Math 31A Discussion Notes
Week 4
October 20 and October 22, 2015
To prepare for the first midterm, we’ll spend this week working examples resembling the
various problems you’ve seen so far this term. In these notes we reference various results
without explaining them; you can find the explanations in the textbook, Dr. Heilman’s
notes, or our old notes. The questions and solutions are listed separately so that you can
work each problem, and then click on (Solution) to see the solution. Finally, understand
that this is just a sampling of problems that you should know how to do, and
should not be considered an indication of the problems you’ll see on the exam.
Limits
x2 + x − 6
, if it exists. (Solution)
1. Evaluate lim
x→2
x−2
3x4 + 2x3 + x2
, if it exists. (Solution)
x→0 3x4 + 5x3 + 7x2
2. Evaluate lim
3. Find two functions f, g such that limx→a f (x) does not exist, limx→a g(x) does not
exist, but such that
lim (f (x) + g(x))
x→a
does exist. (Solution)
sin2 (θ)
, if it exists. (Solution)
θ→0 1 − cos(θ)
4. Evaluate lim
|x|
, if it exists. (Solution)
x→0 x
5. Evaluate lim
x−1
6. Evaluate lim √
, if it exists. (Solution)
x→1
x−1
1
1
, if this limit exists. (Solution)
7. Evaluate lim
−
t→0
t t2 + t
x
8. Evaluate lim √
, if this limit exists. (Solution)
x→0
1 + 3x − 1
Continuity
9. Find values k, m ∈ R so that
 3
x ≤ −1
 2x + x + 7
f (x) := m(x + 1) + k −1 < x ≤ 2

x2 + 5
2<x
1
is continuous, if possible. (Solution)
10. Find two functions f, g such that neither f nor g is continuous at a, but such that
f + g is continuous. (Solution)
The Squeeze Theorem and Limits at Infinity
11. Find lim θ2 cos(1/θ), if this limit exists. (Solution)
θ→0
12. Evaluate lim
x→0
sin(2x) sin(3x)
, if this limit exists. (Solution)
sin(5x) sin(7x)
sin(2x − 4)
, if this limit exists. (Solution)
x→2
x2 − 4
13. Evaluate lim
cos(x2 − 7)
, if this limit exists. (Solution)
x→∞
x
14. Evaluate lim
The (Limit Definition of the) Derivative
15. Use the limit definition of the derivative to find f 0 (a), where f (x) =
(Solution)
√
x and a 6= 0.
16. Use the limit definition of the derivative to find g 0 (t), where g(t) = 10 + 5t − 4.9t2 .
(Solution)
Computing Derivatives
17. Find
dy
x
, where y = 2
. (Solution)
dx
x +1
√
dy
3x3
7x
7
√
, where y = √
−
+
8
x3 . (Solution)
5
3
dx
x2
x4
√
19. Find g 0 (x), where g(x) = 3 4 − 9x. (Solution)
18. Find
20. Find
√
ds
ds
and
, where s = t a2 + t2 . (Solution)
dt
da
dy
a−x
, where y =
. (Solution)
dx
a+x
r
dy
1 − cx
22. Find
, where y =
. (Solution)
dx
1 + cx
21. Find
2
Solutions
x2 + x − 6
, if it exists.
x→2
x−2
1. Evaluate lim
(Solution) Doing some factoring, we have
x2 + x − 6
(x − 2)(x + 3)
= lim
= lim (x + 3) = 5.
x→2
x→2
x→2
x−2
x−2
lim
3x4 + 2x3 + x2
, if it exists.
x→0 3x4 + 5x3 + 7x2
2. Evaluate lim
(Solution) We have
3x4 + 2x3 + x2
x2 (3x2 + 2x + 1)
3x2 + 2x + 1
=
lim
=
lim
x→0 3x4 + 5x3 + 7x2
x→0 x2 (3x2 + 5x + 7)
x→0 3x2 + 5x + 7
limx→0 (3x2 + 2x + 1)
1
=
= .
2
limx→0 (3x + 5x + 7)
7
lim
3. Find two functions f, g such that limx→a f (x) does not exist, limx→a g(x) does not
exist, but such that
lim (f (x) + g(x))
x→a
does exist.
(Solution) There are many different solutions to this problem, but one pair of functions
in particular is given by
f (x) :=
1
x−a
and
g(x) :=
−1
.
x−a
You can check that neither limx→a f (x) nor limx→a g(x) exists, but f (x) + g(x) = 0 for
x 6= a, so limx→a (f (x) + g(x)) = 0.
sin2 (θ)
, if it exists.
θ→0 1 − cos(θ)
4. Evaluate lim
(Solution) We can use the identity sin2 (θ) + cos2 (θ) = 1 to find
sin2 (θ)
1 − cos2 (θ)
(1 − cos(θ))(1 + cos(θ))
= lim
= lim
lim
θ→0 1 − cos(θ)
θ→0 1 − cos(θ)
θ→0
1 − cos(θ)
= lim(1 + cos(θ)) = 2.
θ→0
|x|
, if it exists.
x→0 x
5. Evaluate lim
3
(Solution) Notice that
|x|
=
x
and that
|x|
x
−1, x < 0
,
1, x > 0
is undefined when x = 0. So we see that
lim−
x→0
|x|
= lim− −1 = −1
x→0
x
and
lim+
x→0
Since the left and right limits are not equal, limx→0
|x|
x
|x|
= lim+ 1 = 1.
x→0
x
does not exist.
x−1
6. Evaluate lim √
, if it exists.
x→1
x−1
(Solution) Since both the numerator and denominator approach 0 as x approaches 1,
we must find a way to rewrite the fraction. We do this by rationalizing the denominator:
√
√
x−1
x+1
(x − 1)( x + 1)
x−1
√
=√
·√
=
.
x−1
x−1
x−1
x+1
Then
√
√
x−1
(x − 1)( x + 1)
= lim ( x + 1) = 2.
lim √
= lim
x→1
x→1
x−1
x − 1 x→1
7. Evaluate lim
t→0
1
1
−
, if this limit exists.
t t2 + t
(Solution) Notice that neither quotient has a limit as t → 0, so we find a way to
combine the terms:
2
1
1
t +t−t
t2
lim
−
= lim
= lim 2
t→0
t→0
t→0
t t2 + t
t(t2 + t)
t (t + 1)
1
= lim
= 1.
t→0
t+1
x
8. Evaluate lim √
, if this limit exists.
x→0
1 + 3x − 1
(Solution) Again, we need to rationalize the denominator.
√
√
x
x
1 + 3x + 1
x( 1 + 3x + 1)
= lim √
·√
= lim
lim √
x→0
1 + 3x − 1 x→0 1 + 3x − 1
1 + 3x + 1 x→0 (1 + 3x) − 1
√
1 + 3x + 1
2
= lim
= .
x→0
3
3
4
9. Find values k, m ∈ R so that
 3
x ≤ −1
 2x + x + 7
f (x) := m(x + 1) + k −1 < x ≤ 2

x2 + 5
2<x
is continuous, if possible.
(Solution) Notice that all three pieces of f are continuous functions, so the only points
where we have to worry about continuity are at x = −1 and x = 2. For each of these
points, we’ll need to ensure that the left- and right-limits agree. First we have
lim f (x) = lim − (2x3 + x + 7) = −2 − 1 + 7 = 4,
x→−1−
x→−1
while
lim f (x) = lim + (m(x + 1) + k) = k.
x→−1+
x→−1
So k = 4. Then at x = 2,
lim f (x) = lim− (m(x + 1) + k) = 3m + 4
x→2−
x→2
and
lim f (x) = lim+ (x2 + 5) = 9.
x→2+
x→2
So 3m + 4 = 9, meaning that m = 5/3.
10. Find two functions f, g such that neither f nor g is continuous at a, but such that
f + g is continuous.
(Solution) The two functions we used for the limit version of this question won’t work,
since their sum isn’t defined at x = a, but the following pair of functions will do the
trick:
0, x ≤ a
1, x ≤ a
f (x) :=
and
g(x) :=
1, x > a
0, x > a
11. Find lim θ2 cos(1/θ), if this limit exists.
θ→0
(Solution) We know that cos(1/θ) oscillates as θ → 0, but this won’t be a problem
here. Since −1 ≤ cos(1/θ) ≤ 1 for all θ 6= 0, we have
−θ2 ≤ θ2 cos(1/θ) ≤ θ2 .
But limθ→0 (−θ2 ) = 0 and limθ→0 θ2 = 0, so the squeeze theorem allows us to conclude
that
lim θ2 cos(1/θ) = 0.
θ→0
5
sin(2x) sin(3x)
, if this limit exists.
x→0 sin(5x) sin(7x)
12. Evaluate lim
(Solution) We want to take advantage of the fact that
sin(u)
= 1,
u→0
u
lim
so we rewrite our limit as
sin(2x) sin(3x)
6
lim
= lim
x→0 sin(5x) sin(7x)
x→0 35
6
= .
35
13. Evaluate lim
x→2
sin(2x)
2x
sin(3x)
3x
5x
sin(5x)
7x
sin(7x)
sin(2x − 4)
, if this limit exists.
x2 − 4
(Solution) As with the previous problem, we’ll rewrite this limit to look like something
we know. First, notice that
sin(2x − 4)
1
sin(2x − 4)
= lim
· 1
.
2
x→2
x→2
x −4
2x − 4
x+1
2
lim
Since the argument of sine and the denominator in our first fraction agree and approach
0 linearly, we may use the same rule we used above to find that
sin(2x − 4)
1
1
= 1 · lim 1
= .
2
x→2
x→2 x + 1
x −4
2
2
lim
cos(x2 − 7)
, if this limit exists.
x→∞
x
14. Evaluate lim
(Solution) Since −1 ≤ cos(x2 − 7) ≤ 1, we see that the numerator of this quotient
remains bounded while x → ∞, but the denominator grows without bound. So the
quotient approaches 0 as x → ∞. More rigorously,
−1 cos(x2 − 7) 1
≤
≤ x.
x
x
Since both bounding functions approach 0 as x → ∞, our limit is 0.
15. Use the limit definition of the derivative to find f 0 (a), where f (x) =
6
√
x and a 6= 0.
(Solution) We have
√
0
f (a) = lim
h→0
a+h−
h
√
a
.
Here we’ll rationalize the numerator to find
√
√ √
√
(a + h) − a
a+h− a
a+h+ a
0
f (a) = lim
·√
√ = lim √
√
h→0
h
a + h + a h→0 h( a + h + a)
h
1
= lim √
√ = lim √
√
h→0 h( a + h +
h→0
a)
a+h+ a
1
= √ .
2 a
16. Use the limit definition of the derivative to find g 0 (t), where g(t) = 10 + 5t − 4.9t2 .
(Solution) We have
g(t + h) − g(t)
h→0
h
(10 + 5(t + h) − 4.9(t + h)2 ) − (10 + 5t − 4.9t2 )
= lim
h→0
h
2
10 + 5t + 5h − 4.9t − 9.8ht − 4.9h2 − 10 − 5t + 4.9t2
= lim
h→0
h
5h − 9.8ht − 4.9h2
= lim
= lim (5 − 9.8t − 4.9h)
h→0
h→0
h
= 5 − 9.8t.
g 0 (t) = lim
17. Find
x
dy
, where y = 2
.
dx
x +1
(Solution) According to the quotient rule,
dy
(x2 + 1)(1) − x(2x)
=
dx
(x2 + 1)2
1 − x2
= 2
.
(x + 1)2
18. Find
√
7x
dy
3x3
7
√
, where y = √
−
+
8
x3 .
5
3
2
4
dx
x
x
(Solution) First, we rewrite y as
y = 3x3 x−2/5 − 7x1 x−4/3 + 8x3/7 = 3x13/5 − 7x−1/3 + 8x3/7 .
We may now apply the power rule to find that
dy
39
7
24
= x8/5 + x−4/3 + x−4/7 .
dx
5
3
7
7
19. Find g 0 (x), where g(x) =
√
3
4 − 9x.
(Solution) Notice that g(x) = f (h(x)), where f (x) =
the chain rule gives
√
3
x and h(x) = 4 − 9x. Then
2
1
−3
.
g 0 (x) = f 0 (h(x))h(x) = (4 − 9x)− 3 (−9x) = p
3
3
(4 − 9x)2
20. Find
√
ds
ds
and
, where s = t a2 + t2 .
dt
da
(Solution) First, we’ll use the product and chain rules to see that
ds √ 2
2t
a2 + t2 + t2
a2 + 2t2
= a + t2 + t √
= √
=√
.
dt
2 a2 + t 2
a2 + t2
a2 + t2
On the other hand,
t(2a)
ds
at
= √
=√
.
da
2 a2 + t 2
a2 + t2
21. Find
a−x
dy
, where y =
.
dx
a+x
(Solution) By the quotient rule,
dy
(a + x)(−1) − (a − x)(1)
−2a
=
=
.
2
dx
(a + x)
(a + x)2
dy
, where y =
22. Find
dx
r
1 − cx
.
1 + cx
(Solution) Using the chain rule we find that
−1/2
dy
1 1 − cx
(1 + cx)(−c) − (1 − cx)(c)
=
dx
2 1 + cx
(1 + cx)2
−c (1 − cx)−1/2 (1 + cx) + (1 − cx)1/2
=
.
2
(1 + cx)3/2
8