# 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?

• Can you use that $\lim_{x\to\infty}(\log x)/x=0$? Aug 21, 2020 at 12:41

$$\lim_{x\to \infty }\frac{1}{x}\ln\left(\frac{x+1}{1+x^2}\right)=\lim_{u\to 0}u\ln\left(\frac{u^2+u}{u^2+1}\right).$$

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}

$$\frac{1}{x}\log\left(\frac{x+1}{1+x^2}\right)=\frac{1}{x} \log \frac{1}{x}+\frac{\log \left(1+\frac{1}{x} \right)}{\frac{1}{x}\cdot x^2}-\frac{\log \left(1+\frac{1}{x^2} \right)}{\frac{1}{x^2}\cdot x^3}$$

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 $$0$$

put $$x=\frac{1}{t}$$ or the limit becomes

$$\lim_{t\rightarrow 0}{t\log(1+t)+t\log t-t\log(1+t^2)}$$

=$$0$$

\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}