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 123h(f(h)−1) =
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Detailed Solution
fx+y=fxfy for all x∈R put x=0,y=0 then f0=f0f0 ⇒f0=1 since f0≠0 given that f'0=3 ⇒limh→0f0+h-f0h=3 ⇒limh→0f0fh-f0h=3 ⇒limh→0fh-1h=3