MA 15910 Lesson 10 Notes, Section 3.1 (2nd half of text)
Part 2 of Limits
In lesson 9 (previous lesson), we determined a few ways to find a limit lim f ( x) .
x a



Sometimes, especially with polynomial expressions or polynomial functions, direct
substitution could be used; the limit was the value when a was evaluated in f(x).
Sometimes we could use a table (or a graph) and select values closer and closer to a, from
the left and from the right to see if the approached values were the same.
Sometime we could simplify the expression f(x) and then use direct substitution.
Here is a summary.
Techniques for Evaluating Limits:
1. With a polynomial function (or many other functions), direct substitution sometimes can be
used to find the limit. See the examples below.
a)
lim ( x 2  x)  (2) 2  (2)  4  2  6
x  ( 2)
b)
lim(2n  4)  2(3)  4  6  4  2
c)
lim 2a  6  2(5)  6  16  4
n 3
a 5
1
 2n  5  2(2)  5 1
lim 


or 

n2
2 1
3
3
 n 1 
With a rational function or rational expression, you can sometimes write an equivalent
expression for the function by simplifying, then use ‘direct substitution’ in the
equivalent expression. Direct substitution before simplifying may yield 0/0, the
indeterminant form.
 x2  9 
 ( x  3)( x  3) 
a) lim 
 lim 
x  3)  3  3  6

  lim(
x 3
x3
 x 3
 x  3  x 3 
d)
2.
 c 2  4c  5 
 (c  5)(c  1) 
b) lim 
c  1)  5  1  6
  lim

  lim(
c 5
c

5
c5

 c 5
 c5 
 3( x  x)  2  (3 x  2) 
 3 x  3(x)  2  3 x  2 
c) lim 
 lim 


x  0
x
x

 x  0 

 3(x) 
 lim 
  lim 3  3
x  0
 x  x  0
In this lesson we will discover a few other techniques to find limits.
We have ‘rationalized denominators’ to simplify an expression with a square root in the
denominator. Sometimes this was done by multiplying numerator and denominator by the
conjugate of the denominator as in this expression.
2
2( 5  3)
2( 5  3) 2( 5  3) ( 5  3)
3
5
Simplify:




or 
59
4
2
2 2
5  3 ( 5  3)( 5  3)
1
When finding limits and direct substitution lead to the indeterminant form, it is sometimes
beneficial to rationalize the numerator.
Examine this example:
Ex 1: Find the limit:
 x 4
lim 

x 16
 x  16 
 ( x  4)( x  4) 
 lim 

x 16
 ( x  16)( x  4) 
Numerator and Denominator were multiplied
by ( x  4), the conjugate of the numerator.


x  16
 lim 
 Cancel the common factor
x 16 ( x  16)( x  4)


 1 
 lim 
 Direct substituion.
x 16
 x 4
1
1
1



16  4 4  4 8
Techniques for Evaluating Limits (continued):
3.
 x a
With a limit of the form lim 
 , try rationalizing the numerator.
x a
x

a


Ex. 2: Find each limit.
 12  x 
a) lim 

x 144 144  x 


 a 2
b) lim 

a 4
 a4 
2
Basic limit at infinity principle: For any positive real number n,
 1 
 1 
lim  n   0 and lim  n   0 (for positive values of x only in the 2nd case).
x  x
x   x
 
 
1
Why is the principle above true? Think. A number such as 10000000000000000  0 .
(Limits at infinity correspond to horizontal asymptotes of the graph
of a function.)
Study this example:
 4x  3 
 4 3 
4
 4x  3 
x  40
lim 
 lim  5xx 2x   lim 



x  5 x  2

 x    x  5  2  5  0 5
x 
 x x

∞
Direct substitution of ∞ or -∞ does not yield a number or limit value. Sometimes you get ∞ or
−∞
∞
. These expressions have no meaning. A technique described below must be used.
Techniques for Evaluating Limits (continued):
4.
When finding the limit of a rational expression or rational function when the variable is
approaching ±∞, divide each term of the numerator and denominator by the highest
power of the variable in the denominator, simplify and then evaluate the limit. Use the fact
 1 
 1 
that lim  n   0 and lim  n   0 (basic limit at infinity principle from above).
x  x
x  x
 
 
Ex 3: Find each limit.
 12u 2

a) lim  2

u  5u  3u  1


b)
 9 x 
lim 

x  3 x  7


3
c)
e)
 2 x2  4 
lim  3

x  3 x  4 x 2  x


 2 x2  4 x  3 
lim 

x 
x 5


d)
f)
 4c5  3c3  2c 
lim  5

c  5c  4c 4  6c


 5v3  3v 4 
lim  3

v  2v  3v  2


Note: Our
textbook and
MyMathLab
does accept a
limit answer of
∞ or -∞.
Short-cuts for limits at infinity:
1)
If the degree of the numerator is less than the degree of the denominator, the limit will be
0.
2)
If the degree of the numerator is the same as the degree of the denominator, the limit is
the ratio of the coefficients of the highest powers from numerator and denominator.
3)
If the degree off the numerator is greater than the degree of the denominator, there is no
limit (or the limit would be ∞ or -∞).
3x5  4 x3  12 x
g ) lim

x  6 x  9 x 3  7 x 5
4x  2
lim

x  3x 2  8
4a8  7a 4
lim 4

a  8a  9a 2  2
4
Example 4: This example is problem 84 on page 139 of the textbook. You will need to look
at the graph of this problem.
The graph shows how the postage required to mail a letter in the U. S. has changed in recent
years. Let C(t) be the cost to mail a letter in the year t. Find the following. (You will have to
approximate to the nearest cent.)
a)
c)
lim C (t )
t  2009 
lim C (t )
t  2009
b)
lim C (t )
t  2009
d)
C (2009)
 x  2 if x  3

if 3  x  6
Ex 5: Given: f ( x)  1
2
 x  2 if x  6

Find the following, if they exist.
(a) lim f ( x) 
x 3
(b) lim f ( x) 
x 6
5