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?

$$\lim_{x\to\infty}\frac{\ln(x^2+1)}x\sim\lim_{x\to\infty}\frac{\ln(x^2)}x=\lim_{x\to\infty}\frac{2\cdot\ln x}x=2\cdot\lim_{t\to\infty}\frac t{e^t}=2\cdot0=0.$$
