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 can always brute force it by $\log(x^2+1)\leq \log 2 + 2\log x$ (for $x\geq 1$), replace $y=\log x$ and use a Taylor expansion on $\exp(y)$.
