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(http://www.education.com/study-help/ap-notes-calculus)Limits and Continuity Solutions to Practice Problems
for AP Calculus
Collect It!
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By William Ma
McGraw-Hill Professional
(http://www.education.com/partner/articles/mcgrawhillprof/)Updated on Oct 24, 2011
The practice problems for these solutions can be found at
Limits and Continuity Practice Problems for AP
Calculus
(http://www.education.com/reference/article/practice-problems/).
Part A—The use of a calculator is not allowed.
2.
3.
Since the degree of the polynomial in the numerator is the same as the degree of the polynomial in the
denominator,
4.
Since the degree of the polynomial in the numerator is 2 and the degree of the polynomial in the denominator is
3,
5.
The degree of the monomial in the numerator is 2 and the degree of the binomial in the denominator is 1. Thus,
6.
Divide every term in both the numerator and denominator by the highest power of x. In this case, it is x. Thus,
you have
8.
9.
10.
11.
The graph of f indicates that:
I.
II.
"x =4 is not in the domain of f " is false since f (4)=2.
III.
Part B—Calculators are allowed
Examining the graph in your calculator, you notice that the function approaches the x -axis as x → ∞or as x
→–∞. Thus, the line y =0 (the x -axis) is a horizontal asymptote. As x approaches 1 from either side, the
function increases or decreases without bound. Similarly, as x approaches –2 from either side, the function
increases or decreases without bound. Therefore, x = 1 and x = –2 are vertical asymptotes. (See Figure 5.7-1.)
13.
As x → 5
+, the limit of the numerator (5+[5]) is 10 and as x → 5
+, the denominator approaches 0 through
negative values. Thus, the
14.
Since f (x ) is a rational function, it is continuous everywhere except at values where the denominator is 0.
Factoring and setting the denominator equal to 0, you have (x + 6) (x – 2) = 0. Thus, the points of discontinuity
are at x = – 6 and x = 2. Verify your result with a calculator. (See Figure 5.7-2.)
15.
In order for g (x ) to be continuous at x =3, it must satisfy the three conditions of continuity:
g (3)=3
2+5=14,
1.
2.
3.
16.
Checking with the three conditions of continuity:
f (2)=12,
1.
17.
The graph indicates that
f (3)=4,
a.
b.
c.
d.
18.
(See Figure 5.7-3.) If b =0, then r =0, but r cannot be 0. If b = –3, –2, or –1 f would have more than one root.
Thus b =1. Choice (E).
19.
Substituting x =0 would lead to 0/0. Substitute (1 – cos
2x ) in place of sin
2x and obtain
20.
21.
22.
23.
Therefore, the line y = –1 is a horizontal asymptote. As for vertical asymptotes, f(x) is continuous and defined
for all real numbers. Thus, there is no vertical asymptote.
Excerpted From:
5 Steps to a 5 AP Calculus AB and BC
(http://www.amazon.com/gp/product/0071624759?ie=UTF8&tag=educationcom-20&linkCode=xm2&camp=1789&creativeASIN=0071624759)Buy this book »
(http://www.amazon.com/gp/product/0071624759?ie=UTF8&tag=educationcom-20&linkCode=xm2& camp=1789&creativeASIN=0071624759)From 5 Steps to a 5 AP Calculus AB and BC. Copyright © 2010 by The McGraw-Hill Companies. All Rights
Reserved.
(http://www.mhprofessional.com/)
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