If $f$ is positive and differentiable in $(0,\infty)$, then I want to find the following limit.
$\lim\limits_{n\to \infty}\left(\dfrac{f\left(x+\dfrac{1}{n}\right)}{f(x)}\right)^n$.
I have done as follows:
$\lim\limits_{n\to \infty}\left(\dfrac{f\left(x+\dfrac{1}{n}\right)}{f(x)}\right)^n=\left(\lim\limits_{n\to \infty}\dfrac{f\left(x+\dfrac{1}{n}\right)}{f(x)}\right)^n=1$ as $f$ is continuous at $x=\dfrac{1}{n}$. Am I right? I doubt. Please help!
Answer
$$\log\left[\left(\dfrac{f\left(x+\dfrac{1}{n}\right)}{f(x)}\right)^n
\right]=
n\log \dfrac{f(x) + \frac 1n f'(x)+\epsilon(\frac 1n)\frac 1n}{f(x) } =
n\log \left[1+ \frac 1n\dfrac{ f'(x)}{f(x)}+\epsilon\left(\frac 1n\right)\frac 1n\right]
$$
with $\lim_0\epsilon = 0$.
$$\sim n \frac 1n\dfrac{ f'(x)}{f(x)} = \dfrac{ f'(x)}{f(x)}
$$so the limit is $$\exp\dfrac{ f'(x)}{f(x)}$$
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