<s>[INST] 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.
 [/INST] Alternatively:
$$\lim_{x\to+\infty} x^2 \ln \left(\frac{x^2+1}{x^2+3}\right)=\\
\lim_{x\to+\infty} \ln \left(\frac{x^2+1}{x^2+3}\right)^{x^2}=\\
\ln \left[\lim_{x\to+\infty} \left[\left(1+\frac{-2}{x^2+3}\right)^{\frac{x^2+3}{-2}}\right]^{\frac{-2x^2}{x^2+3}}\right]=\\
\ln e^{\lim_\limits{x\to+\infty} \frac{-2x^2}{x^2+3}}=\\
\ln e^{-2}=-2.$$
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