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?
 A: Hint: Set $x^2=e^t-1$ to get  $\lim \limits_{x\to +\infty}\left(\dfrac{\log(x^2+1)}{x}\right)=\lim \limits_{t\to +\infty}\left(\dfrac{t}{\sqrt{e^t-1}}\right)$.
A: $$\lim_{x\to\infty}\frac{\ln(x^2+1)}x\sim\lim_{x\to\infty}\frac{\ln(x^2)}x=\lim_{x\to\infty}\frac{2\cdot\ln x}x=2\cdot\lim_{t\to\infty}\frac t{e^t}=2\cdot0=0.$$
A: 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 $\exp(y)$.
A: You may write directly, as  $\lim_{x \rightarrow \infty} \frac{\log (x)}{x^a}=0$, $a>0$
$$\frac{\log (1+ x^2)}{x} \cong \frac{\log (x^2)}{x} =\frac{2\log (x)}{x} \overset{ x \rightarrow \infty}{\longrightarrow} 0$$
A: See as the limit is of  (infinity / infinity ) form we can use L'hospital rule on this limit.....
we have to apply the derivative on the numerator and denominator repeatedly untill we have a fraction is of form other than  (infinity / infinity )......
 Then we apply the process for two time and we have a fraction as: 
                                                                                                            lim(x--->infinity)  (2) / (2x + 1) .
and now we put x--> infinity in the limit.....and we have the limiting value as ZERO or 0.
Hope this discussion will help you.....best of luck.
