Calculus I: MATH 1351-030
Fall 2011
Exam 2 (B)
Name:
Instructions. (Part 1) Solve each of the following problems. Choose the best solution to
each problem and clearly mark your choice.
1. Given the following graph of f (x) = bx , what can you say about b?
(a) b = 1
(c) Can’t determine anything about b.
(e) b < 0
(b) 0 < b < 1
(d) b > 1
2. Solve the following exponential equation for x:
4x
(a) x = 0, −1
(c) x = 1
(e) x = 1, −2
3. Evaluate the following limit:
2 +x
= 16
(b) x = 2
(d) x = 4
cos x − 1
x→π
2x
lim
(a) Limit does not exist.
(c) 1/2
(e) −1/π
(b) 0
1
(d) 2π
Calculus I: MATH 1351-030/Exam 2 (B) – Page 2 of 7 – Name:
4. Evaluate the expression log3 35 + ln e0.5 .
√
(a) 35 + e
(c) 0
(e) 7/2
(b) 11/2
(d) 1/2
5. Which conditions, when present, are sufficient to conclude that a function f (x) has a
limit as x approaches some value c?
(a) Either the limit of f (x) as x → c from the left exists or the limit of f (x) as x → c
from the right exists.
(b) The limit of f (x) as x → c from the left exists, the limit of f (x) as x → c from the
right exists, and these two limits are the same.
(c) The limit of f (x) as x → c from the left exists, the limit of f (x) as x → c from the
right exists, and at least one of these limits is the same as f (c).
(d) The limit of f (x) as x → c from the left exists and the limit of f (x) as x → c from
the right exists.
(e) f (c) exists, the limit of f (x) as x → c from the left exists, and the limit of f (x) as
x → c from the right exists.
6. Which of the following statements best summarizes the Intermediate Value Theorem?
f (a)+f (b)
=
.
(a) If a function f is continuous on an open interval (a, b), then f a+b
2
2
(b) If a function f is continuous on a closed interval [a, b], then f (a) < f (x) < f (b) on
the open interval (a, b).
(c) If a function f is defined on an open interval (a, b), then f (x) must take on all values
between f (a) and f (b).
(d) If a function f is continuous on a closed interval [a, b], then f (x) must take on all
values between f (a) and f (b).
7. Solve the following logarithmic equation for x:
logx 16 = 2
(a) x = ±e
(c) x = 4
(e) x = e
(b) x = 10
(d) x = ±4
8. Which of the following functions is not continuous everywhere?
(a) sin x
(b) All of these are continuous everywhere.
(c) cos x
(d) None of these is continuous everywhere.
(e) tan x
Calculus I: MATH 1351-030/Exam 2 (B) – Page 3 of 7 – Name:
9. Identify all suspicious points and determine all points of discontinuity for
f (x) =
3x
−x
x2
(a) Suspicious points are x = 0, 1; both are points of discontinuity.
(b) Suspicious points are x = 0, 1; f (x) is continuous at x = 1 and discontinuous at
x = 0.
(c) There are no suspicious points; f (x) is continuous on (−∞, ∞).
(d) Only suspicious point is x = 1; f (x) is discontinuous at x = 1.
(e) Suspicious points are x = 0, 1; f (x) is discontinuous at x = 1 and continuous at
x = 0.
10. A function f (x) is continuous at a point x = c if
(a) f (c) is defined.
(b) limx→c− f (x) = limx→c+ f (x).
(c) f (c) is defined and limx→c f (x) exists.
(d) f (c) is defined, limx→c f (x) exists, and limx→c f (x) = f (c).
11. Use left and right one-sided limits to find the following limit if it exists:
t2 −t
, t>1
t−1
lim f (t),
where f (t) = √
t→1
1 − t, t ≤ 1
(a) Right-hand limit = 1, left-hand limit = 0, two-sided limit = 1.
(b) Right-hand limit = 1, left-hand limit = 1, two-sided limit = 1.
(c) Right-hand limit = 0, left-hand limit = 0, two-sided limit = 0.
(d) Right-hand limit does not exist, left-hand limit = 0, two-sided limit does not exist.
(e) Right-hand limit = 1, left-hand limit = 0, two-sided limit does not exist.
12. For b > 0 and b 6= 1, the Change of Basis Theorem says that
ln b
(a) logb x = ln
(b) logb x = ln x
x
x
(c) logb x = ln
(d) logb x = logx b
ln b
(e) logb x = (ln x)(ln b)
Calculus I: MATH 1351-030/Exam 2 (B) – Page 4 of 7 – Name:
13. Determine whether the following function is continuous on the closed interval [−5, 5]. If
it is not continuous, redefine the function at x = −2 to make it continuous (if possible).
x2 −x−6
if x 6= −2
x+2
f (x) =
−4
if x = −2
(a) f (x) is not continous on [−5, 5] as given. To make it continuous on [−5, 5], redefine
f (−2) = 0.
(b) f (x) is not continous on [−5, 5] as given. To make it continuous on [−5, 5], redefine
the function so that x = −2 is not in its domain.
(c) f (x) is continuous on [−5, 5] as given.
(d) f (x) is not continous on [−5, 5] as given. To make it continuous on [−5, 5], redefine
f (−2) = −5.
(e) f (x) is not continous on [−5, 5] as given and there is no way to redefine the function
at x = −2 to make it continous on [−5, 5].
14. Evaluate the following limit:
1 − cos x
x→0 x cos x
lim
(a) 0
(c) 1/2
(e) 1
(b) Limit does not exist.
(d) -1
15. Evaluate the following left-hand limit:
lim−
x→0
(a) 1
(c) −x
(e) 0
|x|
x
(b) -1
(d) Left limit does not exist.
16. If b > 0 and b 6= 1, the logarithm of x to the base b is the function y = logb x that
satisfies:
(a) bx = y
(b) xb = y
(c) by = x
(d) ex = y
(e) y b = x
Calculus I: MATH 1351-030/Exam 2 (B) – Page 5 of 7 – Name:
17. Find the limit, if it exists:
x2 + 4x − 12
x→−6 x2 + 2x − 24
lim
(a) -4/5
(c) Limit does not exist.
(e) 4/5
(b) 0
(d) 3/10
18. Evaluate the following limit:
sin(2x)
x→0 sin(5x)
lim
(a) 5/2
(c) 2/5
(e) 0
(b) 1
(d) Limit does not exist.
19. Let limx→9 f (x) = 7 and limx→9 g(x) = −10. Find limx→9
(a) 9
(c) -10/7
(e) -7/10
(b) Limit does not exist.
(d) 7/10
20. Evaluate the following limit:
√
lim
x→0
(a) 1
(c) 2
(e) Limit does not exist.
f (x)
.
g(x)
x2 + 4 − 2
x
(b) -2
(d) 0
Calculus I: MATH 1351-030/Exam 2 (B) – Page 6 of 7 – Name:
Instructions. (Part 2) Solve the following problem. Show your work clearly. You must
write out all relevant steps. Simply having the correct answer does not give you
credit.
1. Evaluate the following limits if they exist. If a limit does not exist, explain why.
|x − 1|
x→1 x − 1
lim
√
ln x
lim
x→e
x
2. Show that the equation
x + sin x = √
1
x+3
has at least one solution on the interval (0, π). Clearly state which theorem you are
using and be sure to show that the requirements of the theorem are met in this problem.
Calculus I: MATH 1351-030/Exam 2 (B) – Page 7 of 7 – Name:
3. Solve for x. Clearly show all of your steps.
1
3x 4(x+ 2 ) = 288
4. (bonus) Prove using the epsilon-delta definition of the limit that limx→2 5x − 3 = 7.