Particular limits To evaluate this type of limits, how can I do, considering $f$ differentiable, and $ f (x_0)> 0 $
$$\lim_{x\to x_0} \biggl(\frac{f(x)}{f(x_0)}\biggr)^{\frac{1}{\ln x -\ln x_0 }},\quad\quad x_0>0,$$
$$\lim_{x\to x_0}  \frac{x_0^n f(x)-x^n f(x_0)}{x-x_0},\quad\quad n\in\mathbb{N}.$$
 A: For the first: for $x\neq x_0$, since $f(x)>0$ in a neighborhood of $x_0$
\begin{align*}
\frac{f(x)}{f(x_0)}^{\frac 1{\ln x-\ln x_0}}&=\exp\left(\frac {\ln f(x)-\ln f(x_0)}{\ln x-\ln x_0}\right)\\
&=\exp\left(\frac {\ln f(x)-\ln f(x_0)}{x-x_0}\frac{x-x_0}{\ln x-\ln x_0}\right),
\end{align*}
and taking the limit $x\to x_0$, since $\ln f$ is differentiable, we get 
$$\lim_{x\to x_0}\frac{f(x)}{f(x_0)}^{\frac 1{\ln x-\ln x_0}}=\exp\left(\frac{f'(x_0)}{f(x_0)}\frac 1{\frac 1{x_0}}\right)=\exp\left(x_0\frac{f'(x_0)}{f(x_0)}\right).$$
For the second question 
\begin{align*}\frac{x_0^nf(x)-x^nf(x_0)}{x-x_0}&=x_0^n\frac{f(x)-f(x_0)}{x-x_0}+f(x_0)\frac{x_0^n-x^n}{x-x_0}\\
&=x_0^n\frac{f(x)-f(x_0)}{x-x_0}-f(x_0)\sum_{k=0}^{n-1}x^kx_0^{n-k-1},
\end{align*}
and taking the limit $x\to x_0$ we get 
$$\lim_{x\to x_0}\frac{x_0^nf(x)-x^nf(x_0)}{x-x_0}=x_0^nf'(x_0)-nf(x_0)x_0^{n-1}.$$
A: Hint: write $x=x_0+h$ with $h\to0$ and expand each term up to order $h$.
For example, $\log(x)-\log(x_0)=\log(1+h/x_0)=h/x_0$ and $f(x)=f(x_0)+hf'(x_0)$ hence $f(x)/f(x_0)=1+hf'(x_0)/f(x_0)$, hence...
A: For the second limit you can also observe that if $x_0 \neq 0$ then
$$\lim_{x\to x_0}  \frac{x_0^n f(x)-x^n f(x_0)}{x-x_0} = \lim_{x \to x_0}x^nx_0^n \frac{ \frac{f(x)}{x^n}- \frac{f(x_0)}{x_0^n}}{x-x_0}= (x_0)^{2n} (\frac{f(x)}{x^n})'(x_0) \,.$$
The case $x_0=0$ is trivial.
A: For the first problem, one may be tempted to use L'Hopital's Rule:
Noting that $f(x)>0$ for $x$ sufficiently close to $x_0$:
$$\ln \biggl[\,\Bigl ({f(x)\over f(x_0)}\Bigr)^{1\over \ln x-\ln x_0}\,\biggr] = {\ln f(x)-\ln f(x_0)\over \ln x-\ln x_0}.$$
We have:
$$\tag{1}\lim_{x\rightarrow x_0} {\ln f(x)-\ln f(x_0)\over \ln x-\ln x_0}

=
\lim_{x\rightarrow x_0} {{f'(x)/ f(x)} \over 1/x}
=\lim_{x\rightarrow x_0} { {xf'(x)\over f(x)}  }={ {x_0f'(x_0)\over f(x_0)}  }.
$$
So $$\lim_{x\rightarrow x_0}\Bigl ({f(x)\over f(x_0)}\Bigr)^{1\over \ln x-\ln x_0}=
\exp\Bigl({ {x_0 f'(x_0)\over f(x_0)}  }\Bigr). $$
But, this line of reasoning is incorrect. We do not know that the last limit appearing in (1) exists.
See the other answers for  correct arguments.
