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Let f(x) = g(x) * cos(1/x) for x ≠ 0, and f(0) = 0, where g(x) is an even function that is differentiable at x = 0 and passes through the origin (g(0) = 0). Then f'(0) equals:
- 1
- 0
- 2
- does not exist
Correct answer: 0
Solution
f'(0) = lim(h->0) g(h)*cos(1/h)/h. Since g is even and differentiable at 0 with g(0) = 0, we have g(-h) = g(h), and g'(0) = lim(h->0) g(h)/h. For even g with g(0) = 0, g'(0) must equal 0 (since g'(0) = lim g(h)/h from both sides must be equal, but evenness forces g'(0) = -g'(0), so g'(0) = 0). Thus g(h)/h -> 0 and cos(1/h) is bounded between -1 and 1, so the product -> 0.
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