Evaluate $\lim_{ x\to \infty} x^2 \times \log \left(x \cot^{-1}x\right)$

Question

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?

Answer 1

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

$$\frac{\log\left(\frac{\tan^{-1}y}{y}\right)}{y^2}=\frac{\log\left(\frac{y-\frac{y^3}{3}+o(y^3)}{y}\right)}{y^2}=\frac{\log\left(1-\frac{y^2}{3}+o(y^2)\right)}{y^2}=\\=\frac{-\frac{y^2}{3}+o(y^2)}{y^2}=-\frac13+o(1)\to-\frac13$$

Answer 2

I think this change of variable might be more helpful: $$u=\cot^{-1}x\to x=\cot u$$ Then the limit would be: $$\lim_{u\to {0^+}}\frac{\ln\frac{u}{\tan u}}{\tan^{2}u}$$by using consecutive L'Hopital's rule we obtain: $$\lim_{u\to {0^+}}\frac{\ln\frac{u}{\tan u}}{\tan^{2}u}=\lim_{u\to\ 0^+}\frac{\tan u-u-u\tan^2 u}{2u\tan^2 u(1+\tan^2 u)}=\lim_{u\to\ 0^+}\frac{1+\tan^2 u-1-\tan^2 u-2u\tan u (1+\tan^2 u)}{2\tan^2 u(1+\tan^2 u)+4u\tan u(1+\tan^2 u)^2+4u\tan^3 u(1+\tan^2 u)}=\lim_{u\to\ 0^+}\frac{-u}{\tan u+2u(1+\tan^2 u)+2u\tan^2 u}=\lim_{u\to\ 0^+}\frac{-1}{\frac{\tan u}{u}+2+4\tan^2 u}=-{1\over 3}$$