Barrons AP Calculus - David Bock
(B) Since |x| = x if x > 0 but equals −x if x < 0, while ...
(E) Note that x can be rewritten as and that, as x → ∞, ...
(A) As x → π, (π − x) → 0. ...
(C) Since f (x) = x + 1 if x ≠ 1, exists (and is equal to 2). ...
(B) for all x ≠ 0. For f to be continuous at x = 0, must equal f (0). ...
(B) Only x = 1 and x = 2 need be checked. Since for x ≠ 1, 2, and = −3 = f (1), f is continuous at x = 1. Since does not exist, ...
(C) As x → ±∞, y = f (x) → 0, so the x-axis is a horizontal asymptote. Also, as x → ±1, y → ∞, so x = ±1 are vertical asymptote ...
(C) As x → ∞, the denominator (but not the numerator) of y equals 0 at x = 0 and at x = 1. ...
(D) The function is defined at 0 to be 1, which is also ...
(D) See Figure N2–1. ...
(E) Note, from Figure N2–1, that ...
(E) As x → ∞, the function sin x oscillates between −1 and 1; hence the limit does not exist. ...
(A) Note that if x ≠ 0 and that ...
32. (A) ...
(E) Verify that f is defined at x = 0, 1, 2, and 3 (as well as at all other points in [−1,3]). ...
(C) Note that However, f (2) = 1. Redefining f (2) as 0 removes the discontinuity. ...
(B) The function is not continuous at x = 0, 1, or 2. ...
36. (B) ...
(E) As x → 0−, arctan The graph has a jump discontinuity at x = 0. (Verify with a calculator.) ...
(D) No information is given about the domain of f except in the neighborhood of x = −3. ...
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