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$$\lim_{x \to \infty} x^2\big(\ln(x\cot^{-1}(x))$$

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.

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  • $\begingroup$ @KaviRamaMurthy I think that here $\cot^{-1}(x)$ means $\text{arccot}(x)=\arctan(1/x)$ which is positive for $x>0$. $\endgroup$ – Robert Z Aug 18 '19 at 12:44
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We have that $$\text{arccot}{(x)}=\arctan{\left(\frac1x\right)}=\frac1x-\frac1{3x^3}+o\left(\frac1{x^3}\right)$$ as $x\to\infty$. Also note that as $y\to0$ we have $$\ln{(1+y)}=y+o(y)$$ Hence our limit is simply $$\begin{align} \lim_{x\to\infty}x^2\ln{(x\text{arccot}{(x)})} &=\lim_{x\to\infty}x^2\ln{(x(1/x-1/(3x^3)+o(1/x^3)))}\\ &=\lim_{x\to\infty}x^2\ln{(1-1/(3x^2)+o(1/x^2))}\\ &=\lim_{x\to\infty}x^2(-1/(3x^2)+o(1/x^2))\\ &=\lim_{x\to\infty}(-1/3+o(1))\\ &=-\frac13\\ \end{align}$$

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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, $$\frac{-x^3}{2}\bigg(\frac{-x}{1+x^2} + \cot^{-1}(x)\bigg)= \frac{-x^3}{2}\bigg(-\frac{1}{x}\frac{1}{1+\frac{1}{x^2}} + \frac{1}{x}-\frac{1}{3x^3}+o(1/x^3)\bigg)\\ =\frac{-x^3}{2}\bigg(-\frac{1}{x}+\frac{1}{x^3} + \frac{1}{x}-\frac{1}{3x^3}+o(1/x^3)\bigg)\to -\frac{1}{3}.$$

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