If f is positive and differentiable in (0,∞), then I want to find the following limit.
lim.
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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