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

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