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CALCULUS 1 FINAL EXAM PROBLEM COLLECTION.pdf

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November 1, 2023
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CALCULUS I: FIU FINAL EXAM PROBLEM COLLECTION: VERSION WITH
ANSWERS

FIU MATHEMATICS FACULTY NOVEMBER 2017




Contents
1. Limits and Continuity 1
2. Derivatives 4
3. Local Linear Approximation and differentials 6
4. Related Rates 7
5. Mean-Value Theorem 8
6. Monotonicity, Convexity, Local and Global Extrema, Graphs of functions 9
7. Optimization 12
8. True/False questions 14
9. Antiderivatives 18



1. Limits and Continuity
Problem 1.1. Use the graph of function f to find the limits




(1) lim f (x) = −5 (7) lim f (x) = −4
x→−3− x→3−
(2) lim f (x) = −5 (8) lim f (x) = +∞
x→−3+ x→3+
(3) lim f (x) = −5 (9) lim f (x) = d.n.e.
x→−3 x→3
(4) lim f (x) = 4 (10) lim f (x) = −6
x→0− x→−4−
(5) lim f (x) = 2 (11) lim f (x) = −6
x→0+ x→−4+
(6) lim f (x) = d.n.e. (12) lim f (x) = −6
x→0 x→−4




Date: 2018/04/14.
Compiled on 2018/04/14 at 02:06:22.
Filaname: output.tex.
1

,2 FIU MATHEMATICS FACULTY NOVEMBER 2017

Problem 1.2. Use the graph of function f to find the limits




(1) lim f (x) = +∞ (7) lim f (x) = −5
x→−3− x→3−
(2) lim f (x) = −4 (8) lim f (x) = −5
x→−3+ x→3+
(3) lim f (x) = d.n.e. (9) lim f (x) = −5
x→−3 x→3
(4) lim f (x) = 2 (10) lim f (x) = −2
x→0− x→6−
(5) lim f (x) = 4 (11) lim f (x) = −2
x→0+ x→6+
(6) lim f (x) = d.n.e. (12) lim f (x) = −2
x→0 x→6


Problem 1.3. Find the limits. If the limit does not exist, then write dne
2x2 −3x+2 e2x
(1) lim 2 = 41 (14) lim 2 = +∞
x→2 x +4x+4 x→∞ x
2 −3x−2 ln(2x)
(2) lim 2x 2 +4x+4 =0 (15) lim 3x = 0
x→2 x x→∞ e
2
(3) lim 2x2 −3x+2 = +∞ (16) lim tan(4x) = 43
x→2 x −4x+4 x→0 sin(3x)
1−x3
(4) lim 1+x 2 = +∞ (17) lim sin(x) 2 = 16 1
x→−∞ x→0 x(x−4)
4
(5) lim 2x −81 = −108 (18) lim arcsin(3x) = 3
x→3 x −7x+12 x→0 sin(x)
2
x2
(6) lim 9−x
3−x =6 (19) lim 1−cos 2x = 1
x→3 x→0
−x
(7) lim 2 x−3 = +∞ (20) lim xe =0
x→6+ x −x−30 x→∞
(8) lim x
= d.n.e. (21) lim xe−x = −∞
x→2 |2−x| x→−∞
√ √
lim x sin πx = π

2x2 −3 (22)
(9) lim x−2 =− 2 x→∞
x→−∞  √  (23) lim (tan(x) − sec(x)) = 0
(10) lim x2 + 2x − x = 1 x→( π2 )

x→∞ 
√ 
(24) limπ (1 − tan x) sec(2x) = 1
(11) lim x2 + 5 − x = 0 x→ 4
x→∞
(12) lim cos−1 (ln x) = (25) lim x1 − csc(x) = 0
π

x→1 2 x→0+
(13) lim 1−cos(4x) =0 (26) lim xx = 1
x→0 x x→0+
(27) lim (1 + 4x)1/x = e4
x→0+

, CALCULUS I: FIU FINAL EXAM PROBLEM COLLECTION: VERSION WITH ANSWERS 3

√ if x ≤ −5
 0,
Problem 1.4. Let f (x) = 25 − x2 , if − 5 < x < 5 . Find the indicated limits
3x , if x ≥ 5


(1) lim f (x) = 0 (5) lim f (x) = 15
x→−5− x→5+
(2) lim f (x) = 0 (6) lim f (x) = d.n.e.
x→−5+ x→5
(3) lim f (x) = 0 (7) Is f continuous at 5? No
x→−5 (8) Is f continuous at -5? Yes
(4) lim f (x) = 0
x→5−
 4x
Problem 1.5. Let f (x) = 5x+1 , if x < 1 000 000
. Find the indicated limits
5x
4x+1 , if x ≥ 1 000 000
4
(1) lim f (x) = 5 (3) Does the function have any asymptotes
x→−∞
5 (horizontal or vertical)? If yes, provide the
(2) lim f (x) = 4
x→+∞ equation(s).
Ans.: yes, horizontal, y = 45 , y = 54 , ver-
tical x = − 15

Problem 1.6. Locate the discontinuities of the given functions. If there are none, then write “none”

(1) f (x) = x2 − 4 none (5) f (x) = 4−3x
none in domain (−∞, 43 ]
x−2
(2) f (x) = x2x−4 x = −2, x = 2 (6) f (x) = √x−2 none in domain (−∞, 43 )
(3) f (x) = x2x+4 none 4−3x
2 + x3 , if x ≤ 1
(4) f (x) = tan x x = (2k + 1) π2 , k integer (7) f (x) =
2x − 1 , if x > 1
x = 0, 1

Problem 1.7. Find the value of k so that the function f is continuous everywhere

3x + k, if x ≤ 2
f (x) =
kx2 , if x > 2
Ans.: k = 2
Problem 1.8. Find the value of k so that the function f is continuous at x = 0
(
tan(6x)
tan(3x) , if x 6= 0
f (x) =
2k − 1 , if x = 0
3
Ans.: k = 2
Problem 1.9. Given that lim f (x) = 3, and lim h(x) = 0, find the following limits
x→1 x→1
h(x)
a) lim =0
x→1 f (x)
b) lim f (x) 2 = +∞
x→1 (h(x))

Problem 1.10. Use the intermediate value theorem to show that the equation x3 + x2 − 2x = 1 has at
least one solution in [−1, 1].
Ans.: L et f (x) = x3 + x2 − 2x for −1 ≤ x ≤ 1. f is continuous at every point of [−1, 1] since
it is defined by a polynomial formula. f (−1) = 2 and f (1) = 0. Since f (1) < 1 < f (−1), by the
intermediate value theorem there is at least one number c in (−1, 1) for which f (c) = 1.

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