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