TITLE: $\lim_{n\to ∞} \left[\frac{f\left( x +\frac1n\right)}{ f(x)}\right]^n$
QUESTION [5 upvotes]: Could anyone solve this problem for me?  
Let f be a positive differentiable function on the internal $\left[\,0,\infty\right)$.
$$\lim_{n\to ∞} \left[\frac{f\left( x +\frac1n\right)}{ f(x)}\right]^n$$
I have been told to take log, but after taking log what to do I am not able to understand.

REPLY [8 votes]: Yes, you can take the logarithm; since $f$ is assumed to be positive, we can consider $g(x)=\log f(x)$. Taking the logarithm of the sequence you want to compute the limit of, we are led to
$$
\lim_{n\to\infty}n(g(x+1/n)-g(x))=
\lim_{n\to\infty}\frac{g(x+1/n)-g(x)}{1/n}=g'(x)=\frac{f'(x)}{f(x)}
$$
because $f$ is differentiable, so $g$ is differentiable as well and we can apply the chain rule.
Therefore your limit is
$$
\lim_{n\to ∞}\biggl[\frac{f\left( x +\frac{1}{n}\right)}{f(x)}\biggr]^n=
\exp\left(\frac{f'(x)}{f(x)}\right)
$$