Section 2.5 Limits involving infinity
I] Infinity, Minus Infinity as a limit as x approaches a:
The line x=a is a vertical asymptote of the function.
Basic limits to remember:
lim
x0
1
x
 ,
Find each limit for the function f ( x ) 
1
lim
x0
x
x 2  3 x  10
  ,
lim ln | x | 
x0
.
x2  4
a) lim f ( x ) This is a rational function and substitution gives the indeterminate form 0/0.
x2
( x  5)( x  2 )
We factor numerator and denominator.
x5
b) lim f ( x )  lim
x  2
x  2
( x  2 )( x  2 )

x5
x2

7
4
as
x  2.
DNE because substitution gives 3/0. x= -2 is a vertical
x2
asymptote. In fact the right side approaches positive infinity, the left side approaches
minus infinity. We can check the sign of the expression for x> -2 and see it is + , check
the sign for x< -2 and see it is negative or follow c) and d) below.
c)
lim
x  2 
f ( x) 
d) Similarly,
We write lim
x0
x  2 
lim
x  2 
1
x2
x5
lim
x2

lim
x23
x  20
x2
 lim
t0
t3
t

f ( x )   .
  since both sides approach positive infinity.
Example: Find lim
x0
ln | x |
x
or show it does not exist.
ln|x| is approaching minus infinity, on both sides of 0. x is changing sign and is negative
on the left, positive on the right.
lim
x0
ln | x |
x
 
lim
x0
ln | x |
x

and lim
x0
ln | x |
x
does not exist.
II] Limits as x approaches positive infinity, and as x approaches minus infinity.
The tendency of a polynomial as x approaches plus or minus infinity is the same as the
tendency of its leading term. You can see this two ways:
Graph Y 1  x 3  5 x 2  3 x  10 Y 2  x 3 in your calculator. Use a window in which Xmin
and Xmax are large, say Xmin=1000 and Xmax=1100. The graphs are nearly identical.
If x=1000, then the leading term is on the order of 1000 times any other term.
lim x 3  
lim Y 1  
so
x
lim x 3  
so
x  
The same works at minus infinity:
x
lim Y 1  
x  
Ex. lim ( 2 x 5  20 x 4  15 x 3  35 )  lim  2 x 5   .
x  
x  
Why is this? Factor out the leading term to get  2 x 5 (1 
10
x

15
2x
2

35
2x5
)
. The term
in parentheses approaches 1 as x approaches +infinity or - infinity.
Any odd power of x will approach minus infinity as x approaches minus infinity because
an odd power of a negative is negative. Then the coefficient -2 makes it positive.
Rational functions:
You only need to consider the tendency of the ratio of the leading terms.
Example: lim
2 x 4  15 x 3  16 x 2
x
Example: lim
x
6 x 4  32 x 3  50 x
 lim
x
3 x 2  4 x  10
2 x 3  5 x 2  10 x  15
2x4
6x4
 lim
x

3x 2
2x3
2
6

1
3
 lim
x
y=1/3 is a horizontal asymptote.
3
2x
0
asymptote.
Example:
lim
x  
4 x 4  3x 2  9
3
5x  2 x
2
 lim
x  
4x4
5x
3
 lim
x  
4x
5
 
y=0 is a horizontal
Other limits: "infinity minus infinity" is indeterminate.
Example:
Find lim  x 2  4 x  x 2  1  Both terms tend to infinity so we have
x  

"infinity minus infinity". The limit could be anything or not exist. We must do algebra.
x2  4x 
x2  1 
4x  1
x2  4x 
x2  1

( x2  4x 
x 2  1 )( x 2  4 x 
x2  4x 
4x
2x
2

( x 2  4 x )  ( x 2  1)
x2  4x 
x2  1
The  means ' has the same limit as'.
Exercises: Show lim ( x 2  6 x  x )  3 and
x
x2  1
x 2  1)
lim ( x 2  6 x  x )   3
x  
Show that every odd degree polynomial has at least one root. Use the Intermediate Value
Theorem.
