Let f : R → R satisfy the equation f(x + y) = f(x) . f(y) for all x, y ∈R and f(x) ≠ 0 for any x∈R. If the function f is differentiable at x = 0 and f'(0) = 3, then
lim _(h → 0) 1/h(f(h) - 1) is equal to ____________.

Solution

Maths sir · 3 steps

Find f(0) first. Once you see it is 1, the limit is just the derivative at 0 in disguise.

1. Pin down f(0). Put x = y = 0. The equation gives f(0) = f(0) · f(0), so f(0)(f(0) - 1) = 0. Since f(x) ≠ 0 for every x, f(0) cannot be 0, so f(0) = 1.
2. Rewrite the limit. Replace the 1 in the limit with f(0). The limit becomes the definition of the derivative of f at x = 0.

   lim(h → 0) (f(h) - 1)/h = lim(h → 0) (f(0 + h) - f(0))/h = f'(0)

3. So. The limit equals f'(0) = 3, so the answer is 3.

The idea to keep. In any functional equation, plug in x = y = 0 first to pin down f(0). A limit like (f(h) - f(0))/h is f'(0) written out by its definition.

Answer from the NTA final answer key. Solution by PrepWiser.