Evaluating Limits Algebraically
We have looked at cases in which we can
use algebra to rigorously determine the
value of the limit.
Substitution (if f is continuous at c).
Continuous extension (if f has a removable
discontinuity at c).
But in all those cases, the limit existed.
How can algebra help us determine if a
limit does not exist?
f(x) = 1/x
5
-5
5
Does the limit exist
as x approaches 0?
-5
lim f(x) = ∞
f(x) = 1/x
5
-5
x→0+
5
Vertical
Asymptote
lim f(x) = –∞
x→0–
-5
Vertical Asymptote
A function f has a vertical asymptote at
x=c if at least one of the one-sided limits
as x approaches c is infinite.
lim f(x) = ∞
lim f(x) = –∞
lim f(x) = ∞
lim f(x) = –∞
x→c–
x→c+
x→c–
x→c+
“1/0 Is Infinity”
The tendency of f(x) = 1/x toward infinity
as x approaches 0 is the reason we say
that “1/0 is infinity”…
“0/0” was indeterminate and could be anything;
the limit exists in that case.
“#/0” (i.e., any other value divided by zero)
indicates that the limit tends toward infinity.
After all, what happens if you divide 1 by a
value that becomes smaller and smaller?
f(x) = 1/x
5
-5
5
Does the limit exist as
x gets infinity large?
-5
f(x) = 1/x
5
-5
5
Horizontal
Asymptote
-5
Horizontal Asymptote
A function f has a horizontal asymptote
at y=L if the limit as x approaches ∞ or –∞
is defined.
lim f(x) = L
x→∞
lim f(x) = L
x→–∞
A graph has zero, one, or two asymptotes.
What happens at ±∞ is known as the end
behavior of the graph.
Asymptotes
f(x) = ex
f(x) = x3
f(x) = arctan x
f(x) = 1/x2
Analyzing for Asymptotes
Consider the function
2
x –x+5
f(x) =
x–2
Let’s find
lim f(x)
x→2
Analyzing for Asymptotes
Substitution yields 7/0, which is not
indeterminate—therefore, the graph likely
contains a vertical asymptote, with the
limits tending toward infinity.
Graphically, we can see the left-hand limit
is –∞ and the right-hand limit is ∞.
How can we prove this algebraically?
Analyzing for Asymptotes
As x approaches 2, the numerator is
positive (and approaches 7).
When x<2, the denominator is negative,
and as x approaches 2 from the left, we
have (positive)/(negative) = negative.
Thus, this one-sided limit is –∞.
Try applying the same logic to the righthand limit.
A Few Notes
If a limit is one of the “infinities” then it
technically doesn’t exist, but to see your
understanding of the concept I would
prefer you indicate that it equals ∞ or –∞.
If both the LH and RH limits are the same
(i.e., both –∞), then it is acceptable to say
that the two-sided limit is also –∞, even
though ∞ technically doesn’t equal ∞. 
End Behavior Models
An end behavior model is a method of
determining what is happening to a
function as x becomes infinitely large (in
either the positive or negative direction).
For example, compare the graphs of the
two functions below.
f(x) = x3
g(x) = x3 – 6x2 + 2x – 5
End Behavior Models
As x becomes larger, the two graphs begin
to look quite similar.
This is because g(x) becomes asymptotic
to f(x).
Consider what happens as x becomes
infinitely large: the process of cubing x
affects the value of y more so than
squaring, subtracting 5, etc.
End Behavior Models
x
x^3
How far
off?
x^3-6x^2+2x-5
1
1
4
400.000%
10
1000
1615
161.500%
100
1000000
1060195
106.020%
1000
1000000000
1006001995
100.600%
10000
1000000000000
1000600019995
100.060%
100000
1000000000000000
1000060000199990
100.006%
1000000
1000000000000000000
1000006000002000000
100.001%
End Behavior of Rational Functions
End behavior models are useful for
determining what happens to rational
functions as x becomes infinitely large.
Consider the function
x10–7x5+2x3–1
h(x) =
3x10–4x7–9x+88
How can we easily determine what
happens as x approaches ∞ and –∞?
End Behavior of Rational Functions
As was the case when comparing the two
polynomials, what really matters is the
highest power of x in both terms.
We simply ignore the rest.
h(x) =
10
5
3
x –7x +2x –1
10
7
3x –4x –9x+88
End Behavior of Rational Functions
Thus, for sufficiently large values of x,
10
x
1
h(x) ~
10
3x
=
3
The new function is an end behavior
model for h, and can be used to determine
1
lim h(x) =
x→±∞
3