Find $\space\ \begin{align*} \lim_ {x \to+\infty} \left [ \frac{\log_{2}(x-1)}{x}\right] \end{align*}$.
After some minutes around this limit I did it this way:
$\log_{2}(x-1)=y \Leftrightarrow 2^y=x-1$
So,$\space x=2^y+1$.
When $x \to +\infty$,$\space y \to +\infty$ also. By substitution:
$\begin{align*} \lim_ {y \to+\infty} \left [ \frac{\log_{2}(2^y+1-1)}{2^y+1}\right]=\lim_ {y \to+\infty} \left [ \frac{\log_{2}(2^y)}{2^y+1}\right]=\end{align*}$
$\begin{align*}\lim_ {y \to+\infty} \left [ \frac{y}{2^y+1}\right]=\lim_ {y \to+\infty} \left [ \frac{1}{\frac{2^y+1}{y}} \right]=\lim_ {y \to+\infty} \left [ \frac{1}{\frac{2^y}{y}+\frac{1}{y}}\right]= \frac{1}{+\infty+0}=0 \end{align*}$
Is this correct?Are there any other easy way to find this limit?Thanks