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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)
Right-Handed Limit: lim+ f (x)
x→a
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+
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Math 151, Benjamin
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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
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Math 151, Benjamin
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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
, then there is a vertical asymptote. If the limit simplifies to a number, then
the limit simplifies to nonzero
0
there is a hole.
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
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Math 151, Benjamin
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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).
x→a
lim (3x2 + 5x + 1)4
x→−2
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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
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Math 151, Benjamin
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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
*
2(t − 2)−1 − 2
,
t−3
t
9
−
t − 3 t(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)
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Math 151, Benjamin
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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
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Math 151, Benjamin
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|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
.
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Math 151, Benjamin
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2.5 Continuity
In Section 2.3 we saw that the limit as x approaches a can sometimes be found by evaluating the function
at a. If this is the case, then the function is continuous.
Definition: A function is continuous at a number a if
lim f (x) = f (a)
x→a
Otherwise, we say the function is discontinuous at a, or that there is a discontinuity at a.
In order for a function to be continuous at a number a:
(1) f (a) must be defined. – So a function will NOT be continuous anywhere it is undefined.
(2) lim f (x) must exist. (The left-handed and right-handed limits must both equal the same value.)
x→a
(3) lim f (x) = f (a)
x→a
Examples of discontinuities: Holes, vertical asymptotes, and jumps.
A “hole” in a graph is also referred to as a removable discontinuity because if we wanted to, we could
just redefine the function at that point to make it continuous. Removable discontinuities occur where the
limit exists at a (left and right limits are equal), but is not equal to f (a).
A vertical asymptote is referred to as an infinite discontinuity.
A jump in the graph is referred to as a jump discontinuity. Jumps occur where the limits from the left
and right exist, but are not equal.
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Math 151, Benjamin
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A function is continuous from the left at a number a if lim f (x) = f (a) and continuous from the
x→a−
right if lim+ f (x) = f (a). A function is continuous if and only if it is continuous from both the right and
the left.
x→a
Examples: Determine where the functions below are discontinuous. State the type of discontinuity and
explain why mathematically using limits. Is the function continuous from the left or right there?
(1) f (x) =
x2 − 25
x−5
(2) f (x) =
(x + 3)(x − 2)
(x − 2)3
(3) f (x) =
(
x2 − 4
x+1
if x ≤ −1
if x > −1
Fact: All polynomials are continuous everywhere!
Fact: A rational function is continuous wherever it is defined, i.e. where the denominator is not 0.
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Math 151, Benjamin
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(4) f (x) =


 2x − 1
6

 x2 − 9
(5) f (x) =
if x < 4
if x = 4
if x > 4

3x + 1




 x2 − 5
x−1

3 − 25


x


x−2
if x < −2
if − 2 ≤ x ≤ 3
if x > 3
What values of a and b would make the following function continuous everywhere? What are the values of
lim f (x) and lim f (x)?
x→−3
x→1

ax2 + bx + 1


 2
x + 6x − 7
f (x) =

x−1


−4x + a + 2b
if x ≤ −3
if − 3 < x < 1
if x ≥ 1
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Math 151, Benjamin
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If f and g are continuous at a and c is any constant, then the functions f + g, f − g, cf , f g, and
g(a) 6= 0) are all continuous functions.
f
g
(where
The Intermediate Value Theorem: Suppose f is continuous on the closed interval [a, b] and let N be any
number strictly between f (a) and f (b). Then there exists a number c in (a, b) such that f (c) = N .
Example: Show that the equation −x3 + 2x + 2 = 0 has a root (solution) on the interval (1, 2).
Example: If f (x) = x4 − x3 + 3x2 + 2, show that there is a number c so that f (c) = 3.
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