Question Number 139949 by mnjuly1970 last updated on 02/May/21
$$\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:………{nice}\:….\:\ast\:….\ast\:….\:\ast\:….{calculus}\left({I}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\:\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \left(\sqrt[{\mathrm{3}}]{{cos}\left(\sqrt{{x}}\:\right.}\:\right)^{{cot}\left({x}\right)} =?? \\ $$$$\:\:\:\:\:\:\:\:{solution}……… \\ $$$$\:\:\:\:\:\:\:{sin}\left({x}\right)\:\approx\:{x}\:\:\:\:\:\:\:\:\left({x}\:\rightarrow\:\mathrm{0}\:\right) \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\phi}:=\:\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \left\{\left({cos}\left(\sqrt{{x}}\:\right)\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \right\}^{{cot}\left({x}\right)} ={lim}_{{x}\rightarrow\mathrm{0}^{+} } \left\{\left(\mathrm{1}−\frac{{x}}{\mathrm{2}}\right)^{{cot}\left({x}\right)} \right\}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\::={lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\left\{\left(\mathrm{1}−\frac{{x}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{{sin}\left({x}\right)\approx{x}}} \right\}^{\frac{{cos}\left({x}\right)}{\mathrm{3}}} \underset{{cos}\left({x}\right)\rightarrow\mathrm{1}\:\:\left({x}\:\rightarrow\mathrm{0}\right)} {\overset{\langle\:{lim}_{{x}\rightarrow\mathrm{0}} \left(\mathrm{1}−{ax}\right)^{\frac{\mathrm{1}}{{x}}} =\frac{\mathrm{1}}{{e}^{\:{a}} }\:\:\rangle} {=}}\:{e}^{\frac{−\mathrm{1}}{\mathrm{6}}} =\:\sqrt[{\mathrm{6}}]{\frac{\mathrm{1}}{{e}}\:}\:\:\:…. \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\: \\ $$