Limits - Unit 1
Starred problems/topics are for BC students only.
Definition & interpretation of a limit:
lim f ( x)  L means that the function f maps x values near a to y values near L. The notation x  a means to let x
x a
approach a from the right (values greater than a) and from the left (values less than a). The y values get infinitely
close to L as the x values get infinitely close to a, either from the right or the left. It does not matter if f (a) = L or if
3 x
3 x 1
f (a) is even defined. As an example, lim 2  since
can be made arbitrarily close to ¼ by evaluating it at x
x2 x
x2
4
values very close to 2. By “arbitrarily close” I mean that no matter how close you’d like a y value to be to ¼, it is
possible to find an x value near 2 such that f (x) is as close or closer to ¼.
1. Enter the function above into a graphing calculator and zoom in near the x value 2. a. Approaching from the left, how
close to 2 do you need to be in order to ensure that the functional values are within 0.01 of 0.25 ?
b. Approaching from the right, how close to 2 do you need to be in order to ensure that the functional values are within
0.01 of 0.25 ?
c. Based on your answers in parts a and b, how close to 2 do you need to be in order to ensure that the y values are within
0.01 of 0.25 ?
d. What interval centered at x = 2 will assure that the y values are in the (0.24, 0.26) ?
To evaluate the limit above we are able simply to plug in 2, but it is not always that simple. Consider the function

5 x  4, x  6 Then,
lim g ( x)  26 even though g(6) = -99. As far as the limit goes as x approaches 6, we
g ( x)  
x 6

 99, x  6
are not interested in 6 itself. We’re only interested in values very close to 6. In fact, if g(6) were left undefined, the limit
would still be 26. However, because lim g ( x)  g (6), we can conclude that g is not continuous at x = 6. (More about
x 6
continuity later.)
2. a. Define a piecewise function whose limit as x  10 is 0 but f (10) = 11.37.
b. Come up with a function whose limit as x  5 is 3 but is undefined at x = 5.
c. Think of a function whose limit as x  -4 is 1 and is continuous at x = -4.
3. Mainly due to tidal forces of the moon, the earth’s day length is increasing at a rate of 1/500 of a second per century.
Our current period of rotation is 23 h, 56 min, 4.1 s. As the duration of time that passes from the present approaches 2
centuries, what does the earth’s period approach? State your answer in the form of a limit.
□ Sample problem I: If f ( x) 
2 x 2  5 x  12
, find lim f ( x ) .
x3 / 2
12 x  18
Solution: Replacing x with 3/2 yields a 0 / 0 situation. (We call this an indeterminant form.) Note that the numerator
approaching zero tends to make the whole fraction zero, but the denominator approaching zero tends to make the
numerator blow up to  . So, we have competing effects, and as far as we know right now, the limit could zero, it
could be  , it could be any constant, or might not even exist. To make a determination we begin by factoring:
lim f ( x )  lim (2 x  3)( x  4) . Clearly the function is undefined at x = 3/2, but this is beside the point when it
x3 / 2
x 3 / 2
6(2 x  3)
x4
comes to evaluating a limit; we are only concerned with x values very near 3/2. Reducing, lim f ( x )  lim
x3 / 2
x 3 / 2
6
3/ 2  4
= 11/12.

6
2x  4
.
x 4 3 x 3  192
4. Factor to find lim
5. Consider the function f ( x) 
2x
14

and the limit lim f ( x ) . If we were to take the limit term by term,
x 7
x7 x7
we’d find that each the limit does not exist for either term. This is because each term approaches   depending on the
direction in which x approaches 7. For either term, since the one-sided limits are different, the overall limit D.N.E. So it
would appear that the limit of the sum of the terms would not exist either. However, the theorem that the limit of a sum is
the sum of the limits only applies when limits exist. Find this limit by rewriting f (x) as a single fraction.
1 5

x
8
6. Find lim8
by simplifying the fraction.
5x  8
x
5
(
x 9
.
x 3
□ Sample problem II: Find lim
x 9
Solution: Here again we have a 0 / 0 indeterminant situation. One way to do this is to multiply numerator and
denominator by the conjugate of the denominator. lim
x 9
( x  9)( x  3)
x 9
( x  9)( x  3)
 lim
= lim
x 9
x9
x  3 x9 ( x  3)( x  3)
 lim ( x  3)  9  3  6 .
x 9
7. The last problem can also be done by factoring and reducing. Do this by forcing the numerator to factor as the
difference of squares.
8. Prove that
lim
x5
x 2  25
  by factoring and
x 5
simplifying. (The Mathematica syntax to find this limit--but
not the proof--is shown at the right.)
e 3 x 4 tan x
f (x) 
, find lim f ( x ) .
x10
sin 2 (3x  4)  2 x
2
□ Sample problem III: Given
Solution: Ostensibly this problem is extremely intimidating. However, as x  10, both the numerator and denominator
e3(10) 4 tan(10)
lim f ( x) 
x10
sin 2 (3  10  4)  2  10
2
approach a nonzero constant. Thus we simply evaluate f (x) at x = 10:
 7.1  10127, an enormous number. (Radian mode is required here.)
9. To get an inkling of just how large this number is, convert this many millimeters into light years. Show your work as
a product of fractions (dimensional analysis). Use the fact that light travels about 186,000 miles in a single second.
10. For f (x) above, find lim f ( x ) , making sure your calculator is in radian mode.
x11
n
□ Sample problem IV (by Raquel Roney, Nayeon Kang, & Ayush Dulguun): Evaluate the limit: lim
n 
3  2n
.
2n


n
n
 3 n 
2n 
 3 
 3
3  2n



 n  = lim    1 = lim    lim (1)
= lim
n    2 n

 n   2  n 
2  n   2 
2n








n
Solution: lim
n 
n
 3
  1  1 . Since
 lim 

n 
2


3 / 2  0.87 < 1, y = ( 3 / 2 )x is an exponential function that decays to zero. (If you
take 87% of a number over and over, you get closer and closer to zero. Another way to think about this limit is to recall
3 / 2 . Since r  1, the
geometric sequences from last year. Here the common ratio between successive terms is r =
x
3  2x
limit of the sequence is zero. This means our limit is 0 + 1 = 1 and that the function f ( x) 
has a horizontal
2x
asymptote of y = 1.
 5
lim
3x
11. Find
 10
.
25 x
x 
□ Sample problem V (by Raquel Roney, Nayeon Kang, & Ayush Dulguun): Evaluate lim
x 0
4x
.
tan x
sin x
 1.
x 0
x
x
4x
4 x cos x
4x
4x
 1
 lim
 lim
 lim cos x  4 lim
lim
= lim
x

0
x 0 sin x / cos x
x 0
x

0
x

0
x 0 tan x
sin x
sin x
sin x
lim (1)
x
1
1
 lim
 x 0
  1.
= 4(1)  1 = 4. Note that lim
x 0 sin x
x 0 sin x
sin x 1
lim
x

0
x
x
Solution: We will make use of the fact that lim
12. Find lim
x 0
5 sec  x tan  x
13 x
Numerical approximation of a limit:
Some limits are difficult, if not impossible, to evaluate analytically. So, we often evaluate the function very close to the
value of interest to approximate the limit.
(3  2 x)e cos 3 x
□ Sample problem VI: Let f (x) =
and determine the limit: lim f ( x ) .
x 5
x5
.
Solution: Note that f (5) is not defined, and there is no way to reduce the fraction, so we suspect the function has a
vertical asymptote at x = 5. If there is a V.A. at x = 5, then our limit would come out to be ∞. To test this let’s
evaluate f (x) at values of x slightly less than 5. I used Mathematica to do this, but you could do it on a calculator with
the table feature. First I defined the function:
f x_ :
3 2 x E ^Cos 3 x
x
5
Mathematica is very particular about syntax. When defining a function use an underscore after the variable on the left
side of the equation and use a colon before the equal sign. All built-in functions, like cosine, as well as user-made
functions enclose the arguments in square brackets, not parentheses. Next I will evaluate f (x) at five different values of
x very close to 5, which are enclosed in braces. (The output is right below this input).
f
4.9, 4.95, 4.9993, 4.999999, 4.9999999999
6
10
39.8819, 71.7562, 4683.59, 3.27469 10 , 3.27469 10
The output shows the corresponding functional values. Since these values appear to be growing without bound, we
conclude that our limit is + ∞. However, this analysis does not represent a proof.
Asymptotes:
Limits underlie the existence or nonexistence of horizontal, vertical, and oblique (slant) asymptotes of graphs of functions.
□ Sample problem VII: Let h( x) 
asymptote.
75 x 4  150 x 2  25
. a. Use a limit to prove that h(x) has no horizontal
5x 3  7 x 2  1
Solution: A function has a H.A. if its graph approaches some constant as x becomes very positive or very negative.
More precisely, if the limit as x    is a constant, the function has a H.A. (It is possible for the limits to be different
constants as x   and x  - , such as with y = arctan x, but with rational functions, such as h(x), if there is a
H.A., there is just one.) You should recall from precalculus that, for a rational function (a polynomial over a
polynomial), whenever the degree of the numerator is larger than that of the denominator, there is no H.A. and the graph
zooms off to   on each side. This is our situation, but let’s use a limit to prove this is indeed the case. Our technique
will be to divide each term, top and bottom, by the highest power of x in the denominator:
75 x  150 / x  25 / x 3
. Next we use the following facts: the limit of a quotient is the quotient of the
x 
x 
5  7 / x  1/ x3
limits; the limit of a sum is the sum of the limits; and the limit of a constant is the constant. Thus, lim h( x )
lim h( x)  lim


lim 75 x  150 / x  25 / x
x 

lim 5  7 / x  1 / x
x 
3

3



lim 75 x  lim 150 / x   lim 25 / x
x 
x 
x 

lim 5  lim 7 / x   lim 1 / x
x 
x 
x 
3

3

x 

lim 75 x  0  0
x 
500
= . We could use the
exact same process to show that lim h( x )   . Therefore, h(x) has no H.A.
x  
b. Use a limit to find the oblique asymptote for h(x).
Solution: You should recall from precalculus that, for a
rational function, whenever the degree of the numerator
exceeds that of the denominator by exactly one, the graph
has an oblique asymptote. The technique for finding it is to
employ long division. Of course, you are expected to be able
to do this by hand, but let’s see how it’s done with
Mathematica. For convenience we define two functions, one
for the numerator and one for the denominator. Then we use
two built-in functions to find the quotient and remainder.
(The quotient has been assigned to q for later reference.)
 3x 2  15 x  4
So, we now know that h( x)  15 x  21 
, from which it is easy to see that the fractional portion of
5x 3  7 x 2  1
 3x 2  15 x  4
 0 , since the degree of the
x  5 x 3  7 x 2  1
h(x) shrinks to zero as x becomes infinitely large. That is, lim
numerator is less than that of the denominator. This means that as x gets larger and larger in magnitude, the graph of
h(x) begins to appear more and more like the graph of y  15 x  21 . Thus this is the equation of our oblique
asymptote.
13. Prove that the above limit is indeed zero with the technique used in part a.
c. Find the vertical asymptotes for h(x) and express them in terms of limits.
Solution: In a reduced rational function,
V.A.’s occur at x values in which the
denominator is zero but the numerator
isn’t. It is appropriate to use technology to
find the roots of the cubic denominator.
The NSolve function is for solving an
equation numerically. This means the
output is approximate rather than in
complex radical form. Note the use of the
double equal sign when solving an
equation. There are three roots of the
denominator, and to rule out the possibility
of holes in the graph, we evaluate each in
the numerator (in the form of a list
enclosed in braces). Since the numerator is
not zero at any of these values, they are
indeed locations of V.A.’s. This means that h(x) zooms off to   at each of these three x values. Let’s look at a
graph to determine whether the graph is heading up or down at these locations. The Plot function was used to graph both
h(x) and the oblique asymptote, which was defined as q above. The graph makes the following limits apparent:
lim
x 1.277 
h( x)   ,
lim
x 1.277 
h( x)   ,
lim
x 0.462 
h( x)   , and
lim
x 0.462 
h( x)   .
14. Write the remaining two one-sided limits for V.A.’s
15. Each student should come up with an example of his/her own of a function with the following properties: two
nonremovable discontinuities (one jump, one infinite); one removable discontinuity; a domain of all real numbers; a
horizontal at y = 3; x-intercepts at 2; a y-intercept at +7. Write the one-sided limits at your removable and jump
discontinuities. Write limits as x   ∞. Write one-sided limits as x approaches the point where you have an infinite
discontinuity.
* Formal definition of a limit:
lim f ( x)  L    0,    0 such that | f (x) - L | <
xa
 whenever 0 < | x - a | < .
 means “for each” or “for every”;  means “there exists”;
 means “if and only if ” or “is logically equivalent to”; | x - a | is the distance between the variable x and the constant
a; | f ( x ) - L | is the distance between the y value corresponding to x and the limit as x approaches a.
This is a formal way of saying that no matter how close you’d like the y values to be to L, this can be always be
accomplished using x values close to a.
 and  are small, positive real numbers;
* □ Sample problem VIII: Do a formal - proof to show that lim (3 x  1)  11.
x4
Solution: Let  > 0 be given. No matter how small this  is, we must show there is a  such that whenever x is within
 units of 4, y will be within  units of -11. We begin with some preliminary work: The distance between the y
values and the proposed limit is given by | (-3x + 1) - (-11) | = | -3x + 12 | = | -3(x - 4) | = |-3|  | x - 4| = 3 | x - 4|.
This is what must be less than , no matter how small  is. So we write 3 | x - 4| < . If the limit really is -11, then
even if  = 0.000000000001, then there should still be a  such that when the distance between x and 4 is less than 
(but not zero), the corresponding y values will be within 0.000000000001 of -11. To show that such a  exists for any
conceivable , we have to find a relationship between  and . 3 | x - 4| <  implies that | x - 4| <  /3. This means
that the distance between x and 4 must be less than  /3. Thus we choose  =  /3. To complete the proof we essentially
work backwards and show that this is the appropriate choice for  : If | x - 4| <  /3, then 3 | x - 4| < , implying
| 3x - 12 | <  and | -(3x - 12) | < . So, | -3x + 12 | < , and | (-3x + 1) - (-11) | < . We have used the definition of
a limit to show that the as x gets infinitely close to 4, y gets infinitely close to -11.
* 16. Do a formal - proof to show that lim (5 x  3)  7.
x2
* 17. Do a formal - proof to show that lim (2 x  10)  8.
x  1
* 18. What choice of  would you make (in terms of ) to show that lim (mx  b)  L ?
x c
* 19. We’ve been making use of the fact that for real numbers a and b, |ab| = |a| |b|. Prove that this is indeed the case.
Hint: Consider the following cases for a and b: both positive; both negative; one positive and one negative; and at least
one of them zero. Also, use the fact that |x| = x when x  0 and |x| = -x when x  0.
* □ Sample problem IX: Use the definition of a limit to argue that lim (3x  1)  20.2.
x 7
Solution: To do this we must show that there is some  such that it is impossible for the y values to be within  of
20.2 for all x values within  of 7, no matter how small  is. Suppose we try  = 0.5. This choice will not suffice
since the interval (20.2 - , 20.2 + ) includes the real limit of 20.0. That is, for a small  like 0.01, all the x values in
the interval (7 - , 7 + ) are mapped to y values contained in (20.2 - , 20.2 + ). So, let’s try a smaller , one that will
not allow any “overlap” between the true limit and the claimed limit. If  = 0.1, then no matter how small  is, at least
some of the values in (7 - , 7 + ) will be not be mapped to (20.2 - , 20.2 +  ). We know this is the case since x
values very close to 7 are mapped to y values very close to 20, closer to 20 than  can “reach” from 20.2.
* 20. Use the definition of a limit to argue that lim (4  x)  15.99.
x  12
* □ Sample problem X: Use the definition of a limit to prove that lim (4 x 2  1)  15.
x 2
Solution: Let  > 0 be given. | f (x) - L | = | 4x 2 - 1 - 15 | = | 4x 2 - 16 | = 4 | x + 2 |  | x - 2 |. Since | x - 2 | = | x - a |, the
quantity 4 | x + 2 | is analogous to the slope in the linear proofs, and we need to get a handle on just how big this can be.
If  < 1, then our x values are in the interval (1, 3). The biggest 4 | x + 2 | can be in this interval is 20 (when x = 3). So
we choose  = min(1,  /20), which means let  be the smaller of 1 and  /20 (remember that  depends on  ). For
values of  over 20, we’d go with 1 for  ; for small values of , we’d go with a value of  twenty times smaller. Now
we show our choice for  works: 0 < |x - 2 | <   |x - 2 | < min(1,  /20)  | x - 2 | <  /20 and | x - 2 | < 1.
| x - 2 | <  /20  20 | x - 2 | <  , and since | x - 2 | < 1, we know x < 3 and 4 | x + 2 | < 20. Together these inequalities
imply 4 | x + 2 |  | x - 2 | < , from which it follow that | 4x 2 - 1 - 15 | < . Thus, lim (4 x 2  1)  15. Note: working in an
x 2
interval of radius 1 around x = 2 was completely arbitrary (the radius, that is); the proof would have been very similar
for any choice of interval.
* 21. Do a formal - proof to show that lim (3x 2  4)  79.
x 5
Definition of Continuity:
.
If lim f ( x)  f (a), then we say f (x) is continuous at a.
xa
We say f (x) has a removable discontinuity at x = a if f (x) is not continuous at a but can be made continuous with an
appropriate definition for f (a). Graphically, they correspond to holes. Example: f (x) = (x 2 - x) / x is discontinuous at
x = 0. By reducing the fraction, the function can be written as f (x) = x - 1, for x  0. Thus the graph is simply a line with
a hole at (0, -1). In this case the discontinuity is removable since it is possible to “plug the hole.” We do so adding to the
definition of f (x). Currently f (0) is undefined since it yields division by zero, but if we define f (0) to be -1, then the
discontinuity vanishes.
We say f (x) has a nonremovable discontinuity at x = a if f (x) is not continuous at a and cannot be made continuous
with any definition for f (a). Graphically, they correspond to vertical asymptotes, jumps, or infinite oscillations (see pg.
76). Example: f (x) = (x + 7) / (x + 5) is discontinuous at x = -5, at which point exists a vertical asymptote. While it is
possible to define f (-5), no definition will make f (x) continuous at x = -5.
□ Sample problem XI (by Raquel Roney, Nayeon Kang, & Ayush Dulguun): Is the function
f ( x) 

2 x3 1,
6 x 3,
x<2
x2
continuous at x = 2 ?
Solution: We must see if f (2) = lim f ( x ). f (2) = 6(2) - 3 = 9, but lim f ( x) D.N.E. since the one-sided limits are not
x 2
x 2
equal. Approaching from the right, we use the top rule: lim f ( x )  2(2)3 - 1 = 15. Approaching from the left we use
x2
the bottom rule: lim1 f ( x)  6(2) - 3 = 9. Thus f (x) is not continuous at x = 2.
x 2
□ Sample problem XII: Determine whether the function f ( x) 
x2  9
is continuous for all values of x, and if
x 2  x  12
not, make a continuous extension by removing any removable discontinuities.
( x  3)( x  3)
. We see that the function is continuous except when x = 3 (a hole) and
( x  4)( x  3)
when x = -4 (a V.A.). We can rewrite the function as: f ( x)  ( x  3) /( x  4), x  3. We now take a limit:
Solution: First factor: f ( x) 
lim
x 3
x3
 6. Note that it doesn’t matter that f (3) is undefined when taking this limit. Thus we must redefine f (x) so
x4
( x  3) /( x  4), x  3
that f (3) will be -6: f ( x)  

 6, x  3
With this definition we have removed the removable
discontinuity, but there is still a nonremovable discontinuity (in the form of a vertical asymptote) at x = -4.
22. Using the correspondence a↔1, b↔2, c↔3,…, find the number associated with absolute value of the difference of
your own first and last initial. Call this number p. For example, for me, with initials L.B., I get p = | 12-2 | = 10. For
you, what is the limit of f (x) = x 2 -3 as x p ? * Prove this using the definition of a limit. Prove that f (x) is
continuous at x = p. * Use the definition of a limit to prove that the limit as x7 of f (x) is not 14. What is the limit of
f (x) = x2 -3 as x  ∞ ? Does f (x) have any horizontal asymptotes? Why or why not? What about vertical
asymptotes?
23. Is it possible for a function with the following properties to exist? (Explain why not or give an example.) The limit as
x  ∞ does not exist but the function is bounded. The function is not periodic. (Bounded means there exists some
number M such that | f (x) | < M, and periodic means that f (x) has a period, T, where f (x + T ) = f (x) for any value of x
such that x and x + T are in the domain of f. )
24. Problem #63 on page 64.
25.. If you haven’t already, download or open the power program that I wrote in Excel: Office XP version Office 2000
version (You’ll need to be on a machine on which Microsoft Excel has been installed.) This program allows you to
explore limits of power functions. When you open it, choose “Enable Macros.”
a. What types of values of A and n will generate the following types of curves: concave up, decreasing; concave up,
increasing; concave down, increasing; and concave down, decreasing? (There are two sets of answers for some cases.)
b. Which of the above types of graphs will yield a “green length” greater than a “red length”? When are the lengths
always equal? Compose a clearly written, detailed paragraph explaining the following: why the light blue lines are always
equally spaced relative to L but not to x0; and * why delta must be chosen to be smaller than the minimum of the red and
green lengths.
* c. When y = sqrt(x) and x0 = 2, use the program to find appropriate delta values for the following values of epsilon:
0.5, 0.4, 0.3, 0.2, and 0.1.
* d. Suppose you thought (incorrectly) that the limit in part c was 1.7. Use the definition of the derivative to prove that
this can’t be true. That is, use the program to find an epsilon such that no delta will assure that the y-values are within
epsilon of 1.7. (Placing the pointer where the blue lines connect with the axes will display their x or y positions.) Explain
your answer.