then = 4 cos t. Thus:NOTE: Since each of the limits in Questions 35–39 yields an indeterminate
form of the type , we can apply L’Hôpital’s Rule in each case, getting
identical answers.
(C) The given limit is the derivative of f (x) = x^6 at x = 1.(B) The given limit is the definition for f ′(8), where (B) The given limit is f ′(e), where f (x) = ln x.(C) The given limit is the derivative of f (x) = cos x at x = 0; f ′(x) = −sin
x.(B)Thus f is discontinuous at x = 1, so it cannot be differentiable.(E) so the limit exists. Because g(3) = 9, g is
continuous at x = 3.Since