Question Number 192115 by mnjuly1970 last updated on 08/May/23
$$ \\ $$$$\:\Omega\:=\:\mathrm{lim}_{\:{x}\rightarrow\mathrm{0}} \:\left(\:\:\frac{\:\mathrm{cot}^{\:−\mathrm{1}} \:\left(\frac{\mathrm{1}}{{x}}\:\right)}{\:{x}}\:\right)^{\frac{\mathrm{1}}{{x}^{\:\mathrm{2}} }} =\:?\:\:\:\:\:\: \\ $$$$\:\: \\ $$$$ \\ $$
Answered by mehdee42 last updated on 08/May/23
$${tip}\:\mathrm{1}:\:{cot}^{−\mathrm{1}} \frac{\mathrm{1}}{{x}}={tan}^{−\mathrm{1}} {x}\:\:\: \\ $$$${tip}\mathrm{2}\::\:{x}\rightarrow\mathrm{0}\Rightarrow{tan}^{−\mathrm{1}} {x}\sim{x}\:\:\&\:{x}\rightarrow\mathrm{0}\:;\:\:{tanx}−{x}\:\sim\:−\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} \:\: \\ $$$${tip}\mathrm{3}\::\:{if}\:\:;\:{f}\rightarrow\mathrm{1}\:\:,\:{g}\rightarrow\infty\:\Rightarrow{limf}^{{g}} ={e}^{{lim}\:\left({f}−\mathrm{1}\right){g}} \\ $$$${SO} \\ $$$$\Omega={lim}_{{x}\rightarrow\mathrm{0}} \:\left(\frac{{tan}^{−\mathrm{1}} {x}}{{x}}\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} ={e}^{{lim}_{{x}\rightarrow\mathrm{0}} \:\left(\:\frac{{tan}^{−\mathrm{1}} {x}}{{x}}−\mathrm{1}\right)\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \\ $$$$={e}^{{lim}_{{x}\rightarrow\mathrm{0}} \:\left(\:\frac{{tan}^{−\mathrm{1}} {x}−\mathrm{1}}{{x}}\right)\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} ={e}^{{lim}_{{x}\rightarrow\mathrm{0}} \left(\frac{−{x}^{\mathrm{3}} }{\mathrm{3}{x}^{\mathrm{3}} }\right)} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{e}}}\:\checkmark \\ $$
Commented by senestro last updated on 08/May/23
$${awsome}. \\ $$
Commented by mnjuly1970 last updated on 08/May/23
$$\:\:{so}\:{nice}\:{sir}\:\:\:\underline{\underbrace{\lesseqgtr}} \\ $$