I wanted to know whether the following statement is true for all functions $f:A\to (0, \infty)$ where $A\subseteq \mathbb R$ and $a\in A$
$$\lim_{x\to a}{f(x)^{m/n}}=(\lim_{x\to a}{f(x)})^{m/n}\ ;\ m,n\in \mathbb N$$
My attempt in proving this..
Let $\lim_{x\to a}{f(x)}=L$. Since $f(x)>0 \ \forall\ x \in A$, we have $L>0$ and $\lim_{x\to a}{f(x)^{1/n}}>0\ \ \ $...(This can be proven..)
We know that
$$f(x) = \underbrace{{f(x)^{1/n}}\cdot {f(x)^{1/n}}\cdots {f(x)^{1/n}}}_{n\text{ times}}$$
Therefore, we have (using the "product rule" for limits)
$$\lim_{x\to a}{f(x)}=(\lim_{x\to a}{f(x)^{1/n}})^n \cdots (*)$$
After this, we can take the $(1/n)$th power on both sides and then we get
$$\lim_{x\to a}{f(x)^{1/n}}=L^{1/n}$$
Then..applying the "product rule" for limits $m$ times..we get the final result..
$$\lim_{x\to a}{f(x)^{m/n}}=L^{m/n}$$
First of all..I would like to know whether my proof or even my statement is correct.. and if it is not, I would like to know the correct statement/proof.
I have a few questions here:
1: Now, in the step that I have labelled (*), the "product rule" for limits can be used only when $\lim_{x\to a}{f(x)^{1/n}}$ exists..which I have implicitly assumed in my proof..and this may or may not be the case...is there any way to fix this?
2: If my proof is correct, then can it be generalised in any way? Because I know that $\lim_{x\to -1}{x^{1/3}}=-1$..but I cannot use the above statement to prove this (as the range of $x^{1/3}$ is $\mathbb R$)
3: Is there a more general version of this statement? If yes, how do we prove it?
Thanks for any answers!!