Limit of a differentiable function and mean value theorem?

Question

Let $f:\mathbb{R}\to \mathbb{R}$ and $a\in \mathbb{R}$. We assume that $f(a)>0$ and that $f$ is differentiable at $a$. I want to calculate the limit $$ \lim_{n\to \infty}\left(\dfrac{f\left(a+\frac{1}{n}\right)}{f(a)}\right)^{1/n}. $$

My attempt at a solution:

I Taylor expand $f$ around $a$ to get $$ f\left(a + \frac{1}{n}\right)=f(a) + f^\prime(a)\frac{1}{n}+ \mathcal{O}\left(\dfrac{1}{n^2}\right). $$ Then $$ \dfrac{f\left(a + \dfrac{1}{n}\right)}{f(a)}=1 + \dfrac{f^\prime(a)}{f(a)}\dfrac{1}{n} + \mathcal{O}\left(\dfrac{1}{n^2}\right). $$ Now as $n\to \infty$ the above expression tends to 1. On the other hand as $n\to \infty$, $\frac{1}{n}$ tends to zero. So by continuity of the function $1^x$, I would argue that the required expression tends to 1.

Coming from a physics background I would like to know how sloppy the above argument is, if correct at all.

Answer 1

$({{f(a+{1\over n})}\over{f(a)}})^{1\over n}$

$=e^{{ln(f(a+{1\over n})-ln(f(a))}\over n}$ and

$lim_{x\rightarrow 0}{{ln(f(a+x))-ln(f(a))}\over x}$ is the derivative of $ln(f(x))$ at $a$.

So $lim_{n\rightarrow +\infty}e^{{ln(f(a+{1\over n})-ln(f(a))}\over n}=lim_{n\rightarrow +\infty}e^{{1\over n^2}{{ln(f(a+{1\over n})-ln(f(a)}\over{1\over n}}}=lim_{n\rightarrow +\infty}e^{{1\over n^2}{{f'(a)}\over{f(a)}}}=1$.

Answer 2

Your approach is correct/rigorous, but more than what is needed. Just mention that $f$ is continuous at $a$ so that $f(a+1/n)\to f(a) $ and thus the ratio $f(a+1/n)/f(a)\to 1$ and the expression tends to $1^{0}=1$.