Calculating limit in parts. Why possible? Let $f$, continuous function, differentiable at $x=1$ and $f(1)>0$.  
Consider the following equation:  
$$\lim \limits_{x\to 1} \left(\frac{\color{Blue}{f(x)-f(1)}(x-1)}{\color{Blue}{(x-1)}f(1)}\right)^\frac{1}{\log x} = \lim \limits_{x\to 1} \left( \frac{\color {Blue}{f'(1)}(x-1)}{f(1)} \right)^\frac{1}{\log x}$$
My question is: Why can you evaluate the blue expression on the LHS as $f'(1)$ before taking the limit? Moreover, the whole expression is powered by $\frac{1}{\log x}$
It seems to me like claiming $$\lim\limits_{n\to\infty}(1+\frac{1}{n})^n = \lim\limits_{n\to\infty}(1+0)^n = 1$$
So what's making the first valid whereas the second is nonsense.
 A: This: $$\lim_{x\rightarrow\infty} f(x)^{g(x)}=(\lim_{x\rightarrow\infty}f(x))^{\lim_{x\rightarrow\infty}g(x)}$$
only if both limits exist (in the narrow sense)! On the second example the outer limit $\lim_{n\rightarrow\infty}n=\infty$, therefore doesn't exist (in the narrow sense).
In your first example I think it should be $$\frac{\textbf ( f(x)-f(1)\textbf{)}(x-1)}{(x-1)f(1)}$$ inside the brackets..
A: Maybe this is not a satisfying way to see that the equality holds (and I hope that the following is correct), but:
Let $$y = \left(\frac{\color{Blue}{[f(x)-f(1)]}(x-1)}{\color{Blue}{(x-1)}f(1)}\right)^\frac{1}{\ln x},$$ then
$$
\ln(y) = \frac{1}{\ln(x)}\ln\left(\frac{\color{Blue}{[f(x)-f(1)]}(x-1)}{\color{Blue}{(x-1)}f(1)}\right).
$$
Now
$$
\lim_{x\to 1} \left(\frac{\color{Blue}{[f(x)-f(1)]}(x-1)}{\color{Blue}{(x-1)}f(1)}\right) = \lim_{x\to 1} \frac{\color{Blue}{[f(x)-f(1)]}}{\color{Blue}{(x-1)}}\lim_{x\to 1}\frac{x-1}{f(1)} = 0,
$$
so
$$
\lim_{x\to 1} \ln\left(\frac{\color{Blue}{[f(x)-f(1)]}(x-1)}{\color{Blue}{(x-1)}f(1)}\right) = -\infty.
$$
Therefore
$$
\lim_{x\to 1} \ln(y) = -\infty.
$$
(You naturally might convince yourself that the right hand limit and the left hand limit indeed both are equal to this.
This shows that the limit of the left hand side is $0$. 
Likewise 
$$
\lim_{x\to 1}\left( \frac{\color {Blue}{f'(1)}(x-1)}{f(1)} \right)^\frac{1}{\ln x} = 0.
$$
