Calculus: Early Transcendentals (6th) -- Student Solutions by James Stewart

By James Stewart

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X→1 17. Given ε > 0, we need δ > 0 such that if 0 < |x − (−3)| < δ, then |(1 − 4x) − 13| < ε. But |(1 − 4x) − 13| < ε ⇔ |−4x − 12| < ε ⇔ |−4| |x + 3| < ε ⇔ |x − (−3)| < ε/4. So if we choose δ = ε/4, then 0 < |x − (−3)| < δ ⇒ |(1 − 4x) − 13| < ε. Thus, lim (1 − 4x) = 13 by the definition of x→−3 x a limit. 19. Given ε > 0, we need δ > 0 such that if 0 < |x − 3| < δ, then So choose δ = 5ε. Then 0 < |x − 3| < δ of a limit, lim x→3 x 3 − <ε ⇔ 5 5 ⇒ |x − 3| < 5ε ⇒ 1 5 |x − 3| < ε ⇔ |x − 3| < 5ε. |x − 3| <ε ⇒ 5 x 3 − < ε.

X→1 (d) f (x) approaches 4 as x approaches 5 from the left and from the right, so lim f (x) = 4. x→5 (e) f(5) is not defined, so it doesn’t exist. 7. (a) lim g(t) = −1 (b) lim g(t) = −2 t→0− t→0+ (c) lim g(t) does not exist because the limits in part (a) and part (b) are not equal. t→0 (d) lim g(t) = 2 (e) lim g(t) = 0 t→2− t→2+ (f ) lim g(t) does not exist because the limits in part (d) and part (e) are not equal. t→2 (g) g(2) = 1 (h) lim g(t) = 3 t→4 9. (a) lim f (x) = −∞ (b) lim f(x) = ∞ (d) lim f (x) = −∞ (e) lim f (x) = ∞ x→−7 x→6− (c) lim f (x) = ∞ x→−3 x→0 x→6+ (f ) The equations of the vertical asymptotes are x = −7, x = −3, x = 0, and x = 6.

Remember that |a| = a if a ≥ 0 and that |a| = −a if a < 0. Thus, x + |x| = + 2x if x ≥ 0 0 if x < 0 and y + |y| = + 2y 0 if y ≥ 0 if y < 0 We will consider the equation x + |x| = y + |y| in four cases. (1) x ≥ 0, y ≥ 0 2x = 2y (2) x ≥ 0, y < 0 2x = 0 x=y x=0 (3) x < 0, y ≥ 0 0 = 2y (4) x < 0, y < 0 0=0 0=y Case 1 gives us the line y = x with nonnegative x and y. Case 2 gives us the portion of the y-axis with y negative. Case 3 gives us the portion of the x-axis with x negative. Case 4 gives us the entire third quadrant.

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