<s>[INST] Evaluate $\lim_{ x\to \infty} x^2 \times \log \left(x \cot^{-1}x\right)$ Evaluate $$L=\lim_{ x\to \infty} x^2 \times \log \left(x  \cot^{-1}x\right)$$
My Try:we have by change of variable $y=\frac{1}{x}$
$$L=\lim_{ y\to 0}\frac{\log\left(\frac{\tan^{-1}y}{y}\right)}{y^2}=\lim_{ y\to 0}\frac{\log\left(\tan^{-1}y\right)-\log y}{y^2} $$
We have for $y$ very very small$$\tan^{-1}y=y-\frac{y^3}{6}$$
$$L=\lim_{ y\to 0}\frac{\log\left(y-\frac{y^3}{6}\right)-\log y}{y^2}$$ $\implies$
$$L=\lim_{ y\to 0}\frac{\log y+\log\left(1-\frac{y^2}{6}\right)-\log y}{y^2}$$ $\implies$
$$L=\lim_{ y\to 0}\frac{\log\left(1-\frac{y^2}{6}\right)}{y^2}$$ $\implies$
$$L=\frac{-1}{6}$$
Is this approach fine?
 [/INST] You need to fix the expansion for $\tan^{-1}y$ and be more rigorous using little-o notation
$$\tan^{-1}y=y-\frac{y^3}{3}+o(y^3)$$
then
$$\frac{\log\left(\frac{\tan^{-1}y}{y}\right)}{y^2}=\frac{\log\left(\frac{y-\frac{y^3}{3}+o(y^3)}{y}\right)}{y^2}=\frac{\log\left(1-\frac{y^2}{3}+o(y^2)\right)}{y^2}=\\=\frac{-\frac{y^2}{3}+o(y^2)}{y^2}=-\frac13+o(1)\to-\frac13$$
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