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.

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*}
