# 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?

• Why is the continuity of the logarithm not convincing to you? If $\lim_{x \to \infty} f(x) \in \mathbb{R}^+$, then this is nothing more than the statement that $\ln$ is sequentially continuous.
– user296602
Feb 13 '18 at 4:40
• My apologies, but I'm too familiar with the term "sequentially continuous." Could you explain?
– user465188
Feb 13 '18 at 4:43
• en.wikipedia.org/wiki/…
– user296602
Feb 13 '18 at 4:44
• If $g$ is continuous then $g (\lim f (x)) = \lim g (f(x))$ in a dead certain lead cinch to me. Have you ever read the proof of that. Feb 13 '18 at 6:06
• The definition of continuity says that a function is continuous if one can swap the functional operation with limit operation on in symbols $$\lim_{x\to a} f(x) =f(\lim_{x\to a} x)$$ and this can be generalized further (for example in current context). Feb 13 '18 at 8:38

## 2 Answers

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.

$\lim f (x)=k$ 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.