Solve $\lim_{x\rightarrow +\infty}\frac{1}{x}\log\left(\frac{x+1}{1+x^2}\right) = 0$ without L'Hôpital's rule How would you solve the limit
$$\lim_{x\rightarrow +\infty}\frac{1}{x}\log\left(\frac{x+1}{1+x^2}\right) = 0$$
without using L'Hôpital's rule?
 A: $$\lim_{x\to \infty }\frac{1}{x}\ln\left(\frac{x+1}{1+x^2}\right)=\lim_{u\to 0}u\ln\left(\frac{u^2+u}{u^2+1}\right).$$
Now,
$$\frac{1}{1+u^2}=1+o(1)\implies \frac{u^2+u}{u^2+1}=u+o(u).$$
Therefore
\begin{align}
u\ln\left(\frac{u^2+u}{u^2+1}\right)&=u\ln(u)+u\ln(1+o(u))\\
&=u\ln(u)+uo(1)\\
&=u\ln(u)+o(u)\underset{u\to 0}{\longrightarrow }0.
\end{align}
A: $$\frac{1}{x}\log\left(\frac{x+1}{1+x^2}\right)=\frac{1}{x} \log \frac{1}{x}+\frac{\log \left(1+\frac{1}{x}  \right)}{\frac{1}{x}\cdot x^2}-\frac{\log \left(1+\frac{1}{x^2}  \right)}{\frac{1}{x^2}\cdot x^3}$$
A: Since $\log (1+t) \leq t$ for all $t \geq 0$ we get $\frac  1 x \log (1+x)=\frac  2 x \log \sqrt {1+x} \leq \frac  2 x (\sqrt {1+x}-1) \to 0$.  Similarly, $\frac  1 x \log (1+x^{2}) \leq \frac  3 x ( ({1+x^{2}})^{1/3}-1) \to 0$. So  difference between these  two tends to $0$
A: put $x=\frac{1}{t}$ or the limit becomes
$\lim_{t\rightarrow 0}{t\log(1+t)+t\log t-t\log(1+t^2)}$
=$0$
A: \begin{align}
\lim_{x\to+\infty}\frac{1}{x}\log\left(\frac{x+1}{1+x^2}\right) 
  &= \lim_{x\to+\infty}\left(\frac{1}{x}\cdot\frac{1+x^2}{x+1}\right)\left[\frac{x+1}{1+x^2}\log\left(\frac{x+1}{1+x^2}\right)\right]=\\
  &= \lim_{x\to+\infty}\frac{1+x^2}{x^2+x}\cdot\lim_{y\to0}y\log y=1\cdot0=0\\
\end{align}
