How to find the limit $\lim \limits_ {x \to+\infty} \left [ \frac{4 \ln(x+1)}{x}\right]$ 
Solve $\space \begin{align*} \lim_ {x \to+\infty} \left [ \frac{4 \ln(x+1)}{x}\right]  
\end{align*}$.

I did this way:
$$\begin{align*}
\lim_ {x \to+\infty} \left [ \frac{4 \ln(x+1)}{x}\right] & = 4\lim_ {x \to+\infty} \left [\frac{1}{x} \ln(x+1) \right]= \\\\=4\lim_ {x \to+\infty} \left [ \ln(x+1)^{\frac{1}{x}}\right] &= 4 \ln \left[\lim_ {x \to+\infty}(x+1)^{\frac{1}{x}}\right] =4 \cdot \ln(1)=0
\end{align*}$$
What is the rule behind the shift that I made between the $\ln$ and the $limit$?
I'm in the high school.Thanks
 A: As has been mentioned, since $\ln(x)$ is continuous at $x=1$,
$$
\lim_{x\to1}\ln(x)=\ln\left(\lim_{x\to1}x\right)=\ln(1)=0\tag{1}
$$
However, how do you know that
$$
\lim_{x \to+\infty}(x+1)^{\frac{1}{x}}=1\tag{2}
$$
Usually the way one shows $(2)$ is by proving the limit you started with; that is,
$$
\lim_{x\to\infty}\frac{\ln(1+x)}{x}\tag{3}
$$
Let $x=e^t-1$, then $(3)$ becomes
$$
\lim_{t\to\infty}\frac{t}{e^t-1}\tag{4}
$$
Let $f(t)=\dfrac{t}{e^t-1}$. Then
$$
\begin{align}
\lim_{t\to\infty}\frac{f(t+1)}{f(t)}
&=\lim_{t\to\infty}\frac{t+1}{t}\frac{e^t-1}{e^{t+1}-1}\\
&=\lim_{t\to\infty}\left(1+\frac1t\right)\left(\frac{1-e^{-t}}{e-e^{-t}}\right)\\
&=\frac1e\tag{5}
\end{align}
$$
Limit $(5)$ says that for some $T$, when $t\ge T$, $\dfrac{f(t+1)}{f(t)}<\dfrac12$. Therefore, since $f$ is continuous on $[T,T+1]$, there is some $C$ so that
$$
f(t)<C2^{-t}\tag{6}
$$
Thus,
$$
0\le\lim_{x\to\infty}\frac{\ln(1+x)}{x}=\lim_{t\to\infty}\frac{t}{e^t-1}\le\lim_{t\to\infty}C2^{-t}=0\tag{7}
$$
and therefore, by the Sandwich Theorem, $\lim\limits_{x\to\infty}\frac{\ln(1+x)}{x}=0$.
A: It is that $\ln$ is a continuous function.
However, I have a problem with your last step.  You've got the indeterminate form $\infty^0$.  The limit is such cases is not always $1$.  One of the simplest cases where that fails is
$$
\lim_{x\to\infty} \left( (x+1)^{1/\ln(x+1)} \right) = e.
$$
A: I would call it the "chain rule" or just simply "continuity," specifically because
$$\lim_{x\to a} f\big(g(x)\big) =f\left(\lim_{x\to a} \,g(x)\right) $$
whenever the limits exist and $f$ is continuous in a neighborhood of $\lim_{x\to a}g(x)$.
A: Just use l'Hopital's Rule to show that the limit is equal to the limit of $$\frac{4}{x+1},$$ which is simply zero.
