Infinite Limits
Lesson 1.5
Infinite Limits
Two Types of infinite limits.
Either the limit equals infinity or the limit is
approaching infinity.
We are going to take a look at when the limit
equals infinity, for now.
1.5 Infinite Limits
• Vertical asymptotes at x = c will give you
infinite limits
• Take the limit at x = c and the behavior of
the graph at x = c is a vertical asymptote
then the limit is infinity
• Really the limit does not exist, and that it
fails to exist is b/c of the unbounded
behavior (and we call it infinity)
The function f(x) will have a vertical
asymptote at x = a if we obtain any of
the following limits:
lim f ( x)  
x a
lim f ( x)  
xa
lim f ( x)  
xa
Definition of Infinite Limits
f(x) increases without
bound as x  c
NOTE: may decrease
without bound ie: go to
negative infinity!!
M --------------
Vertical Asymptotes
• When f(x) approaches
infinity as x → c
– Note calculator often
draws false asymptote
• Vertical asymptotes generated by
rational functions when g (x) = 0
f ( x)
h( x ) 
g ( x)
c
Theorem 1.14
Finding Vertical Asymptotes
• If the denominator = 0 at x = c AND the
numerator is NOT zero, we have a vertical
asymptote at x = c!!!!!!! IMPORTANT
• What happens when both num and den
are BOTH Zero?!?!
A Rational Function with Common Factors
(Should be x approaching 2)
• When both numerator and denominator are both
zero then we get an indeterminate form and we
have to do something else …
x  2x  8
lim
x 2
x2  4
2
– Direct sub yields 0/0 or indeterminate form
– We simplify to find vertical asymptotes but how do we
solve the limit? When we simplify we still have
indeterminate form.
x4
lim
, x  2
x 2 x  2
A Rational Function with Common
Factors, cont….
• Direct sub yields 0/0 or indeterminate form. When we
simplify eliminate indeterminate form and we learn that
there is a vertical asymptote at x = -2 by theorem 1.14.
• Take lim as x-2 from left and right
x  2x  8
lim

2
x 2
x 4
2
x  2x  8
lim
 
2
x 2
x 4
2
• Take values close to –2 from the right and values close
to –2 from the left … Table and you will see values go to
positive or negative infinity
Determining Infinite Limits
• Denominator = 0 when x =
1 AND the numerator is
NOT zero
– Thus, we have vertical
asymptote at x=1
• But is the limit +infinity or
–infinity?
• Let x = small values close
to c
• Use your calculator to
make sure – but they are
not always your best
friend!
x 2  3x
x 2  3x
Find lim
and lim
x 1
x 1
x 1
x 1
Infinite Limits:
4
3
1
f  x 
x
2
1
-4
As the denominator approaches
zero, the value of the fraction gets
very large.
-3
-2
-1
0
If the denominator is negative then
the fraction is negative.
2
3
4
-1
-2
-3
-4
If the denominator is positive then the
fraction is positive.
1
vertical
asymptote
at x=0.
1
lim  
x 0 x
1
lim  
x 0 x

Example 4:
1
lim 2  
x 0 x
1
lim 2  
x 0 x
The denominator is positive
in both cases, so the limit is
the same.
1
 lim 2  
x 0 x

Properties of Infinite Limits
• Given
lim f ( x)   and lim g ( x)  L
x c
Then
• Sum/Difference
x c
lim  f ( x)  g ( x)  
x c
• Product
lim  f ( x)  g ( x)  
lim  f ( x)  g ( x)  
x c
• Quotient
x c
 g ( x) 
lim 
0

x c
 f ( x) 
L0
L0
Find each limit, if it exists.
1.
 1 
lim 

x 4  x  4 
6
4
2
-5
5
-2
-4
-6
Find each limit, if it exists.
1.
1
 1 
lim 


x 4  x  4 
3.999  4
One-sided limits
will always exist!
1

VS 
 
Very small
negative #
6
4
2
-5
5
-2
-4
-6
 1 
2. lim 

x 1 x  1


6
4
2
-5
5
-2
-4
-6
 1 
2. lim 
 DNE

x 1 x  1


This time we only care if the two
sides come together—and where.
Can’t do Direct Sub, need to go to our LAST resort…
check the limits from each side.
1
 1 
lim 


x 1  x  1 
0.999  1
1

VS 
 
6
1
 1 

lim 

x 1  x  1 
1.001  1
1

VS 

4
2
-5
5
-2
-4
-6
3.
Find any vertical asymptotes of
x 2  2x  8
f (x) 
x2  4
6
4
2
-5
5
-2
-4
-6
3.
Find any vertical asymptotes of
x 2  2x  8
f (x) 
x2  4
x  4  x  2


 x  2 x  2
6
4
2
-5
-2
Discontinuous at x = 2 and -2.
x4

x2
V.A. at x = -2
5
-4
-6
Hole at  2, 32 
Try It Out
2 x
g ( x)  2
x 1  x 
• Find vertical asymptote
2
x
lim 2
x 4 x  16
• Find the limit
• Determine the one sided limit
x 1
f ( x)  2
x  x 1
3
lim f ( x)
x 1
Methods
• Visually: Graphing
• Analytically: Make a table close to “a”
• Substitution: Substitute “a” for x
If Substitution leads to:
1) A number
L, then L is
the limit
4) 0/0, an
indeterminant form,
you must do more!
2) 0/k, then
the limit is
zero
3) k/0, then the
limit is ±∞, or
dne