Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 2.5 CONTINUITY ¤ 57
25. By Theorem 7, the exponential function e −5t and the trigonometric function cos 2πt are continuous on (−∞, ∞).
By part 4 of Theorem 4, L(t) =e −5t cos 2πt is continuous on (−∞, ∞).
27. By Theorem 5, the polynomial t 4 − 1 is continuous on (−∞, ∞). By Theorem 7, ln x is continuous on its domain, (0, ∞).
By Theorem 9, ln t 4 − 1 is continuous on its domain, which is
t | t 4 − 1 > 0 = t | t 4 > 1 = {t ||t| > 1} =(−∞, −1) ∪ (1, ∞)
29. The function y =
1
is discontinuous at x =0because the
1+e1/x left- and right-hand limits at x =0are different.
31. Because we are dealing with root functions, 5+ √ x is continuous on [0, ∞), √ x +5is continuous on [−5, ∞), sothe
quotient f(x) = 5+√ x
√ 5+x
is continuous on [0, ∞). Sincef is continuous at x =4, lim
x→4
f(x) =f(4) = 7 3 .
33. Because x 2 − x is continuous on R, the composite function f(x) =e x2 −x is continuous on R, so
lim
x→1 f(x) =f(1) = e1 − 1 = e 0 =1.
x
2
if x