Regarding limits: $\lim\limits_{n\to \infty}\left(\frac{f\left(x+\frac{1}{n}\right)}{f(x)}\right)^n$ 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!
 A: Hint: take logarithms, and compare what you get to the difference-quotient definition of the derivative. 
A: If $f$ is differentiable, we can write $f(x+{1 \over n}) = f(x) + {1 \over n} f'(x) + r({1 \over n})$, where $\lim_n {r({1 \over n}) \over {1 \over n}} = 0$.
In particular, for any $\epsilon>0$ we can find some $N$ such that if $n \ge N$ then
$-\epsilon < {r({1 \over n}) \over {1 \over n}} < \epsilon$, and so
 $ {1 \over n} (f'(x) - \epsilon) < {1 \over n} (f'(x) + {r({1 \over n}) \over {1 \over n}}) <{1 \over n} (f'(x) + \epsilon) $.
Hence we have $\left( 1+ {1 \over n}({ f'(x) - \epsilon \over f(x)}) \right)^n < \left( { f(x+{1 \over n}) \over f(x)}  \right)^n \le \left( 1+ {1 \over n}({ f'(x) + \epsilon \over f(x)})  \right)^n$, and taking limits gives
$e^{{ f'(x) - \epsilon \over f(x)}} \le \liminf_n \left( { f(x+{1 \over n}) \over f(x)}  \right)^n  \le \limsup_n \left( { f(x+{1 \over n}) \over f(x)}  \right)^n \le e^{{ f'(x) + \epsilon \over f(x)}  } $. Letting $\epsilon \downarrow 0$ yields
$\lim_n \left( { f(x+{1 \over n}) \over f(x)}  \right)^n = e^{{ f'(x)  \over f(x)}}$.
A: $$\log\left[\left(\dfrac{f\left(x+\dfrac{1}{n}\right)}{f(x)}\right)^n
\right]=
n\log \dfrac{f(x) + \frac 1n f'(x)+\epsilon(\frac 1n)\frac 1n}{f(x) } =
n\log \left[1+ \frac 1n\dfrac{ f'(x)}{f(x)}+\epsilon\left(\frac 1n\right)\frac 1n\right]
$$
with $\lim_0\epsilon = 0$.
$$\sim n \frac 1n\dfrac{ f'(x)}{f(x)} = \dfrac{ f'(x)}{f(x)}
$$so the limit is $$\exp\dfrac{ f'(x)}{f(x)}$$ 
