<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] 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.
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