Another way to compute $\lim\limits_{x\to+\infty}x^2\log\left(\frac{x^2+1}{x^2+3}\right)$ I need to compute as a title a limit with $x\to+\infty$. The only way I found to obtain a result is to use the L'Hôpital's rule:
$$\lim\limits_{x\to+\infty}x^2\log\bigg(\frac{x^2+1}{x^2+3}\bigg)=\lim\limits_{t\to 0}\frac{1}{t^2}\log\bigg(\frac{1+t^2}{1+3t^2}\bigg)\stackrel{H}{=}\lim\limits_{t\to 0}\frac{1}{2t}\bigg(\frac{2t}{1+t^2}-\frac{6t}{1+3t^2}\bigg)=\lim\limits_{t\to 0}\bigg(\frac{1}{1+t^2}-\frac{3}{1+3t^2}\bigg)=-2$$
It seems the result is the difference between the two functions inside the log, but how this can be possible? Is there another way to compute these type of limits? I mean a calculus way without using theorems.
 A: You can write 
$$x^2 \ln \left( \frac{x^2+1}{x^2+3} \right) = x^2 \ln \left( 1- \frac{2}{x^2+3} \right) \sim x^2 \times \frac{-2}{x^2+3}$$
So the limit is $-2$.
A: If it is OK for you to calculate directly a derivative then you may proceed as follows:


*

*Set $x^2 = \frac{1}{t}$ and consider $t\to 0^+$
So, you get
\begin{eqnarray*}x^2\log\bigg(\dfrac{x^2+1}{x^2+3}\bigg)
& \stackrel{x^2 = \frac{1}{t}}{=} & \frac{\log \frac{1+t}{1+3t}}{t}\\
& = &\frac{\log (1+t) - 0}{t} - \frac{\log (1+3t) - 0}{t} \\
& \stackrel{t \to 0^+}{\longrightarrow} & \left. \frac{d}{dt}\log(1+t) \right|_{t=0} - \left. \frac{d}{dt}\log(1+3t) \right|_{t=0} \\
& = & \frac{1}{1+0} - \frac{3}{1+3 \cdot 0} = \boxed{-2}
\end{eqnarray*}
Another way is using


*

*$(1-y)^{\frac{1}{y}}\stackrel{y \to 0}{\longrightarrow} e^{-1}$
So, 
\begin{eqnarray*}x^2\log\bigg(\dfrac{x^2+1}{x^2+3}\bigg)
& \stackrel{x^2 = \frac{1}{t}}{=} & \frac{\log \frac{1+t}{1+3t}}{t}\\
& = & \log \left(1 - \frac{2t}{1+3t}\right)^{\frac{1}{t}} \\
& = & \log \left( \left(1 - \frac{2t}{1+3t}\right)^{\frac{1+3t}{2t}}\right)^{\frac{2t}{1+3t}\cdot \frac{1}{t}} \\
& \stackrel{t \to 0^+}{\longrightarrow} & \log \left(e^{-1}\right)^2  = \boxed{-2}\\
\end{eqnarray*}
A: Hint
Use $$\lim_{h\to0}\dfrac{\ln(1+h)}h=1$$
for $h=t^2,3t^2$ to get the limit $=1-3$
A: Alternatively:
$$\lim_{x\to+\infty} x^2 \ln \left(\frac{x^2+1}{x^2+3}\right)=\\
\lim_{x\to+\infty} \ln \left(\frac{x^2+1}{x^2+3}\right)^{x^2}=\\
\ln \left[\lim_{x\to+\infty} \left[\left(1+\frac{-2}{x^2+3}\right)^{\frac{x^2+3}{-2}}\right]^{\frac{-2x^2}{x^2+3}}\right]=\\
\ln e^{\lim_\limits{x\to+\infty} \frac{-2x^2}{x^2+3}}=\\
\ln e^{-2}=-2.$$
A: \begin{align}\lim_{x\to\infty}x^2\log\left(\frac{x^2+1}{x^2+3}\right)&=\lim_{x\to\infty}x\log\left(\frac{x+1}{x+3}\right)\\&=\lim_{x\to0}\frac{\log\left(\frac{x+1}{3x+1}\right)}x.\end{align}So, compute $f'(0)$, where $f(x)=\log\left(\frac{x+1}{3x+1}\right)$.
A: Let $z:=x^2+3$ and
$$\lim_{z\to\infty}\left((z-3)\log\frac{z-2}{z}\right)=\lim_{z\to\infty}\left(z\log\left(1-\frac2{z}\right)\right)
=\log\left(\lim_{z\to\infty}\left(1-\frac2{z}\right)^z\right)
\\=\log e^{-2}.$$

The term $-3$ can be ignored because the $\log$ vanishes.
A: Write
$$x^2\log\left(x^2+1\over x^2+3\right)=x^2\int_{x^2+3}^{x^2+1}{dt\over t}=-x^2\int_{x^2+1}^{x^2+3}{dt\over t}$$
and note that, since $1/t$ is strictly decreasing for $t\gt0$, we have
$${(x^2+3)-(x^2+1)\over x^2+3}\lt \int_{x^2+1}^{x^2+3}{dt\over t}\lt {(x^2+3)-(x^2+1)\over x^2+1}$$
Simplifying the numerators and remembering to reverse the directions of inequalities when multiplying by a negative quantity (namely $-x^2$), we have
$${-2x^2\over x^2+3}\gt-x^2\int_{x^2+1}^{x^2+3}{dt\over t}\gt{-2x^2\over x^2+1}$$
and the Squeeze Theorem does the rest.
A: $$x^{2}\ln\left(\frac{x^{2}+1}{x^{2}+3}\right)$$$$=\ln\left(\lim\limits_{x\to+\infty}\left(1+\left(\frac{x^{2}+1}{x^{2}+3}-1\right)\right)^{x^{2}}\right)$$$$=\ln\exp\left(-2\lim\limits_{x\to+\infty}\frac{x^{2}}{x^{2}+3}\right)=-2$$
