )  = b ⋅ lim  f ( x )  = b ⋅ L
x→a
x→a
HOLLOMAN’S COLLEGE ALGEBRA &
INTRODUCTION TO CALCULUS
CAICNOTES03, PAGE #4
Operations can be distributed: lim  f ( x ) ± g ( x )  = lim f ( x ) ± lim g ( x ) = L ± K ,
x→a
x→a
x→a
lim  f ( x ) ⋅ g ( x )  = lim f ( x ) ⋅ lim g ( x ) = L ⋅ K , lim
x→a
x→a
x→a
x→a
f ( x)
g ( x)
=
lim f ( x )
x→a
lim g ( x )
x→a
=
L
(provided that K ≠ 0),
K
and lim  f ( x )  = L .
x→a
n
n
Strategies
Remember that substitution works, as long as the location of the limit (a, in the previous
examples) is in the domain. Thus, the only limits that will cause you any grief are those with
domain problems. To date, you know of two potential domain problems: dividing by zero, and
taking an even-indexed root of a negative number. Of these, the one that comes up most often is
the dividing by zero issue. Let's look at how you can take care of those with some examples.
x−2
x−2
. Now let's find lim 2
.
Here's a function we looked at earlier: y = 2
x→2 x − 4
x −4
We can't just let x = 2; we've got a domain problem. We can factor:
x−2
x−2
= lim
. Now, remember: when we take a limit, we're letting x get really
lim 2
x→2 x − 4
x → 2 ( x − 2 )( x + 2 )
close (but at) the value where we're taking the limit. So, in this case, we want to know what
happens when x is really close to (but not quite at) 2. Since we're not letting x be exactly 2, what
x−2
do the factors
do to the value of the function?
x−2
Nothing! As long as x isn't exactly 2, these factors cancel out. As far as the limit is concerned,
these factors do not contribute to the function value. Cancel them out!
x−2
1
lim 2
= lim
. Now, there is no domain issue with x = 2, so we can simply plug in!
x→2 x − 4
x→2 x + 2
x−2
1
1
1
lim 2
= lim
=
= .
x→2 x − 4
x→2 x + 2
2+2 4
(a − x)
lim
2
− a2
.
x
Once again, we've got a domain problem at the desired value of x—we can't simply let x = 0.
Let's try this one (let a be a real number):
Let's work out that numerator:
(a − x)
2
−a
(a − x)
lim
x →0
x →0
2
x
2
− a2
a 2 − 2ax + x 2 − a 2
. Hey, look at that! Stuff
x →0
x
= lim
−2ax + x
. Now, we can factor an x out of the numerator, which
x →0
x →0
x
x
can then be canceled out (just like the last example).
cancels! lim
(a − x)
lim
x →0
2
x
− a2
= lim
x →0
2
= lim
x ( −2 a + x )
x
= lim ( −2a + x ) = −2a .
HOLLOMAN’S COLLEGE ALGEBRA &
INTRODUCTION TO CALCULUS
x →0
CAICNOTES03, PAGE #5
Continuity and One-Sided Limits
The function f(x) is continuous at x = c provided lim f ( x ) = f ( c ) .
x →c
The function f(x) is continuous on the open interval (a, b) provided it is continuous at every point
within the interval (a, b).
The function f(x) is continuous on the closed interval [a, b] provided it is continuous on the open
interval (a, b), lim+ f ( x ) = f ( a ) , and lim− f ( x ) = f ( b ) .
x→ a
x →b
It turns out that continuity is terribly important in Calculus. Happily, most of the functions that
we know and work with are (mostly) continuous.
Infinite Limits
Sometimes, as we approach a particular value of x, the function value doesn't approach a finite
1
value—instead, the function shoots off towards infinity. For example, the function y =
near
x −1
x = 1.
Earlier, we noted that on the left of 1, the function heads down towards negative infinity, and that
it approaches positive infinity on the right. Our definition of limits is for approaching real
numbers—infinity isn't a real number. Technically, the left- and right-hand limits don't exist.
Nonetheless, we will write lim− f ( x ) = −∞ and lim+ f ( x ) = ∞ .
x →1
HOLLOMAN’S COLLEGE ALGEBRA &
INTRODUCTION TO CALCULUS
x →1
CAICNOTES03, PAGE #6