TITLE: Find the following limit: $\lim\limits_{x \to 1} \left(\frac{f(x)}{f(1)}\right)^{1/\log(x)}$
QUESTION [6 upvotes]: find the following limit
$f(x)$ is differentiable at $x=1$ and $f(1)>0$
$\lim\limits_{x\to 1}\left(\dfrac{f(x)}{f(1)}\right)^{\frac{1}{\log(x)}}$

why can i just substitute $x$ for $1$ and thats will be the limit (since $f$ is differentiable, it is also continuous...)

REPLY [4 votes]: $$\lim_{x \to 1} \left(\dfrac{f(x)}{f(1)}\right)^{\frac{1}{\log(x)}}= e^{\lim_{x \to 1}(\frac{f(x)}{f(1)}-1)\frac{1}{\ln x}} = e^{\frac{1}{f(1)}\lim_{x \to 1}\frac{f(x)-f(1)}{x-1}\frac{x-1}{\ln x}}=e^{\frac{1}{f(1)}\cdot f'(1)\cdot1}=e^{\frac{f'(1)}{f(1)}}$$