Honors Calculus AB
Chapter 2 Review Problems
Name:______ANSWER KEY_____
24 September 2015
Topics for Chapter 2
Limits
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evaluation techniques (substitution, algebraic manipulation, etc.)
properties (ex. lim( f (x) + g(x)) = lim f (x) + lim g(x) )
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one-sided vs. two-sided ( lim+ f (x) vs lim f (x) )
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Sandwich Theorem
lim €
f (x) – end behavior (constant?)
•
lim f (x) = ±∞ €
– what is a? €
x→a
x→a
x→a
x→a
x→a
x → ±∞
x→a
Continuity
•
Limit definition lim f (x) = f (a)
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4 types of discontinuities (removable, jump, infinite, oscillating)
Application of continuity to the Intermediate Value Theorem
x→a
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Rates of Change
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average vs. instantaneous
role of limit in instantaneous rate-of-change
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slope of curve and the difference quotient ( lim
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tangent to a curve
normal to a curve
x→a
f (x + h) − f (x)
f (x) − f (a)
or lim
)
h
→
0
(x + h) − x
x −a
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Directions: Use these problems to ascertain your current understanding of the material and to
familiarize youslef with the kind of questions that will be asked on the end-of-chapter test.
Make sure you show work that indicates you understanding, even for the multiple choice
questions. The goal is to figure out what you know and how well you understand it.
1. Determine lim ( 5 − 2x + x 2 ) = 5 − 2(3) + 32 = 5 − 6 + 9 = 8
1. ______B_________
x→3
(A) 2
D. Haupt
(B) 8
(C) 10
(D) 12
(E) 20
Questions are primarily from the supplementary materials for the text.
Chapter 2 Review Problems Answer Key
page 2
5 − 6x + x 2
(5 − x)(1− x)
= lim
= lim(1− x) = −4
x→5
x→5
x→5
5− x
5− x
2. Determine lim
(A) –4
3. Find lim+
x→3
(B) 0
(C) 4
(D) 6
(E) Does not exist
x+3
6
≈
=∞
x − 3 tiny, positive
(A) –∞
(B) –6
(C) 0
3. ______E_________
(D) 6
(E) ∞
$& x 2 − 2 x < 1
1
1
4. Let f (x) = % 1
. What is lim+ f (x) = − (1) +1 =
x→1
&' − 2 x +1 x ≥1
2
2
(A) –1
(B)
1
2
(C) 1
2. ______A_________
(D) 1.73
4. ______B_________
(E) Does not exist
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The graph is
[–4, 4] by [–3, 3]
Use the graph at the right for questions 5 through 7.
5. For the function y = f(x) in the graph, which statement is false?
(A)
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(D)
lim f (x) = 0
(B) lim− f (x) = −1
lim f (x) = −1
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(E) lim− f (x) = lim+ f (x)
x →1
x →1
€
x →0 −
x →−1+
x →3
(C) lim f (x) = ∞ (only true from the right)
x →−3
€6. The function f(x) is continuous
€
at which of the following points?
(A) x = –3 (B) x = –1
(C) x = 1
(D) x = 3
(C) x = 1
(D) x = 3
6. _______C________
(E) All of these
7. The function f(x) has a non-removable discontinuity at which of
the following points or sets of points?
(A) x = –3 (B) x = –1
5. _______C________
7. _______E________
(E) x = –3 and x = –1
Chapter 2 Review Problems Answer Key
page 3
6x +1
x →∞ 6 − 2x
8. Find lim
(A) –3
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8. _______D________
(B) 0
(C) 1
6 – 2x < 0 for x →∞ so lim
x→∞
(D) 3
(E) Does not exist
6x +1
6x +1
6x
= lim
≈ lim
=3
x→∞
x→∞
6 − 2x
−(6 − 2x)
2x
9. Sketch a possible graph for a function f that has the stated properties:
f(–2) exists, lim f (x) exists, f is not continuous at x = –2, and
x →−2
lim f (x) does not exist.
x →1
There
€ are lots of possibilities. Here is a possible solution.
€
10. Find the average rate of change of the function f(x) = 100 – 16x2 over the interval [0, 2].
f (2) − f (0) [100 −16 ⋅ (2)2 ]−[100 −16 ⋅ (0)2 ] −64
=
=
= –32
(2) − (0)
2
2
11. Find the slope of the curve y = x2 + x at x = 3.
lim
h→0
7
f (x + h) − f (x)
[(3+ h)2 + (3+ h)]− (32 + 3)
[9 + 6h + h 2 + 3+ h]− (9 + 3)
= lim
= lim
= lim(6 + h +1)
h→0
h→0
h→0
(x + h) − (x)
h
h
(
( 1x )) .
12. Show how the Sandwich Theorem can be used to find lim 3 + x 2 sin
x →0
3
−1 ≤ sin ( 1x ) ≤ 1
−x 2 ≤ x 2 sin ( 1x ) ≤ x 2
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3− x 2 ≤ (3+ x 2 sin ( 1x )) ≤ 3+ x 2
lim 3− x 2 ≤ lim (3+ x 2 sin ( 1x )) ≤ lim 3+ x 2
x→0
x→0
x→0
3 ≤ lim (3+ x 2 sin ( 1x )) ≤ 3
x→0
it is ok to multiply by x2 because it is always positive
Chapter 2 Review Problems Answer Key
page 4
13. Find the points of discontinuity of the function y =
the type of discontinuity.
x 2 − 3x − 4
. For each discontinuity, identify
x 2 − 7x +12
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y=
(x − 4)(x +1)
(x − 4)(x − 3)
Removeable discontinuity at x = 4, Infinite discontinuity at x = 3
14. Assume that lim− f (x) = 2 , lim+ f (x) = −4 , lim− g(x) = −2 , and lim+ g(x) = 3 .
x →2
x →2
x →2
x →2
Find the value of lim+ f (x)⋅ g(x)
x →2
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€ g(x) = lim+ f (x)⋅ lim+ g(x) = (−4)⋅ (3) = –12
lim f (x)⋅
x→2+
x→2
x→2
€