computing a limit of a function that is positive defined

Question

given a function $f(x)$ positive and continuous at $x=a$, with $f(a)\ne0$, compute:

$$\lim_{n\to+\infty}\left[\frac{f\left(a+\frac{1}{n}\right)}{f(a)}\right]^n$$

a friend asked me that, though I could not answer in time, after a while I got a probable answer which I believe is correct.

the answer is correct?

Answer 1

Correct and there's more simple. By the Taylor series

$$f\left(a+\frac1n\right)\sim_\infty f(a)+\frac{f'(a)}{n}\iff\frac{f\left(a+\frac1n\right)}{f(a)}\sim_\infty1+ \frac{f'(a)}{f(a)}\frac1n$$ so $$\ln\left(\frac{f\left(a+\frac1n\right)}{f(a)}\right)\sim_\infty \frac{f'(a)}{f(a)}\frac1n\iff \left(\frac{f\left(a+\frac1n\right)}{f(a)}\right)^n\sim_\infty e^{\frac{f'(a)}{f(a)}}$$

Answer 2

we have that $\displaystyle\lim_{n\to+\infty}\left[\frac{f\left(a+\frac{1}{n}\right)}{f(a)}\right]^n$ is a limit of $1^{+\infty}$ since $\frac{f\left(a+\frac{1}{n}\right)}{f(a)}\mapsto\frac{f(a)}{f(a)}\mapsto1$ as $n\mapsto+\infty$, then using $a^b=e^{b\ln a}$

$$\begin{align} \lim_{n\to+\infty}\left[\frac{f\left(a+\frac{1}{n}\right)}{f(a)}\right]^n& =\lim_{n\to+\infty}\exp\left[n\ln\frac{f\left(a+\frac{1}{n}\right)}{f(a)}\right]\\ & =\exp\left[\lim_{n\to+\infty}n\ln\frac{f\left(a+\frac{1}{n}\right)}{f(a)}\right] &\left(\lim+\infty\cdot0\right)\\ & =\exp\left[\lim_{n\to+\infty}\frac{\ln\frac{f\left(a+\frac{1}{n}\right)}{f(a)}}{\frac{1}{n}}\right] & \left(\lim\frac{0}{0}\right)\\ & =\exp\left[\lim_{n\to+\infty}\frac{-\frac{1}{n^2}\frac{f'\left(a+\frac{1}{n}\right)}{f(a)}\frac{f(a)}{f\left(a+\frac{1}{n}\right)}}{-\frac{1}{n^2}}\right]\\ & =\exp\left[\lim_{n\to+\infty}\frac{f'\left(a+\frac{1}{n}\right)}{f\left(a+\frac{1}{n}\right)}\right]\\ & =\exp\left[\frac{f'(a)}{f(a)}\right]=e^{\frac{f'(a)}{f(a)}} \end{align}$$

Answer 3

$$\lim_{n\rightarrow\infty}\left[\frac{f\left(a+\frac{1}{n}\right)}{f\left(a\right)}\right]^{n}=\lim_{n\rightarrow\infty}e^{\left[\frac{g\left(a+\frac{1}{n}\right)-g\left(a\right)}{\frac{1}{n}}\right]}=e^{g'\left(a\right)}$$ for $g\left(x\right)=\ln f\left(x\right)$.

Here $g'\left(a\right)=\frac{f'\left(a\right)}{f\left(a\right)}$.

Answer 4

The problem did not specify that $f$ is differentiable at $a$. And if it is not, there can be trouble.

Let $f(x)=1+x\sin\left(\frac{\pi}{2x}\right)$ when $x\ne 0$, and let $f(0)=1$. Then $f$ is everywhere continuous.

However, our limit does not exist at $a=0$. For if $n$ is an even integer, then we are looking at $1^n$. But if $n$ is of the form $4k+1$, then we are looking at $\left(1+\frac{1}{n}\right)^n$.