Question Number 149958 by mathdanisur last updated on 08/Aug/21
$$\underset{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\sqrt[{\boldsymbol{\mathrm{x}}}]{\mathrm{cos}\sqrt{\mathrm{x}}}\:=\:? \\ $$
Commented by dumitrel last updated on 08/Aug/21
$${e}^{−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$
Commented by mathdanisur last updated on 08/Aug/21
$$\mathrm{Thankyou}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{but}\:\mathrm{how}\:\mathrm{please} \\ $$
Answered by john_santu last updated on 08/Aug/21
$$\:\mathrm{ln}\:{L}=\mathrm{ln}\left(\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{cos}\:\sqrt{{x}}\right)^{\frac{\mathrm{1}}{{x}}} \right) \\ $$$$\mathrm{ln}\:{L}\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{ln}\:\left(\mathrm{cos}\:\sqrt{{x}}\right)}{{x}} \\ $$$$\mathrm{ln}\:{L}\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{d}\left(\mathrm{cos}\:\sqrt{{x}}\right)}{\mathrm{cos}\:\sqrt{{x}}}=\:\frac{−\mathrm{sin}\:\sqrt{{x}}}{\mathrm{2}\sqrt{{x}}\:\mathrm{cos}\:\sqrt{{x}}}\: \\ $$$$\mathrm{ln}\:{L}=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}−\frac{\mathrm{1}}{\mathrm{2cos}\:{x}}\:=\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${L}={e}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:=\:\frac{\mathrm{1}}{\:\sqrt{{e}}}\: \\ $$
Commented by mathdanisur last updated on 08/Aug/21
$$\mathrm{Thankyou}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\:\infty\:\mathrm{or}\:\:\frac{\sqrt{\mathrm{e}}}{\mathrm{e}}\:.? \\ $$
Commented by dumitrel last updated on 08/Aug/21
$$\left({lncos}\sqrt{{x}}\right)'=\frac{\left({cos}\sqrt{{x}}\right)'}{{cos}\sqrt{{x}}}=\frac{−{sin}\sqrt{{x}}\centerdot\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}}{{cos}\sqrt{{x}}} \\ $$
Commented by mathdanisur last updated on 08/Aug/21
$$\mathrm{Thank}\:\mathrm{you}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$
Answered by gsk2684 last updated on 08/Aug/21
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\sqrt[{{x}}]{\mathrm{cos}\:\sqrt{{x}}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}\left(\mathrm{cos}\:\sqrt{{x}}\right)^{\frac{\mathrm{1}}{{x}}} } \\ $$$$={e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{{x}}\left(\mathrm{cos}\:\sqrt{{x}}\:−\mathrm{1}\right)} ={e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\sqrt{{x}}}{\mathrm{2}}\right)}{{x}}} \\ $$$$={e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\sqrt{{x}}}{\mathrm{2}}\right)}{\:\left(\sqrt{{x}}\right)^{\mathrm{2}} }} ={e}^{−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} } ={e}^{−\frac{\mathrm{1}}{\mathrm{2}}} =\frac{\mathrm{1}}{\:\sqrt{{e}}} \\ $$
Commented by mathdanisur last updated on 08/Aug/21
$$\mathrm{Thank}\:\mathrm{you}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$
Answered by mathmax by abdo last updated on 08/Aug/21
$$\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{cos}\sqrt{\mathrm{x}}\right)^{\frac{\mathrm{1}}{\mathrm{x}}} \:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{x}}\mathrm{log}\left(\mathrm{cos}\left(\sqrt{\mathrm{x}}\right)\right)} \\ $$$$\mathrm{cos}\left(\sqrt{\mathrm{x}}\right)\sim\mathrm{1}−\frac{\mathrm{x}}{\mathrm{2}}\:\Rightarrow\mathrm{log}\left(\mathrm{cos}\left(\sqrt{\mathrm{x}}\right)\right)\sim\mathrm{log}\left(\mathrm{1}−\frac{\mathrm{x}}{\mathrm{2}}\right)\sim−\frac{\mathrm{x}}{\mathrm{2}}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{x}}\mathrm{log}\left(\mathrm{cos}\left(\sqrt{\mathrm{x}}\right)\right)\sim−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}^{+} } \:\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{e}}} \\ $$
Commented by mathdanisur last updated on 08/Aug/21
$$\mathrm{Thank}\:\mathrm{you}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$
Answered by mnjuly1970 last updated on 08/Aug/21
$$\:\:\:\:{lim}_{\:{x}\rightarrow\mathrm{0}^{\:+} } \left({cos}\:\left(\sqrt{{x}}\:\right)\right)^{\frac{\mathrm{1}}{{x}}\:} ={lim}_{{x}\rightarrow\mathrm{0}^{+} } \left(\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\sqrt{{x}}}{\mathrm{2}}\right)\right)^{\frac{\mathrm{1}}{{x}}} \\ $$$$\:\:\overset{{equivallance}} {=}\:{lim}_{{x}\rightarrow\mathrm{0}^{\:+} } \left(\mathrm{1}−\mathrm{2}\left(\frac{{x}}{\mathrm{4}}\right)\right)^{\frac{\mathrm{1}}{{x}}} ={e}^{\:\frac{−\mathrm{1}}{\mathrm{2}}} \\ $$$$ \\ $$
Commented by mathdanisur last updated on 08/Aug/21
$$\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{thank}\:\mathrm{you} \\ $$