<s>[INST] Is this limit correct: $\lim_{x \to+\infty} \frac{\log_{2}(x-1)}{x} = 0$? 
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
 [/INST] Your way is fine, as an alternative we have that
$$ \frac{\log_{2}(x-1)}{x}= \frac{\log_{2}(x-1)}{x-1}\frac{x-1}{x}\to 0\cdot 1 =0$$
indeed eventually $\log_{2}(x-1) \le \sqrt{x-1}$ and then
$$\frac{\log_{2}(x-1)}{x-1} \le \frac{\sqrt{x-1}}{x-1} \to 0$$
</s>