Question Number 129211 by bramlexs22 last updated on 13/Jan/21
$$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{cos}\:\left(\sqrt{\mathrm{x}}\:\right)^{\frac{\mathrm{1}}{\mathrm{x}}} }\:=?\: \\ $$
Answered by liberty last updated on 13/Jan/21
![L = lim_(x→0) (1/(cos ((√x) )^(1/x) )) ln L = lim_(x→0) ln ((1/(cos ((√x))^(1/x) ))) ln L = lim_(x→0) −((ln cos ((√x)))/x) ; [ L′Hopital ] ln L = lim_(x→0) (1/(cos (√x))). ((sin (√x) )/(2(√x))) ln L = 1. (1/2) ⇔ L = e^(1/2) = (√e)](https://www.tinkutara.com/question/Q129220.png)
$$\:\mathcal{L}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{cos}\:\left(\sqrt{\mathrm{x}}\:\right)^{\frac{\mathrm{1}}{\mathrm{x}}} } \\ $$$$\:\mathrm{ln}\:\mathcal{L}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{ln}\:\left(\frac{\mathrm{1}}{\mathrm{cos}\:\left(\sqrt{\mathrm{x}}\right)^{\frac{\mathrm{1}}{\mathrm{x}}} }\right) \\ $$$$\:\mathrm{ln}\:\mathcal{L}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}−\frac{\mathrm{ln}\:\mathrm{cos}\:\left(\sqrt{\mathrm{x}}\right)}{\mathrm{x}}\:;\:\left[\:\mathrm{L}'\mathrm{Hopital}\:\right] \\ $$$$\:\mathrm{ln}\:\mathcal{L}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{cos}\:\sqrt{\mathrm{x}}}.\:\frac{\mathrm{sin}\:\sqrt{\mathrm{x}}\:}{\mathrm{2}\sqrt{\mathrm{x}}} \\ $$$$\underline{\:\boldsymbol{\mathrm{ln}}\:\boldsymbol{\mathcal{L}}\:=\:\mathrm{1}.\:\frac{\mathrm{1}}{\mathrm{2}}\:\Leftrightarrow\:\boldsymbol{\mathcal{L}}\:=\:\boldsymbol{\mathrm{e}}^{\frac{\mathrm{1}}{\mathrm{2}}} \:=\:\sqrt{\boldsymbol{\mathrm{e}}}\:}\: \\ $$
Answered by mnjuly1970 last updated on 14/Jan/21
![l=lim_(x→0^+ ) (1/([(1−sin^2 ((√x) ))^(1/x) ]^(1/2) )) =_(sin((√x) )≈(√x)) ^(x→0^+ ) lim_(x→0^+ ) ((1/((1−x)^(1/x) )))^(1/2) = (√e) ...✓✓](https://www.tinkutara.com/question/Q129264.png)
$$\:\:{l}={lim}_{{x}\rightarrow\mathrm{0}^{+} \:} \frac{\mathrm{1}}{\left[\left(\mathrm{1}−{sin}^{\mathrm{2}} \left(\sqrt{{x}}\:\right)\right)^{\frac{\mathrm{1}}{{x}}} \right]^{\frac{\mathrm{1}}{\mathrm{2}}} } \\ $$$$\:\:\:\underset{{sin}\left(\sqrt{{x}}\:\right)\approx\sqrt{{x}}} {\overset{{x}\rightarrow\mathrm{0}^{+} } {=}}\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \left(\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\frac{\mathrm{1}}{{x}}} }\right)^{\frac{\mathrm{1}}{\mathrm{2}}} =\:\sqrt{{e}}\:…\checkmark\checkmark \\ $$