$\lim_{n\to ∞} \left[\frac{f\left( x +\frac1n\right)}{ f(x)}\right]^n$

Question

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.

Answer 1

Write $f(x+\frac{1}{n}) = f(x)+\frac{1}{n}f'(x)+O(\frac{1}{n^2})$, and you get that $$\lim \bigg(\frac{f(x+\frac{1}{n})}{f(x)}\bigg)^n = \lim\bigg(\frac{f(x)+\frac{1}{n}f'(x)}{f(x)}\bigg)^n = \lim \bigg(1+\frac{\frac{f'(x)}{f(x)}}{n}\bigg)^n = e^{\frac{f'(x)}{f(x)}}$$

Positivity of $f$ is needed to ensure that the denominator is non-vanishing.

Answer 2

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 $$ \lim_{n\to ∞}\biggl[\frac{f\left( x +\frac{1}{n}\right)}{f(x)}\biggr]^n= \exp\left(\frac{f'(x)}{f(x)}\right) $$