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

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