Under what condition on $f$, does $\lim_{x\to1}\log(f(x))=\log(\lim_{x\to1}f(x))$ hold?

Question

My question may look like duplicate, but my query is different.
This question hits my mind when I am proving the limit $\lim_{x\to1}x^{\frac{1}{x-1}}=e$ in the following manner-
$\lim_{x\to1} \log x^{\frac{1}{x-1}}$
$=\lim_{x\to1}\frac{\log x}{x-1}$
$=\lim_{x\to1}\frac{1/x}{1}$ (by L'Hôpital's rule)
$=1$
So, we get $\lim_{x\to1} \log x^{\frac{1}{x-1}}=1$ from this can we write $\log(\lim_{x\to1} x^{\frac{1}{x-1}})=1$ i.e. can I write $\lim_{x\to1} \log x^{\frac{1}{x-1}}=\log(\lim_{x\to1} x^{\frac{1}{x-1}})$?
There is a theorem which says- If $f,\ g$ be two real valued function and $a\in\Bbb{R}$. $f$ is continuous at $a$ and $\lim_{x\to a}g(x)$ exists, then $\lim_{x\to a}f(g(x))=f(\lim_{x\to a} g(x))$
Here in my problem, $f(x)=\log x$ and $g(x)=x^{\frac{1}{x-1}}$. Now, I don't know anything about $\lim_{x\to 1} g(x)$ here (whether the limit exists or not), I have to prove it $\lim_{x\to 1} g(x)=e$
Hope everybody will understand my query. Can anybody clear my doubts analytically?
Thanks for assistance in advance.

Answer 1

Since $x>0$ we have that

$$x^{\frac{1}{x-1}}=e^{\frac{\log x}{x-1}}$$

and since exponential function is continuous we have that

$$\lim_{x\to1} x^{\frac{1}{x-1}}=e^{\lim_{x\to1}\frac{\log x}{x-1}}$$

therefore your way is completely correct.