How to prove that the limit $\lim_{\delta \to 0}(\frac{f(x+\delta x)}{f(x)})^{\frac{1}{\delta}}$ exists and it is nonzero $f:(0, \infty) \to (0, \infty)$ is a differentiable function and I want to prove that for any $x>0$ limit
$\lim_{\delta \to 0}(\frac{f(x+\delta x)}{f(x)})^{\frac{1}{\delta}}$
exists and is nonzero.
I was thinking of first getting the logarithm of a limit in order to deal with the exponent more easily, and then proving the existence of the limit from $\epsilon-\delta$ definition. I get that as $\delta \to 0$ this limit is 1. So something along the lines of
$\lim_{\delta \to 0}(\frac{1}{\delta}(lnf(x+\delta x)-lnf(x))$
But then, since we don't have the analytical form I have little information to understand what $\epsilon$ and $N$ (number after which we can choose arbitrary epsilon) to choose.
Also, I was thinking that it might be useful to know that $f$ is differentiable, so
$\lim_{x \to x_0}\frac{f(x)-f(x_0)}{x-x_0}$
exists, but I don't see how would it be useful.
What am I missing out? How can I prove this? Thank you.
 A: By the mean value theorem, $f(x + \delta x) = f(x) + f'(t)\delta x$, for some $t \in [x, x + \delta x]$. Then:
$$(\frac{f(x + \delta x)}{f(x)})^{1/\delta} = (1 + \frac{f'(t)x}{f(x)}\delta)^{1/\delta} $$
Substitute $y = \frac{f'(t)x}{f(x)}\delta \Rightarrow \frac{1}{\delta} = \frac{xf'(t)}{yf(x)}$ to get the above as:
$$(1 + y)^{(xf'(t)/f(x))/y} $$
As $y \to 0 \Leftrightarrow \delta \to 0$, we get:
$$\lim_{y \to 0}(1 + y)^{(xf'(t)/f(x))/y} = \boxed{e^{xf'(x)/f(x)}} $$
Clearly, the result is nonzero.
A: Write
\begin{align}
\left(\frac{f(x+\delta x)}{f(x)}\right)^{\frac{1}{\delta}} &= \exp\left(\frac{1}{\delta}\ln\left(\frac{f(x+\delta x)}{f(x)}\right)\right)\\
&= \exp\left(\frac{\ln\left[f(x+\delta x)\right]-\ln\left[f(x)\right]}{\delta}\right)\\
&= \exp\left(x\cdot\frac{\ln\left[f(x+\delta x)\right]-\ln\left[f(x)\right]}{\delta x}\right)
\end{align}
It was assumed that $f$ is differentiable over $(0,\infty)$, so the composition $\ln\circ f$ is differentiable over $(0,\infty)$. Since $\delta x\to0$ as $\delta\to 0$, it follows that the limit
$$\lim_{\delta\to 0}\frac{\ln\left[f(x+\delta x)\right]-\ln\left[f(x)\right]}{\delta x}$$
exists, and equals the derivative of $\ln\circ f$ evaluated at $x$, namely $f'(x)/f(x)$. We conclude from the continuity of the exponential function that
$$\lim_{\delta\to 0}\left(\frac{f(x+\delta x)}{f(x)}\right)^{\frac{1}{\delta}}=\lim_{\delta\to 0}\exp\left(x\cdot\frac{\ln\left[f(x+\delta x)\right]-\ln\left[f(x)\right]}{\delta x}\right)=\exp\left(x\frac{f'(x)}{f(x)}\right)$$
