Right Screen 6 for LC B4
Theorem
1
1
 0 and lim p  0 where p is a real number strictly greater than
p
x

x
x
zero.
lim
x 
EXAMPLE 3
5 x3  x  5
lim
 lim
x  7 x 2  4 x
x 
 lim
x 



5 x3
x
5
x3
x3
x3
7 x2
4x
x3
x3

5
1
x2
 x53
7
x

4
x2
lim 5  lim
x 
divide by highest power of x
simplify
 lim
5
1
2
3
x  x
x  x
7
4
2
x
x 
x  x
lim  lim
apply limit laws
500
evaluate
00
5

(this is an indeterminate form)
0
c
Since this is indeterminate form , c  0 , we know this limit is ∞ or –∞. A little analysis
0
will help determine the correct answer. If we think about substituting in large negative
5 x3  x  5
values of x into
we can make the following conclusions:
7 x2  4 x



The numerator would be a negative number.
The denominator would be a positive number.
Since a negative number divided by a positive number is a negative number, we conclude
5 x3  x  5
lim
  .
x  7 x 2  4 x