Limit calculation I know that $f'(x)$ exists at $x = 1$, and moreover that $f(1) = K > 0$.
I need to calculate the limit 
$$\lim_{x\rightarrow1}  \left(\frac{f(x)}{f(1)}\right) ^ {1/\log(x)}$$
How could I calculate this ?
Thanks in advance ! 
 A: You can bring it into the form $0/0$ via
$$ \lim_{x\to1}\left(\frac{f(x)}{f(1)}\right) ^ {1/\log(x)}
 = \exp \left(\lim_{x\to1} \frac{\log [f(x)/f(1)]}{\log x} \right)$$
De l'Hospital then gives
$$\lim_{x\to1}\left(\frac{f(x)}{f(1)}\right) ^ {1/\log(x)} 
= \exp \left( \lim_{x\to1}\frac{x f'(x)}{f(x)}  \right)$$
The limit can then be evaluated and we have
$$\lim_{x\to1}\left(\frac{f(x)}{f(1)}\right) ^ {1/\log(x)}  =\exp\left(\frac{f'(1)}{f(1)}\right).$$
A: Expand the function around $x=1$ using the derivative and the series expansion of $\log (x+1)$
$$\lim_{\Delta x\rightarrow 0}  \left(\frac{f(1)+\Delta x f'(1)}{f(1)}\right) ^ {1/log(1+\Delta x)} = \lim_{\Delta x\rightarrow 0}  \left(1+\Delta x\frac{f'(1)}{f(1)}\right) ^ {1/\Delta x}$$
$$=e^{f'(1)/f(1)}$$
A: Let $$y=\left(\frac{f(x)}{f(1)}\right) ^ {1/\log(x)}$$
So, $$\log y=\frac{\log f(x)-\log f(1)}{\log x}=\frac{\log f(x)-\log f(1)}{f(x)-f(1)}\frac{f(x)-f(1)}{x-1}\frac{x-1}{\log x-\log 1}$$
$$\lim_{x\to1}\log y=\left(\frac{d\log f(x)}{d f(x)}\frac{d f(x)}{dx}\frac1{\frac{d\log x}{dx}}\right)_{x=1}=\left(\frac1{f(x)}\cdot{f'(x)}\cdot x\right)_{x=1}=\frac{f'(1)}{f(1)}$$
So, $$\lim_{x\to1} \left(\frac{f(x)}{f(1)}\right) ^ {1/\log(x)}=\lim_{x\to1} y=e^{\frac{f'(1)}{f(1)}}$$
