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# Limits and Continuity Solutions to Practice Problems for AP Calculus

based on 8 ratings
By McGraw-Hill Professional
Updated on Oct 24, 2011

The practice problems for these solutions can be found at Limits and Continuity Practice Problems for AP Calculus.

Part A—The use of a calculator is not allowed.

1. Since the degree of the polynomial in the numerator is the same as the degree of the polynomial in the denominator,
2. Since the degree of the polynomial in the numerator is 2 and the degree of the polynomial in the denominator is 3,
3. The degree of the monomial in the numerator is 2 and the degree of the binomial in the denominator is 1. Thus,
4. Divide every term in both the numerator and denominator by the highest power of x. In this case, it is x. Thus, you have
5. The graph of f indicates that:
1. "x =4 is not in the domain of f " is false since f (4)=2.

Part B—Calculators are allowed

1. 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.)
2. 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
3. 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.)
4. In order for g (x ) to be continuous at x =3, it must satisfy the three conditions of continuity:
1. g (3)=32 +5=14,
5. Checking with the three conditions of continuity:
1. f (2)=12,
6. The graph indicates that
1. f (3)=4,
7. (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).
8. Substituting x =0 would lead to 0/0. Substitute (1 – cos2 x ) in place of sin2 x and obtain
9. Verify your result with a calculator. (See Figure 5.7-4)

1. 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.

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