Natural Log of a Limit Equals Limit of a Natural Log Say we have $\ln \lim\limits_{x \to \infty} f(x)$. This is equivalent--unless I'm mistaken, in which case this question should be removed--to $\lim\limits_{x \to \infty} \ln f(x)$. The best argument I've heard for why this is the case is that the natural log function is continuous, which isn't very convincing to me. 
If this is in fact true, why is this the case? 
 A: Let $L:=\lim_{x\rightarrow\infty}f(x)$ be real and strictly greater than zero. We are to prove that $\lim_{x\rightarrow\infty}\ln f(x)=\ln L$. Use the fact that $\ln$ is continuous on $(0,\infty)$, in particular, at $L$, then given that $\epsilon>0$, there is some $\delta>0$ such that if $u>0$ and $|u-L|<\delta$, then $|\ln u-\ln L|<\epsilon$. Now there is some $M>0$ such that $|f(x)-L|<\delta$ for all $x\geq M$. We also assume that for large $x$, $f(x)>0$. For such an $x$, we have $|\ln f(x)-\ln L|<\epsilon$, this completes the proof.
A: $\lim f (x)=k $ means that for any $\epsilon >0$ there is an $M $ so that $x>M $ implies $|f (x)-k|< \epsilon $
Which means for any $x,y>M $ we know $|f (x)-f (y)|\le |f (x)-k|+|f (y)-k|<\epsilon$.
So $1-\epsilon <\frac {1-\epsilon}{1+\epsilon} =1-\frac {\epsilon}{1-\epsilon}<\frac {f (x)}k<\frac {1+\epsilon}{1-\epsilon}= 1+\frac {\epsilon}{1+\epsilon} <1+\epsilon $.
So $\ln (1-\epsilon)<\ln\frac {f (x)}{k}=\ln f (x)-\ln k<\ln (1+\epsilon) $
So $|\ln f (x)-\ln k|< \ln 1+\epsilon $.  [Note:$\ln 1\pm \epsilon $ are a negative and positive number very close to $0$ and $\ln 1+\epsilon <|\ln 1-\epsilon|$]
So for an $\delta >0$,  let $\epsilon =e^{\delta}-1$.  Let $M $ be the value that makes $x,>M $ have to be that $|f (x)-k|<\epsilon $.
Then if $x>M $ we have $|\ln f (x)-\ln k|<\delta $.
So $\lim \ln f(x)=\ln \lim f(x) $.
This assumed $k\ne 0,\infty $.  I'll leave those cases to you.
