$f$ is differentiable at x=1, with $f(1)>0$. Find $\lim \limits_{n\to \infty}\left( \frac{f\left(1+\frac{1}{n}\right)}{f(1)} \right)^\frac{1}{n} $ $f$ is differentiable at $x=1$, with $f(1)>0$. Find $\lim \limits_{n\to \infty}\left( \frac{f\left(1+\frac{1}{n}\right)}{f(1)} \right)^\frac{1}{n}$.
Here's what I got so far:
$$\left( \frac{f(1+\frac{1}{n})}{f(1)} \right)^\frac{1}{n} = \exp\left(\frac{1}{n}\left(\log(f\left(1+\frac{1}{n}\right) - \log f(1)\right)\right) = \\
= \exp\left(\frac{1}{n}\left(\log f\left(1+\frac{1}{n}\right) - \log f(1)\right) \cdot \frac{\frac{1}{n}}{\frac{1}{n}}\right)$$
Now, if $f$ is differentiable at $x=1$, then so it $\log(f)$. Therefore, as $n\rightarrow\infty$:
$$\lim \limits_{n\to \infty}\left(\frac{f\left(1+\frac{1}{n}\right)}{f(1)} \right)^\frac{1}{n} = \exp\left(\lim \limits_{n\to \infty}\frac{1}{n^2} \cdot(\log f)'(1)\right) = e^0 = 1$$ 
I'm not too sure about that fact that the derivative cancels out, as I could have gotten that zero exponent in a much simpler way. Am I missing something?

Just a suggestion (if I may) to make life easier and to get more than the limit.
Consider $$A=\left( \frac{f\left(1+\frac{1}{n}\right)}{f(1)} \right)^{n^a}\implies \log(A)=n^a \log\left( \frac{f\left(1+\frac{1}{n}\right)}{f(1)} \right)$$ Assuming that $f(.)$ is continuoulsy differentiable at $x=1$, Taylor series gives for infinitely large values of $n$
$$f\left(1+\frac{1}{n}\right)=f(1)+\frac{f'(1)}{n}+\frac{f''(1)}{2 n^2}+O\left(\frac{1}{n^3}\right)$$
$$\frac{f\left(1+\frac{1}{n}\right)}{f(1)}=1+\frac{f'(1)}{f(1) n}+\frac{f''(1)}{2 f(1)
   n^2}+O\left(\frac{1}{n^3}\right)$$ Continuing with Taylor
$$\log\left( \frac{f\left(1+\frac{1}{n}\right)}{f(1)} \right)=\frac{f'(1)}{f(1) n}+\frac{f(1) f''(1)-f'(1)^2}{2 f(1)^2
   n^2}+O\left(\frac{1}{n^3}\right)$$ If $\color{red}{a=1}$, then 
$$\log(A)=\frac{f'(1)}{f(1)}+\frac{f(1) f''(1)-f'(1)^2}{2 f(1)^2
   n}+O\left(\frac{1}{n^2}\right)$$ Taylor again using $A=e^{\log(A)}$
$$A=e^{\frac{f'(1)}{f(1)}}\left(1+\frac{f(1) f''(1)-f'(1)^2}{2 f(1)^2 n}+O\left(\frac{1}{n^2}\right)\right)$$ which shows the limit and also how it is approached.
If $\color{red}{a=-1}$, then 
$$\log(A)=\frac{f'(1)}{f(1) n^2}+O\left(\frac{1}{n^3}\right)$$
$$A=1+\frac{f'(1)}{f(1) n^2}+O\left(\frac{1}{n^3}\right)$$ which shows the limit and also how it is approached.
