Solutions to Math 41 First Exam — October 18, 2012
1. (12 points) Find each of the following limits, with justification. If the limit does not exist, explain
why. If there is an infinite limit, then explain whether it is ∞ or −∞.
2
1
(a) lim
−
x→1
x2 − 1 x − 1
(4 points)
lim
x→1
(b)
lim
x→−2+
2
1
−
2
x −1 x−1
2
1
−
x→1 (x + 1)(x − 1)
x−1
2 − (x − 1)
= lim
x→1 (x + 1)(x − 1)
−x + 1
= lim
x→1 (x + 1)(x − 1)
−(x − 1)
= lim
x→1 (x + 1)(x − 1)
−1
= lim
x→1 x + 1
1
= −
2
= lim
x ln(2 + x)
x2 + 2x + 2
(4 points) We have to compute
lim
x→−2+
x ln(2 + x)
x2 + 2x + 2
We know that limx→−2+ ln(2 + x) = −∞ since limx→0+ ln(x) = −∞. So as x goes to −2 from
the right, ln(x + 2) becomes a very large negative number.
We also know that limx→−2+ x = −2 by direct substitution. Thus x ln(x + 2) becomes a very
large positive number as x approaches −2 from the right. Since limx→−2+ x2 + 2x + 2 = 2, and
dividing a large positive number by a positive number close to 2 still gives us a large positive
number, we have that xx2ln(2+x)
becomes a very large positive number as well. Therefore,
+2x+2
lim
x→−2+
x ln(2 + x)
=∞
x2 + 2x + 2
Math 41, Autumn 2012
(c) lim (sin2 x) 2
Solutions to First Exam — October 18, 2012
Page 2 of 13
cos(1/x)
x→0
(4 points) Since the limit of cos(1/x) as x goes to 0 doesn’t exist, but cos(x) is always between
-1 and 1, we will bound the function above and below and then use the Squeeze Theorem.
−1
2−1
≤
cos(1/x)
≤
2cos(1/x)
1
≤
2cos(1/x)
2
1
2
2
cos(1/x)
2 sin (x) ≤ sin (x)2
≤
1
≤
21
≤
2
≤ 2 sin2 (x)
Note that we can write the last inequality because sin2 (x) is always greater than or equal to 0.
From the last inequality, we see that
lim
x→0
1
sin2 (x) ≤ lim sin2 (x)2cos(1/x) ≤ lim 2 sin2 (x)
x→0
x→0
2
We have that limx→0 21 sin2 (x) = limx→0 2 sin2 (x) = 0 by direct substitution. Therefore, by the
Squeeze Theorem,
lim sin2 (x)2cos(1/x) = 0
x→0
A common mistake was to try to find an inequality with sin2 (x) in the middle. This doesn’t
work because the limit of cos(1/x) as x approaches 0 does not exist.
Math 41, Autumn 2012
Solutions to First Exam — October 18, 2012
Page 3 of 13
2. (10 points) Mark each statement below as true or false by circling either TRUE or FALSE. No justification is necessary.
(a) For any positive and any x, we have
TRUE
FALSE
√
|(x2 − 2x + 5) − 4| < whenever 0 < |x − 1| < .
√
(2 points) If |x − 1| < , then |x − 1|2 < . By algebra, the latter statement is equivalent to
the statement |(x2 − 2x + 5) − 4| < .
(Remark: Think about how this statement could be
2
used to prove that lim (x − 2x + 5) = 4 using the delta-epsilon definition of limits.)
x→1
(b) For any x, we have
TRUE
2
|x − 1| < 1
whenever
0 < x −
FALSE
1 1
< .
2 4
(2 points) If 0 < |x − 21 | < 14 , then x lies within 1/4 unit of 1/2 (though also x 6= 1/2); that is, we
know x lies in the interval (1/4, 3/4). For such x, x2 lies in (1/16, 9/16), so x2 − 1 lies between
−15/16 and −7/16. Since this interval lies within (−1, 1), we have |x2 − 1| < 1.
(c) For any positive , there is a corresponding positive δ such that
TRUE
FALSE
1 2
|x − 1| < whenever 0 < x − < δ.
2
(2 points) Note that by our delta-epsilon definition of limits, the above sentence is equivalent to
the statement
lim x2 = 1,
x→ 21
which is false! Thus, the original statement must be false. (To see this more directly, pick = 0.5.
Assume such a δ exists. Pick x = 12 + γ where γ is less than δ and less than 0.1 but strictly
larger than 0. We have that 0 < |x − 12 | < δ, but x2 − 1 will be less than 0.36 − 1 = −0.64. This
ensures that |x2 − 1| > 0.64, which is a contradiction; thus there is no possible δ for this value
of , and the original statement is indeed false.)
Remark: Observe that part (b) can be true even though part (c) is false. (Part (b) is a special
case of part (c), for one particular , that just happens to be true.)
(d) For any function f and any a, if
lim f (x)
x→a+
TRUE
and
FALSE
lim f (x)
x→a−
both exist and are equal, then f is continuous at x = a.
(2 points) We also need f (a) to exist and be equal to lim f (x).
x→a
TRUE
(e) For any odd function f , if
FALSE
f (0) = lim f (x),
x→0+
then f is continuous at x = 0.
(2 points) For such an f , since f is odd and defined at 0, we must have f (0) = 0. Thus, we have
lim f (x) = 0.
x→0+
Also, again using the oddness of f , we have lim f (x) = lim f (−x) = lim −f (x) = −0 = 0.
x→0−
x→0+
x→0+
Thus, we conclude lim f (x) = 0 = f (0), and so f is continuous at x = 0 by definition.
x→0
Math 41, Autumn 2012
Solutions to First Exam — October 18, 2012
Page 4 of 13
ex + e−x
.
ex − e−x
(a) Find the equations of all vertical asymptotes of f , or explain why none exist. As justification for
each asymptote x = a, calculate both the one-sided limits lim f (x) and lim f (x), showing your
3. (12 points) Let f (x) =
x→a+
x→a−
reasoning.
x
−x
+e
(6 points) To find the vertical asymptotes of f (x) = eex −e
−x we must first find where it is
discontinuous — and in the case of the above quotient of continuous functions, it is therefore
sufficient to determine where f is undefined; this will occur only when the denominator is 0, and:
ex − e−x = 0 ⇐⇒ ex = e−x ⇐⇒ x = −x ⇐⇒ x = 0
Thus the denominator is 0 only if x = 0. To find out whether f (x) really has an asymptote at
x
−x
x +e−x
0, we must take limx→0+ eex +e
and limx→0− eex −e
−x . As x approaches 0 from either side, the
−e−x
x
−x
numerator, e + e approaches 2 and the denominator approaches 0. Thus the limits are either
∞ or −∞. To determine the signs, we must find out when ex − e−x is positive and negative.
ex − e−x > 0
ex > e−x
x > −x
x>0
Thus ex − e−x is positive when x > 0. Likewise, it is negative when x < 0. Thus,
x = 0 is an asymptote, and lim
x→0+
ex + e−x
ex + e−x
= ∞ and lim x
= −∞
x
−x
e −e
x→0− e − e−x
(b) Find the equations of all horizontal asymptotes of f , or explain why none exist. Justify using
limit computations.
(6 points) The function f (x) has a horizontal asymptote when at least one of its limits as x
approaches positive and negative infinity is finite. So we compute:
ex + e−x
= lim
x→∞ ex − e−x
x→∞
lim
= lim
x→∞
1
x
−x
ex (e + e )
1
x
−x
ex (e − e )
1 + e−2x
1 − e−2x
=1
since limx→∞ e−x = 0. Note that we had to divide the top and bottom by ex because limx→∞ ex =
∞. Next we compute:
ex + e−x
= lim
x→−∞ ex − e−x
x→−∞
lim
= lim
x→−∞
1
(ex + e−x )
e−x
1
(ex − e−x )
e−x
e2x + 1
e2x − 1
= −1
since limx→−∞ ex = 0. Note that this time we had to divide top and bottom by e−x since
limx→−∞ e−x = −∞. Thus, the horizontal asymptotes are y = 1 and y = −1 .
Math 41, Autumn 2012
Solutions to First Exam — October 18, 2012
Page 5 of 13
4. (8 points) Prove, using precise statements, that there is a real number x between −1 and 1 which is a
solution to the equation
πx sin
= ln(x2 ).
2
The main difficulty of this problem is that ln(x2 ) has a vertical asymptote at x = 0, so it is not
continuous on [−1, 1]. Arguments that did not take this into account received a maximum of 4 out
of 8 points.
2
Let f (x) = sin( πx
2 )−ln(x ). Since it is a combination of polynomials, trig functions, and log functions,
it is continuous on its domain, which is (−∞, 0) ∪ (0, ∞). We are looking for a value of x between
−1 and 1 for which f (x) = 0. With the picture in mind, we focus on the interval [−1, 0). We have
f (−1) = sin(−π/2) − ln(1) = −1 − 0 = −1, and
h πx i
lim f (x) = lim sin
− ln(x2 ) = 0 − (−∞) = ∞.
2
x→0−
x→0−
Since −1 < 0 < ∞, the Intermediate Value Theorem says that there must be some x in the interval
2
(−1, 0) for which f (x) = 0, and therefore sin( πx
2 ) = ln(x ).
Technically, we only learned the Intermediate Value Theorem for closed intervals. To be more rigorous,
we’d have to pick a small negative number b to be the right endpoint of our interval. One possible
choice is b = −1/e. Then
−π
−π
−2
− ln(e ) = sin
− (−2) ≥ −1 + 2 = 1.
f (b) = sin
2e
2e
Since f (−1) = −1 < 0 and f (b) ≥ 1 > 0, the Intermediate Value Theorem on the closed interval
[−1, b] says that there must be some x in (−1, b) for which f (x) = 0.
Math 41, Autumn 2012
Solutions to First Exam — October 18, 2012
Page 6 of 13
1
√
. Find a formula for f 0 (x) using the limit definition of the derivative.
2+ x+1
Show the steps of your computation.
5. (8 points) Let f (x) =
We compute
f (x + h) − f (x)
h→0
h
!
p
√
√ 1
− 2+√1x+1
(2 + (x + h) + 1)(2 + x + 1)
2+ (x+h)+1
p
= lim
×
√
h→0
h
(2 + (x + h) + 1)(2 + x + 1)
√
√
(2 + x + 1) − (2 + x + h + 1)
√
√
= lim
h→0 h(2 + x + h + 1)(2 + x + 1)
√
√
√
√
x+1− x+h+1
x+1+ x+h+1
√
√
√
= lim
× √
h→0 h(2 + x + h + 1)(2 + x + 1)
x+1+ x+h+1
(x + 1) − (x + h + 1)
√
√
√
√
= lim
h→0 h(2 + x + h + 1)(2 + x + 1)( x + 1 + x + h + 1)
−h
√
√
√
√
= lim
h→0 h(2 + x + h + 1)(2 + x + 1)( x + 1 + x + h + 1)
−1
√
√
√
√
= lim
.
h→0 (2 + x + h + 1)(2 + x + 1)( x + 1 + x + h + 1)
f 0 (x) = lim
We can plug h = 0 into this last expression, getting
f 0 (x) =
(2 +
√
√
−1
√
√
x + 1)( x + 1 + x + 1)
x + 1)(2 +
−1
√
√
=
.
2(2 + x + 1)2 x + 1
Math 41, Autumn 2012
Solutions to First Exam — October 18, 2012
Page 7 of 13
6. (12 points) The reproduction pattern of a certain species of fruit fly, grown in bottles in a laboratory,
depends on the number p of female flies in the bottle. A researcher determines values of S(p), the
number of daily offspring per female; the chart below shows a few values of S(p):
p
S(p)
10
5.3
12
4.9
14
4.5
16
4.2
18
4.0
(a) Give your best estimate for the value of S 0 (14), showing your reasoning, and make sure to specify
the units of this quantity.
(14)
(4 points) By the definition of limit, S 0 (14) = limq→14 f (q)−f
. However, we are only given the
q−14
values of S for discrete values of p, so for the best estimate we should average the values of the
difference quotient for the two values of q nearest to 14: q = 12 and q = 16.
1 f (12) − f (14) f (14) − f (16)
1 4.9 − 4.5 4.5 − 4.2
−0.7
0
S (14) =
=
=
+
+
2
12 − 14
14 − 16
2
−2
−2
4
The units of S 0 are offspring per female fly per total number of female flies.
(b) What is the practical meaning of the quantity S 0 (14)? Give a brief but complete one- to twosentence explanation that is understandable to someone who is not familiar with calculus.
(3 points) S 0 (14) describes the instantaneous rate of change of the number of daily offspring per
female fly with respect to the number of female flies, when the number of female flies is 14. In
practice this means that if we have a bottle with 14 female flies and we were to add a small
quantity a of female flies, then the change in the production of daily offspring per female fly
would be −0.7
4 a.
A lot of students answered this question by saying that if we were to add 1 female fly to a
bottle where we initially have 14 female flies, then the production of daily offspring per female
fly would decrease by 0.7
4 offspring per female fly. This not entirely correct, because 1 female
fly is not considered a small quantity when compared to the 14 female flies in the bottle. (One
reasonable example of “adding a small quantity”: having 100 bottles with 14 female flies in each
bottle, and then adding one female fly to exactly one of these bottles, which could be regarded
as adding 0.01 female flies. In such a case, one could argue that we’d expect to observe about
S(14) + 0.01 · S 0 (14) daily offspring per female.)
Math 41, Autumn 2012
Solutions to First Exam — October 18, 2012
Page 8 of 13
(Problem 6 continued) For easy reference, here again is the setup of the problem: The reproduction
pattern of a certain species of fruit fly, grown in bottles in a laboratory, depends on the number p of
female flies in the bottle. A researcher determines values of S(p), the number of daily offspring per
female; the chart below shows a few values of S(p):
p
10 12 14 16 18
S(p) 5.3 4.9 4.5 4.2 4.0
(c) For this and part (d), let g(p) = p · S(p). Compute a formula for g 0 (p). (Express your answer in
terms of quantities such as p, S(p), and S 0 (p).)
(2 points) g(p) = p · S(p). We are asked to compute the derivative of g with respect to p! Using
the product rule we get:
g 0 (p) = p0 · S(p) + p · S 0 (p) = S(p) + p · S 0 (p)
NOTE: p is not a constant! g is a function of p only, so when we are asked for the derivative of
g, we compute the derivative with respect to p.
(d) State a practical implication of the statement “g 0 (14) is positive” using only terminology of the
laboratory setting. (You don’t have to estimate the value of g 0 (14); just assume it is positive for
the purposes of answering this part.)
(3 points) The function g represents the number of offspring per female fly times the number of
female flies which is the total number of offpring(per bottle per day). g 0 (14) > 0 means that the
function g is increasing near the point p = 14. The practical meaning of that is the following: if
we start with a bottle with 14 femaly flies and we add a small quantity of female flies(see point
b) for an explanation), the total number of offspring produced in this bottle will increase(even
if the number of offspring per female decreases according to part b).
Math 41, Autumn 2012
Solutions to First Exam — October 18, 2012
Page 9 of 13
7. (8 points) Find the derivative, using any method you like. You do not need to simplify your answers.
(a) h(x) =
x3 + πx1/4 − (e/x)
√
x
(4 points) First, simplify h(x) as
1
x3 + πx 4 − (e/x)
√
h(x) =
x
3
1
= x3 + πx 4 − ex−1 x− 2
1
1
1
1
=x3− 2 + πx 4 − 2 − ex1− 2
5
1
3
=x 2 + πx− 4 − ex− 2 .
Use the power rule (xn )0 = nxn−1 to take derivative:
5
5 3 π 5 3
h0 (x) = x 2 − x− 4 + ex− 2 .
2
4
2
sin x − 2 cos x
ex
(4 points) Using the quotient rule, we have
(b) f (x) =
(sin x − 2 cos x)0 ex − (sin x − 2 cos x) (ex )0
(ex )2
((sin x)0 − 2(cos x)0 ) ex − (sin x − 2 cos x) (ex )0
=
e2x
(cos x + 2 sin x) ex − (sin x − 2 cos x) (ex )
=
.
e2x
f 0 (x) =
There is no need to simplify the solution further.
Math 41, Autumn 2012
Solutions to First Exam — October 18, 2012
Page 10 of 13
8. (7 points) The figure below shows the graph of a function f that has continuous first and second
derivatives. The dashed lines are tangent to the graph of y = f (x) at (1, 1) and (5, 1).
List the following quantities in increasing order (from smallest to largest). No justification is necessary.
f (1)
f (3) − f (2)
1
3
(f (5) − f (2))
The number −2
f 0 ( 12 )
f 0 (1)
f 0 (5)
We first analyze the sign of the seven quantities. Since the graph has a horizontal tangent at x = 5,
we know f 0 (5) = 0. 31 (f (5) − f (2)) is the slope of secant line between x = 2 and x = 5, which is
positive. Similarly, f (3) − f (2) is the slope of secant line between x = 2 and x = 3, also positive. We
also have f (1) = 1, positive. As for f 0 (1), f 0 ( 12 ), since the slope of tangent lines at x = 1 and x = 12
are negative, f 0 (1), f 0 ( 12 ) are both negative. So far, we know
f 0 ( 12 )
f 0 (1)
−2
< f 0 (5) = 0 <
f (1)
f (3) − f (2)
1
3 (f (5) − f (2))
We next analyse the three negative terms. From the graph, the slope of tangent line at x = 1 is
−2
1
2 = −1. Although we do not know the exact slope of tangent line at x = 2 , between x = 0 and
x = 1 the curve becomes steeper as we move forward alone the x-axis. This tells us f 0 (1) is ‘more
negative’ than f 0 ( 12 ): f 0 (1) < f 0 ( 12 ). We therefore have
−2 < f 0 (1) < f ( 12 ) < f 0 (5) <
f (1)
f (3) − f (2)
1
3 (f (5) − f (2))
We now move on to the three positive terms. From the graph, f (3) − f (2) is greater than 4, and
1
3 (f (5) − f (2)) is less than 1/3. f (1) = 1. Therefore, we conclude
1
−2 < f 0 (1) < f 0 ( 12 ) < f 0 (5) < (f (5) − f (2)) < f (1) < f (3) − f (2)
3
Math 41, Autumn 2012
Solutions to First Exam — October 18, 2012
Page 11 of 13
9. (12 points) The following is a graph of the function g:
(a) Consider the function g 0 , the derivative of g. Based on estimating g 0 from the graph of g above,
on what intervals is g 0 increasing? decreasing? (No justification is necessary.)
(3 points) g 0 is increasing if the graph of g is concave up, and g 0 is decreasing if the graph of
g is concave down. Referring to the graph, g is concave up on (−∞, 0) and (2, ∞). Between
x = 1 and x = 2, g is concave down from x = 0 up until an inflection point x = a, and then g
becomes concave up all the way toward x = 2. From the graph, one might estimate that a ≈ 43 ;
at minimum one should find a value satisfying 12 ≤ a ≤ 1. Thus, the answer is
g 0 is increasing on (−∞, 0), (a, 2), (2, ∞),
g 0 is decreasing on (0, a),
where we gave full credit to any estimated a falling in the range
1
2
≤ a ≤ 1.
(b) Based on the picture of g above, which of the following expressions is a plausible formula for g(x)?
Circle your answer; no justification is necessary. (You may take it as a given that exactly one of
these formulas is the best answer.)
x−1
2
x (x − 2)
(x − 1)3
x2 (x − 2)3
x−1
1/3
x (x − 2)2
(3 points) The four choices are of the form
(x−1)m
,
xb (x−2)n
(x − 1)2
x1/3 (x − 2)2
where m, n are integers and b is either 2 or
1
3.
We determine m, n, b by analysing the sign of g. When x is close to two, g is always positive.
This says that moving from x > 2 to x < 2 does not change the sign of g(x). Therefore, n must
be an even integer. Similarly, when moving from x > 1 to x < 1, the function g(x) changes its
sign from positive to negative, so m must been an odd integer. At this point there is only one
candidate left:
x−1
1
x 3 (x − 2)2
Let’s check near x = 0 to see if this candidate really fits the given graph. Moving from x > 0
and x < 0 changes the sign of g(x). This is consistent with the choice b = 13 .
Math 41, Autumn 2012
Solutions to First Exam — October 18, 2012
Page 12 of 13
(c) On the set of axes below, sketch a plausible graph of a function f satisfying all of the following:
f is continuous on (−∞, 2) ∪ (2, ∞),
f has vertical asymptote x = 2,
and
f 0 (x) = g(x) for all x in the domain of g,
f (0) = 0.
(6 points) Since g(x) = f 0 (x) we have: (i) f is increasing when g > 0, decreasing when g < 0. (ii)
f is concave up when g = f 0 is increasing, concave down when g = f 0 is decreasing. Summarizing,
we have the following table.
x
in/decrease
concavity
0
%
up
1
&
up
2
%
up
%
down
We now focus on the three points x = 0, 1, 2. f has to be continuous at x = 0 while |f 0 (x)|
goes to infinity as x approaches zero. Thus we know the graph of f has a cusp at x = 0. Put
it another way, although the function is continuous at x = 0, the curve becomes steeper and
steeper as we move toward x = 0 from both sides, and the tangent line finally becomes vertical
at the point x = 0. Also, f (0) = 0. At the point x = 1, since f 0 (1) = g(1) = 0, the curve has a
horizontal tangent line. The line x = 2 is set to be a vertical asymptote. Hence we must have
|f (x)| approaching infinity as x approaches 2.
Remark: The x > 2 part of the graph does not have to be the same as the one drawn here. It
can be translated vertically by any amount. The specification f (0) = 0 will not be affected by
translating the red curve since the domain has already been broken into two parts: x < 2 and
x > 2.