Clarification on proof of $\lim_{n\to +\infty}\left(\frac{f(a+\frac{1}{n})}{f(a)}\right)^{\frac{1}{n}}\to 1$ 
Let $f:\mathbb{R}\to \mathbb{R}$ be differentiable at $a\in \mathbb{R}$ such that $f(a)>0$. Evaluate:
  $$\lim_{n\to +\infty}\left(\frac{f(a+\frac{1}{n})}{f(a)}\right)^{\frac{1}{n}}.$$

Attempt. A proof would go like:
\begin{eqnarray} \left(\frac{f(a+\frac{1}{n})}{f(a)}\right)^{\frac{1}{n}}&=&\exp\left\{\ln\left(\frac{f(a+\frac{1}{n})}{f(a)}\right)^{\frac{1}{n}}\right\}=
\exp\left\{\frac{\ln f(a+\frac{1}{n})-\ln f(a)}{n}\right\}\nonumber\\
&=&\exp\left\{\frac{1}{n^2}\,\frac{\ln f(a+\frac{1}{n})-\ln f(a)}{\frac{1}{n}}\right\}\nonumber\\
&\to & \exp\left\{0\cdot \frac{f'(a)}{f(a)}\right\}=1,~n\to +\infty,\nonumber
\end{eqnarray}
where we use the definition of derivative and the chain rule. So far, so good.
My question is: 
one claims that since $f$ is differentible at $a,$ 
then $f$ is continuous at $a$ and so the limit becomes $1^0=1$, according to the algebra of sequential limits.
Is such an approach also correct? ($1^0$ is not an indeterminate form). 
Thanks for the help.
 A: Want
$\lim_{n\to +\infty}\left(\dfrac{f(a+\frac{1}{n})}{f(a)}\right)^{\frac{1}{n}}.
$
I would write,
since
$f(a+h)
\approx f(a)+hf'(a)$,
$\begin{array}\\
\dfrac{f(a+\frac{1}{n})}{f(a)}
&\approx \dfrac{f(a)+\frac1{n}f'(a)}{f(a)}\\
&= 1+ \dfrac{f'(a)}{nf(a)}\\
\text{so}\\
\left(\dfrac{f(a+\frac{1}{n})}{f(a)}\right)^{\frac{1}{n}}
&\approx \left(1+ \dfrac{f'(a)}{nf(a)}\right)^{\frac{1}{n}}\\
&\approx 1+ \dfrac{f'(a)}{n^2f(a)}\\
&\to 1\\
\end{array}
$
Question:
Is that exponent
$\dfrac1{n}$ or $n$?
If it is $n$,
we have a much more
interesting result:
$\begin{array}\\
\left(\dfrac{f(a+\frac{1}{n})}{f(a)}\right)^{n}
&\approx \left(1+ \dfrac{f'(a)}{nf(a)}\right)^{n}\\
&\approx \left(\left(1+ \dfrac{f'(a)}{nf(a)}\right)^{\dfrac{nf(a)}{f'(a)}}\right)^{\dfrac{f'(a)}{f(a)}}\\
&\to e^{\dfrac{f'(a)}{f(a)}}\\
\end{array}
$
A: Your argument at the end is correct. The expression under parentheses tends to $1$ and the exponent $1/n$ tends to $0$ and hence the desired limit is $1^0=1$.
Note the following rule:

If $f(x) \to a>0$ and $g(x) \to b$ as $x\to c$ then $\{f(x) \} ^{g(x)} \to a^b$ as $x\to c$.

Try to prove the above using exponential and logarithmic functions. 
For current question you only need that $f$ is continuous as $a$ and $f(a) \neq 0$.
