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 limₕ→₀ (1/h)(f(h)-1) is equal to......
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 limₕ→₀ (1/h)(f(h)-1) is equal to......
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1 Answer
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The functional equation f (x+y) = f (x)f (y) implies f (x) = a? for some constant a.
Then f' (x) = a? ln (a).
Given f' (0) = 3, we have a? ln (a) = 3 ⇒ ln (a) = 3 ⇒ a = e³.
So, f (x) = e³?
We need to evaluate the limit: lim (x→0) (f (x)-1)/x = lim (x→0) (e³? -1)/x.
Using the standard limit lim (u→0) (e? -1)/u = 1, we can write:
lim (x→0) 3 * (e³? -1)/ (3x) = 3 * 1 = 3.
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