Evaluating a limit (involving derivative) Evaluate the following limit:  $$\mathop {\lim }\limits_{x \to 1} {\left( {{{f(x)} \over {f(1)}}} \right)^{{1 \over {\log x}}}}$$
My work:
$$\eqalign{
  & \mathop {\lim }\limits_{x \to 1} f(x) = \mathop {\lim }\limits_{x \to 1} {{f(x) - f(1)} \over {x - 1}} \cdot (x - 1) + f(1) = \mathop {\lim }\limits_{x \to 1} f'(1) \cdot (x - 1) + f(1)  \cr 
  & \mathop {\lim }\limits_{x \to 1} {\left( {{{f'(1) \cdot (x - 1) + f(1)} \over {f(1)}}} \right)^{{1 \over {\log x}}}} = \exp \left( {\ln \left( {{{f'(1) \cdot (x - 1) + f(1)} \over {f(1)}}} \right) \cdot {1 \over {\log x}}} \right) \cr} $$
Now, I have an expreesion which is essentially $0\cdot \infty$ which is problamatic.
I think I'm on the right way, but missing something here.  
Can you help me? 
Update:
$f$ is differentiable at $x=1$ and $f(1) > 0$
 A: We define,
$$L:=\mathop {\lim }\limits_{x \to 1} {\left( {{{f(x)} \over {f(1)}}} \right)^{{1 \over {\log x}}}}$$
Then,
$$\log{L}=\lim_{x\to 1}\frac{\log{\frac{f(x)}{f(1)}}}{\log{x}}\\
=\lim_{x\to 1}\frac{\frac{f^\prime (x)}{f(x)}}{\frac{1}{x}}\\
=\lim_{x\to 1}\frac{xf^\prime (x)}{f(x)}\\
=\frac{f^\prime (1)}{f(1)},$$
assuming $f(1)\neq 1$. Then,
$$L=e^{\frac{f^\prime (1)}{f(1)}}.$$
A: Notice that $\log x \rightarrow (x - 1)$ when $x \rightarrow 1$.
$\displaystyle \lim_{x\rightarrow 1} \left(\frac{f(x)}{f(1)}\right)^\frac{1}{\log x} = \lim_{x\rightarrow 1} \left( \frac{f(1) + f^\prime(1)(x-1)}{f(1)} \right)^\frac{1}{\log x} = \lim_{x\rightarrow 1} \left( 1 + \frac{f^\prime(1)(x-1)}{f(1)}\right)^\frac{1}{x-1}$.
Defining $y=x-1$, this becomes
$\displaystyle \lim_{y\rightarrow 0} \left( 1 + \alpha y \right)^\frac{1}{y}$,
with $\alpha = \frac{f^\prime(1)}{f(1)}$.
This is almost immediate, since
$\displaystyle \lim_{y\rightarrow 0} \left( 1 + \alpha y \right)^\frac{1}{y} = \lim_{y\rightarrow 0} e^{\frac{1}{y}\log(1+\alpha y)} = \lim_{y\rightarrow 0} e^{\frac{1}{y} \alpha y} = e^\alpha = \exp{\left( \frac{f^\prime(1)}{f(1)}\right)}$.
Of course, all of this is only valid if $f(x)$ is differentiable at $x=1$.
