c
Math 171, Benjamin
Aurispa
2.2 The Limit of a Function
Introductory Example: Consider the function f (x) = √
x is near 0.
x
−1
−0.5
−0.1
−0.05
−0.01
−0.001
−0.0001
f (x)
3.7320508
3.8708287
3.9748418
3.9874607
3.9974984
3.999750
3.999975
x
1
0.5
0.1
0.05
0.01
0.001
0.0001
x
. Examine the values of the function when
x+4−2
f (x)
4.236068
4.1213203
4.0248457
4.0124612
4.0024984
4.000250
4.000025
We can see that as the values of x get closer to 0 from both sides, the values of f (x) are getting closer to 4.
x
=4
This is written mathematically as: lim √
x→0
x+4−2
We write
lim f (x) = L
x→a
read as “the limit of f (x) as x approaches a is L” if we can make the values of f (x) arbitrarily close to L
(as close to L as we want) by letting x get closer and closer to a, but not equal to a.
0
Note that f (0) is undefined in the above example since f (0) = √0+4−2
= 00 . When dealing with limits, we
are examining values as x approaches a, but not equal to a. We’ll see later that the value of the function at
a may or may not equal the value of the limit. For all three figures below:
Left-Handed Limit: lim− f (x)
x→a
Right-Handed Limit: lim+ f (x)
x→a
The limit exists if and only if the left-handed and right-handed limits both exist and are equal.
lim f (x) = L if and only if lim f (x) = L and lim f (x) = L
x→a
x→a−
x→a+
1
c
Math 171, Benjamin
Aurispa
Infinite Limits: Calculate lim
x→0
x
−1
−0.5
−0.1
−0.05
−0.01
−0.001
−0.0001
f (x)
1
4
100
400
10, 000
1, 000, 000
100, 000, 000
1
x2
x
1
0.5
0.1
0.05
0.01
0.001
0.0001
f (x)
1
4
100
400
10, 000
1, 000, 000
100, 000, 000
We can see that as the values of x approach 0 from both sides, that f (x) gets larger and larger without
1
bound. In this case, we say that lim 2 = ∞.
x→0 x
1
1
1
Example: Calculate lim+ , lim− , and lim
x→0
x
x→0 x x→0 x
Definition: The line x = a is a vertical asymptote if the limit from the left, right, or both is ∞ or −∞.
Example: Consider the graph of f below. Find the indicated limits.
lim f (x)
x→4−
lim f (x)
x→4+
lim f (x)
x→4
8
lim f (x)
x→2−
lim f (x)
x→2+
lim f (x)
x→2
6
lim f (x)
4
x→−1
2
lim f (x)
x→−6
−8
lim f (x)
x→−3−
lim f (x)
x→−3+
−6
−4
−2
2
lim f (x)
x→−3
−2
lim f (x)
−4
x→6
−6
What are the vertical asymptotes of f (x)?
−8
2
4
6
8
c
Math 171, Benjamin
Aurispa
Vertical Asymptotes vs Holes
When the limit of a function at x = a is of the form nonzero
, then there is a vertical asymptote at x = a.
0
Why?
When the limit of a function at x = a is of the form 00 we say the limit is indeterminate. Two things could
be happening. There is either a vertical asymptote or a hole in the graph at x = a. If after using algebra the
, then there is a vertical asymptote. If the limit simplifies to a number, then there
limit simplifies to nonzero
0
is a hole. In other words, if a factor in the denominator can be cancelled completely from the denominator,
then there is a hole not a vetical asymptote.
Find all vertical asymptotes of the function f (x) =
at all such values of x.
x2
x−2
and calculate the left and right-hand limits
− 6x + 8
Calculate the following limits:
lim −
x→−5
x−1
x+5
lim +
x→−5
x−1
x→−5 x + 5
x−1
x+5
lim
x3
x→3 (x − 6)(x − 3)2
lim
3
c
Math 171, Benjamin
Aurispa
2.3 Calculating Limits Using the Limit Laws
Limit Laws: Also found on pp. 91-93 of your textbook. Suppose lim f (x) and lim g(x) exist and that c is
x→a
x→a
any constant. Then:
1. lim (f (x) ± g(x)) = lim f (x) ± lim g(x)
x→a
x→a
x→a
2. lim cf (x) = c lim f (x)
x→a
x→a
3. lim f (x)g(x) = lim f (x) lim g(x)
x→a
x→a
f (x)
x→a g(x)
4. lim
x→a
lim f (x)
=
x→a
lim g(x)
x→a
provided that lim g(x) 6= 0
x→a
n
5. lim (f (x))n = lim f (x)
x→a
6. lim
x→a
p
n
f (x) =
In particular,
x→a
q
n
lim f (x). If n is even, then we must have that lim f (x) > 0
x→a
x→a
7. lim c = c
x→a
8. lim x = a
x→a
9. lim xn = an
x→a
Given that lim f (x) = 16, calculate lim
x→3
x→3
p
(x2 − 4)f (x)
f (x) + x + 2
If f is a polynomial or rational function and a is in the domain of f , then lim f (x) = f (a). If you can
x→a
evaluate the function at a and you don’t get division by zero, then that value is the limit!!
lim (3x2 + 5x + 1)4
x→−2
4
c
Math 171, Benjamin
Aurispa
We saw in the previous section what happens when only the limit of the denominator is 0. There is an
infinite limit which yields a vertical asymptote.
If the limits of both the numerator AND denominator are 0, the limit is indeterminate, so you must USE
ALGEBRA to determine the limit. Some methods are expanding, factoring, or multiplying by the conjugate
of a radical.
x2 − x − 12
x→4
x2 − 16
lim
(h − 4)2 − 16
h→0
h
lim
lim
t→2
t−2
(t − 2)3
√
x+3−2
lim
x→1
x−1
5
c
Math 171, Benjamin
Aurispa
For vector functions, if r(t) =< f (t), g(t) >, then
lim r(t) =
t→a
lim f (t), lim g(t)
t→a
t→a
provided the limits of the component functions exist.
Calculate lim r(t), where r(t) =
t→3
*
t−3
√
t2 + 7 − 4
9t−1 − 3(t − 2)−1
,
t−3
+
Find a number a so that the following limit exists and then find the value of the limit.
x2 + ax + a + 5
lim
x→−3 (x + 3)(x − 5)
6
c
Math 171, Benjamin
Aurispa
Let f (x) =

x




2

 x
8−x



−2



x−2
if
if
if
if
if
x<0
0≤x<2
2≤x<5
x=5
x>5
Calculate lim f (x), lim f (x), and lim f (x) or explain why the limit does not exist.
x→0
x→2
x→5
Recall the definition of the absolute value function: |x| =
(
x2 + x
or explain why the limit does not exist.
x→0
|x|
Calculate lim
7
c
Math 171, Benjamin
Aurispa
|x − 3|
or explain why the limit does not exist.
x→3 6 − 2x
Calculate lim
Squeeze Theorem: If g(x) ≤ h(x) ≤ f (x) for all x in an interval that contains a (except possibly at a) and
lim g(x) = lim f (x) = L
x→a
x→a
then lim h(x) = L
x→a
Example: If 4x − 2 ≤ f (x) ≤ x2 + 2 for 0 ≤ x ≤ 3, find lim f (x).
x→2
Example: Find lim x2 sin
x→0
1
x
.
8