Math 31A Discussion Notes
Week 2
October 6 and October 8, 2015
This week we’ll continue discussing continuity and limits, with special attention to the
types of problems you will see on quizzes and homeworks. Time permitting, we will discuss
tangent lines to graphs of continuous functions.
Continuity
Definition. We say that a function f is continuous at a point x = a if:
1. f (a) is defined;
2. limx→a f (x) exists;
3. limx→a f (x) = f (a).
We say that f is continuous on the interval (c, d) if f is continuous at each point a ∈ (c, d).
Example.
1. The function f (x) := |x| is continuous at x = 0, but the function
|x|, x 6= 0,
f (x) :=
42, x = 0
is not. (Why?)
2. The function
sin(1/x), x 6= 0,
g(x) :=
k,
x=0
is continuous on (−∞, 0) and (0, +∞), but is not continuous at x = 0, no matter the
value of k. (Why?)
3. Neither
nor
0, x < 0
f (x) :=
1, x ≥ 0
1, x < 0
g(x) :=
0, x ≥ 0
is continuous at x = 0, but h(x) := f (x) + g(x) is continuous on all of R. (Why?)
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Example. Find constants b and c so that the function

x < −2
4x + b,
cx2 , −2 ≤ x ≤ 2
f (x) :=

12,
2<x
is continuous.
(Solution) Certainly f is continuous when x 6= −2 and x 6= 2; we need to make f continuous
at these two points as well. First, let’s make limx→2− f (x) = limx→2+ f (x). We have
lim f (x) = lim− cx2 = 4c
x→2−
and
lim f (x) = lim+ 12 = 12,
x→2+
x→2
x→2
so in order for limx→2 f (x) to exist and equal f (2), we need c = 3. Then
lim f (x) = lim + 3x2 = 12.
x→−2+
x→−2
But
lim f (x) = lim − (4x + b) = b − 8,
x→−2−
x→−2
so in order for limx→−2 f (x) to exist and equal f (−2), we must have b = 20.
As with limits, we have some important arithmetic properties for continuous functions:
Proposition 1. If f and g are continuous at x = a, then
1. f + g is continuous at x = a;
2. f − g is continuous at x = a;
3. f · g is continuous at x = a;
4. f /g is continuous at x = a, provided g(a) 6= 0.
A very useful property of continuous functions is that they play nicely with limits:
Proposition 2. If f is continuous at x = g(a) and limx→a g(x) exists, then
lim f (g(x)) = f lim g(x) .
x→a
x→a
√
x is continuous on its domain, so
q
√
√
lim x2 + 7 = lim (x2 + 7) = 16 = 4.
Example. The function f (x) =
x→3
x→3
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The Squeeze Theorem
As useful as the limit laws are, there many limits which simply will not fall to these simple
rules. One helpful tool in tackling some of the more complicated limits is the Squeeze
Theorem:
Theorem 3. Suppose f, g, and h are functions so that
f (x) ≤ g(x) ≤ h(x)
near a, with the exception that this inequality might not hold when x = a. Then
lim f (x) ≤ lim g(x) ≤ lim h(x).
x→a
x→a
x→a
In particular, if limx→a f (x) = L = limx→a h(x), then
lim g(x) = L.
x→a
Example.
1. Evaluate the limit limx→0 (x · cos(1/x)), if it exists.
(Solution) We know that −1 ≤ cos(1/x) ≤ 1 for all x 6= 0. Then −x ≤ x·cos(1/x) ≤ x,
so
lim (−x) ≤ lim (x · cos(1/x)) ≤ lim x.
x→0
x→0
x→0
Since limx→0 (−x) = 0 = limx→0 x, we see that
lim (x · cos(1/x)) = 0.
x→0
2. Use the fact that limx→0
sin(x)
x
= 1 to evaluate
sin(4x)
,
x→0 sin(6x)
lim
if this limit exists.
(Solution) For x 6= 0 we can rewrite this quotient as
sin(4x)
x · sin(4x)
4
6x
sin(4x)
=
=
.
sin(6x)
x · sin(6x)
6
sin(6x)
4x
Then
6x
sin(4x)
lim
lim
x→0 sin(6x)
x→0
4x
−1 4
sin(6x)
sin(4x)
=
lim
lim
x→0
6 x→0 6x
4x
4
2
= ·1·1= .
6
3
sin(4x)
4
lim
=
x→0 sin(6x)
6
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Limits at Infinity
We’ll carry out two illustrative examples of limits at infinity.
Example.
8x5 + 3x2 − 4
1. Find lim
, if it exists.
x→∞
4 − 9x5
(Solution) Neither limx→∞ (8x5 + 3x2 − 4) nor limx→∞ (4 − 9x5 ) exists, so we cannot
very well consider a ratio of these limits. What we can do, however, is rewrite this
quotient so that the numerator and denominator limits exist. For x 6= 0, we have
8x5 + 3x2 − 4
8x5 + 3x2 − 4 x−5
8 + 3x−3 − 4x−5
,
=
·
=
4 − 9x5
4 − 9x5
x−5
4x−5 − 9
so
8x5 + 3x2 − 4
8 + 3x−3 − 4x−5
limx→∞ (8 + 3x−3 − 4x−5 )
=
lim
=
x→∞
x→∞
4 − 9x5
4x−5 − 9
limx→∞ (4x−5 − 9)
8
8
=
=− .
−9
9
lim
This technique of writing the denominator as a constant term plus terms with negative
exponents is a good general strategy for determining the end behavior of rational
functions.
2. Consider f (x) :=
exists.
sin(2x + 7) cos(x2 ) + cos2 (4 − x3 )
. Find limx→∞ f (x), if this limit
x
(Solution) This limit may look daunting, but we need only recall that the sine and
cosine functions are bounded. Since sine and cosine take values between −1 and 1, the
values of the product sin(2x + 7) cos(x2 ) will be between −1 and 1, and the values of
cos2 (4 − x3 ) will be between 0 and 1. Thus, the values in the numerator of f will be
between −1 and 2. Since the denominator grows without bound as x → ∞, we see
that limx→∞ f (x) = 0.
Tangent Lines
Given a continuous function f , suppose we want to find the equation of the line which lies
tangent to the graph of y = f (x) at the point (a, f (a)). At first glance this seems like a
difficult problem to approach naively (that is, without derivatives). But we can tackle it
with relative ease by considering secant lines.
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Definition. Given a continuous function f , the secant line passing through (a, f (a)) and
(b, f (b)) (where a 6= b) to the graph of y = f (x) is defined by
y=
f (b) − f (a)
(x − a) + f (a).
b−a
(1)
Intuitively, we can see that the secant line passing through (a, f (a)) and (b, f (b)) should
begin to look more and more like the tangent line at (a, f (a)) as b gets closer to a. So we
should be able to obtain the equation of the tangent line by taking the limit of equation (1).
Example. Find the equation of the tangent line to the graph of y = x2 + 4 at (3, 13).
(Solution) For some b 6= 3, the equation of the secant line passing through (3, 13) and
(b, f (b)) is given by
y=
(b2 + 4) − 13
f (b) − f (3)
(x − 3) + 13 =
(x − 3) + 13.
b−3
b−3
(2)
2
−9
Notice that the only part of this equation which depends on b is the slope, bb−3
. So we
should be able to find the tangent line by finding the limit of this slope as b approaches 3,
and replacing the slope in (2) with this limit. We have
(b − 3)(b + 3)
b2 − 9
= lim
= lim(b + 3) = 6,
b→3
b→3
b→3 b − 3
b−3
lim
so the equation of the tangent line is given by
y = 6(x − 3) + 13 = 6x − 5.
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