$$\lim_{x \to \infty} x^2\big(\ln(x\cot^{-1}(x))$$
I tried using the Series Expansion of the $\ln(x)$ but then got stuck in between. I also tried using the L'Hopital but the expression got quickly messy.
After applying L'Hopital for the first time, I got
$$\lim_{x \to \infty} \frac{-x^3}{2}\bigg(\frac{-x}{1+x^2} + \cot^{-1}x\bigg)$$
The expression is still in the undefined form. Unless the question maker wants to torture the problem solver, this method would not be the way to go.
I have got no other clue for solving this problem.
Any help would be appreciated.
Best Answer
Note that as $x\to +\infty$, $$\cot^{-1}(x)=\arctan(1/x)=\frac{1}{x}-\frac{1}{3x^3}+o(1/x^3).$$ Hence, from your work, $$\frac{-x^3}{2}\bigg(\frac{-x}{1+x^2} + \cot^{-1}(x)\bigg)= \frac{-x^3}{2}\bigg(-\frac{1}{x}\frac{1}{1+\frac{1}{x^2}} + \frac{1}{x}-\frac{1}{3x^3}+o(1/x^3)\bigg)\\ =\frac{-x^3}{2}\bigg(-\frac{1}{x}+\frac{1}{x^3} + \frac{1}{x}-\frac{1}{3x^3}+o(1/x^3)\bigg)\to -\frac{1}{3}.$$