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Question Number 192115 by mnjuly1970 last updated on 08/May/23

$$$$$$\mathrm{\Omega }={\lim }_{x\to 0}{\left(\frac{{\cot }^{-1}\left(\frac{1}{x}\right)}{x}\right)}^{\frac{1}{x^2}}=?$$$$$$$$$$

Answered by mehdee42 last updated on 08/May/23

$$tip1:{cot}^{-1}\frac{1}{x}={tan}^{-1}x$$$$tip2:x\to 0\Rightarrow {tan}^{-1}x\sim x\mathrm{\&}x\to 0;tanx-x\sim -\frac{1}{3}{x}^3$$$$tip3:if;f\to 1,g\to \mathrm{\infty }\Rightarrow {limf}^g=e^{lim(f-1)g}$$$$SO$$$$\mathrm{\Omega }={lim}_{x\to 0}{\left(\frac{{tan}^{-1}x}{x}\right)}^{\frac{1}{x^2}}=e^{{lim}_{x\to 0}(\frac{{tan}^{-1}x}{x}-1)\frac{1}{x^2}}$$$$=e^{{lim}_{x\to 0}\left(\frac{{tan}^{-1}x-1}{x}\right)\frac{1}{x^2}}=e^{{lim}_{x\to 0}\left(\frac{-x^3}{3x^3}\right)}$$$$=\frac{1}{\sqrt[3]{e}}✓$$

Commented by senestro last updated on 08/May/23

$$awsome.$$

Commented by mnjuly1970 last updated on 08/May/23

$$sonicesir\underset{―}{\underbrace{⋚}}$$

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