Why $\lim_{u\to 0}\ln\left((1+u)^\frac{1}{u}\right)=\ln\left(\lim_{u\to 0}(1+u)^\frac{1}{u}\right)$? I came across this issue while analysing $\frac{d}{dx}\ln(x)$ and need a little help understanding the logic:
$\displaystyle\lim_{Δx\to0}\frac{\ln(x+Δx)-\ln(x)}{Δx}$ 
$\displaystyle\lim_{Δx\to0}\frac{\ln\left(\frac{x+Δx}{x}\right)}{Δx}$ (logarithm properties)
$\displaystyle\lim_{Δx\to0}\frac{1}{Δx}·\ln\left(1+\frac{Δx}{x}\right)$ 
$\displaystyle\lim_{Δx\to0}\ln\left(\left(1+\frac{Δx}{x}\right)^\frac{1}{Δx}\right)$, 
at this point, to clear things up, the substitution $u = \frac{Δx}{x}$ is made, (hence, $ux = Δx$) so:
$\displaystyle\lim_{u\to0}\ln\left(\left((1+u)^\frac{1}{u}\right)^\frac{1}{x}\right)$
$\displaystyle\frac{1}{x}·\lim_{u\to0}\ln\left((1+u)^\frac{1}{u}\right)$
$\displaystyle\frac{1}{x}·\ln\left(\lim_{u\to0}(1+u)^\frac{1}{u}\right)$
$\displaystyle\frac{1}{x}·\ln(e) = \frac{1}{x}$
The question on the table is, I know according to limit properties that $\lim_{x\to a}c\, f(x)$ = $c\lim_{x\to a}f(x)$, but I don't understand why $\lim_{u\to 0}\ln\left((1+u)^\frac{1}{u}\right)$ = $\ln\left(\lim_{x\to a}(1+u)^\frac{1}{u}\right)$. I've been told its because the function is continuous, but seeing as I'm self taught and never actually sat through a Calc I class, that didn't mean much to me. Any help is appreciated.
 A: The definition of $f$ being a continuous function at a real number $c$ is
$$\lim_{y \to c} f(y) = f(c).$$
[Note that there are other equivalent ways to define continuity.]
In particular if $\lim_{x \to a} g(x)=c$, then
$$\lim_{x \to a} f(g(x)) = f(c).$$
So if the limit $\lim_{x \to a} (1+x)^{1/x} = c$ exists, then
$$\lim_{x \to a} \ln((1+x)^{1/x}) = \ln(c).$$
A: When there is a discontinuity, the limit and the function do not commute:
$$f\left(\lim_{x\to a}x\right)\ne\lim_{x\to a}f(x).$$
For example, let $$f(x)=\begin{cases}x<0\to0,\\x\ge0\to1.\end{cases}.$$
Then
$$f\left(\lim_{x\to0}x\right)=f(0)=1$$ while
$$\lim_{x\to0}f(x)$$ does not exist (left and right limits differ).
On the opposite, as can be show by an $\epsilon,\delta$ argument, a continuous function and a limit do commute.
A: The reason that
$$
\lim_{u\to0}\log\left((1+u)^{1/u}\right)=\log\left(\lim_{u\to0}(1+u)^{1/u}\right)\tag{1}
$$
is that
$$
\lim_{u\to0}(1+u)^{1/u}=e\tag{2}
$$
and $\log(x)$ is continuous at $x=e$. That is, because of $(2)$, $(1)$ is the same as
$$
\lim_{x\to e}\log\left(x\right)=\log\left(\lim_{x\to e}x\right)\tag{3}
$$
and $(3)$ is true because it is essentially the definition of continuity.
