Simple limit of function How do I show that $\lim_{x \rightarrow \infty } \frac {\log(x^{2}+1)}{x}=0$?I was able to do that using L'Hôpital's rule. But is there any other way?

You may write directly, as  $\lim_{x \rightarrow \infty} \frac{\log (x)}{x^a}=0$, $a>0$
$$\frac{\log (1+ x^2)}{x} \cong \frac{\log (x^2)}{x} =\frac{2\log (x)}{x} \overset{ x \rightarrow \infty}{\longrightarrow} 0$$
