Find the following limit: $\lim\limits_{x \to 1} \left(\frac{f(x)}{f(1)}\right)^{1/\log(x)}$ 
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...)
 A: We have by the definition of the derivative:
$$\lim_{x \to 1} \left(\dfrac{f(x)}{f(1)}\right)^{\frac{1}{\log(x)}}=\lim_{x \to 1}\exp\left(\frac{\log f(x)-\log f(1)}{x-1}\frac{x-1}{\log(x)}\right)\\=\exp\left(\frac{(\log f(x))'\big|_{x=1}}{(\log x)'\big|_{x=1}}\right)=\exp\left(\frac{f'(1)}{f(1)}\right)$$
A: $$\lim_{x \to 1} \left(\dfrac{f(x)}{f(1)}\right)^{\frac{1}{\log(x)}}= e^{\lim_{x \to 1}(\frac{f(x)}{f(1)}-1)\frac{1}{\ln x}} = e^{\frac{1}{f(1)}\lim_{x \to 1}\frac{f(x)-f(1)}{x-1}\frac{x-1}{\ln x}}=e^{\frac{1}{f(1)}\cdot f'(1)\cdot1}=e^{\frac{f'(1)}{f(1)}}$$
A: $\newcommand{\+}{^{\dagger}}%
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*
*
\begin{align}\color{#0000ff}{\large%
\lim_{x \to 1}\bracks{\fermi\pars{x} \over \fermi\pars{1}}^{1/\ln\pars{x}}}
&=
\lim_{x \to \infty}
\bracks{\fermi\pars{1 + 1/x} \over \fermi\pars{1}}^{1/\ln\pars{1 + 1/x}}
=
\lim_{x \to \infty}
\bracks{1 + {\fermi'\pars{1} \over \fermi\pars{1}}\,{1 \over x}}^{x}
\\[3mm]&=
\lim_{x \to \infty}
\exp\pars{x\ln\pars{1 + {\fermi'\pars{1} \over \fermi\pars{1}}\,{1 \over x}}}
=
\lim_{x \to \infty}
\exp\pars{x\bracks{{\fermi'\pars{1} \over \fermi\pars{1}}\,{1 \over x}}}
\\[3mm]&= \color{#0000ff}{\large\expo{\fermi'\pars{1}/\fermi\pars{1}}}
\end{align}

*
\begin{align}
\lim_{x \to 1}\ln\pars{\bracks{\fermi\pars{x} \over \fermi\pars{1}}^{1/\ln\pars{x}}}
&=
\lim_{x \to 1}{\ln\pars{\fermi\pars{x}/\fermi\pars{1}} \over \ln\pars{x}}
=
\lim_{x \to 1}{\fermi'\pars{x}/\fermi\pars{x} \over 1/x}
=
{\fermi'\pars{1} \over \fermi\pars{1}}
\\[3mm]&\quad\imp\quad
\color{#0000ff}{\large%
\lim_{x \to 1}\bracks{\fermi\pars{x} \over \fermi\pars{1}}^{1/\ln\pars{x}}}
=
\color{#0000ff}{\large\expo{\fermi'\pars{1}/\fermi\pars{1}}}
\end{align}


A: Let $L$ be the limit you want to find. Then using L'Hospital's rule,
\begin{align*}
\ln L &= \lim_{x \to 1} \ln \left(\frac{f(x)}{f(1)}\right)^{1/\ln{x}} \\
&= \lim_{x \to 1} \dfrac{\ln f(x) - \ln f(1)}{\ln x} \\
&= \lim_{x \to 1} \dfrac{\dfrac{f'(x)}{f(x)}}{\dfrac 1 x} \\
&= \lim_{x \to 1} \frac{x f'(x)}{f(x)}
\end{align*}
Can you fill in the details, and find the limit from here?
