Let f: R → R be such that f(1) = 3 and f'(1) = 6. Then $\underset{x→0}{\lim}\begin{pmatrix}\frac{f(1+x)}{f(1)}\end{pmatrix}^{1/x}$ equals
Answer & explanation
Correct answer: option 3
Let $l=\underset{x→0}{\lim}\begin{pmatrix}\frac{f(1+x)}{f(1)}\end{pmatrix}^{1/x}$
$∴\log l=\underset{x→0}{\lim}\frac{1}{x}\log\begin{pmatrix}\frac{f(1+x)}{f(1)}\end{pmatrix}=\underset{x→0}{\lim}\frac{\log f(1+x)-\log f(1)}{x}=\frac{d}{dx}[\log f(x+1)]_{At\,x=1}$
$=\frac{1}{f(1)}f'(1)=\frac{6}{3}=2$
$∴l=e^2$