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Question Number 82494 by niroj last updated on 21/Feb/20

Find the limit _(x→0) ^(lim) (((tan x)/x))^(1/x^( 2) )

$$\:\: \\ $$$$\:\:\boldsymbol{\mathrm{Find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{limit}} \\ $$$$\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}} {\overset{\boldsymbol{\mathrm{lim}}} {\:}}\:\left(\frac{\boldsymbol{\mathrm{tan}}\:\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{x}}}\right)^{\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}^{\:\mathrm{2}} }} \\ $$

Commented by abdomathmax last updated on 21/Feb/20

let f(x)=(((tanx)/x))^(1/x^2 ) ⇒f(x)=e^((1/x^2 )ln(((tanx)/x))) u(x)=tanx =u(0)+(x/(1!))u^((1)) (o) +(x^2 /(2!))u^((2)) (0)+(x^3 /(3!))u^((3)) (0)+... u^′ (x)=1+tan^2 x ⇒u^′ (0)=1 u^((2)) (x)=2tanx(1+tan^2 x) ⇒u^((2)) (0)=0 u^((3)) (x) =2(1+tan^2 x)(1+tan^2 x)+ +(2tanx)(2tanx(1+tan^2 x)) ⇒u^((3)) (0)=2 ⇒ tanx =x +(x^3 /3) +o(x^5 ) ⇒((tanx)/x) =1+(x^2 /3) +o(x^4 ) ⇒ ln(((tanx)/x)) =ln(1+(x^2 /3)+o(x^4 )) ∼(x^2 /3) ⇒ (1/x^2 )ln(((tanx)/x)) ∼ (1/3) ⇒lim_(x→0) f(x) =e^(1/3) =^3 (√e)

$${let}\:{f}\left({x}\right)=\left(\frac{{tanx}}{{x}}\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \:\Rightarrow{f}\left({x}\right)={e}^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{ln}\left(\frac{{tanx}}{{x}}\right)} \\ $$$${u}\left({x}\right)={tanx}\:={u}\left(\mathrm{0}\right)+\frac{{x}}{\mathrm{1}!}{u}^{\left(\mathrm{1}\right)} \left({o}\right)\:+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}{u}^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right)+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}{u}^{\left(\mathrm{3}\right)} \left(\mathrm{0}\right)+... \\ $$$${u}^{'} \left({x}\right)=\mathrm{1}+{tan}^{\mathrm{2}} {x}\:\Rightarrow{u}^{'} \left(\mathrm{0}\right)=\mathrm{1} \\ $$$${u}^{\left(\mathrm{2}\right)} \left({x}\right)=\mathrm{2}{tanx}\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right)\:\Rightarrow{u}^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right)=\mathrm{0} \\ $$$${u}^{\left(\mathrm{3}\right)} \left({x}\right)\:=\mathrm{2}\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right)\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right)+ \\ $$$$+\left(\mathrm{2}{tanx}\right)\left(\mathrm{2}{tanx}\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right)\right)\:\Rightarrow{u}^{\left(\mathrm{3}\right)} \left(\mathrm{0}\right)=\mathrm{2}\:\Rightarrow \\ $$$${tanx}\:={x}\:+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\:+{o}\left({x}^{\mathrm{5}} \right)\:\Rightarrow\frac{{tanx}}{{x}}\:=\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{3}}\:+{o}\left({x}^{\mathrm{4}} \right)\:\Rightarrow \\ $$$${ln}\left(\frac{{tanx}}{{x}}\right)\:={ln}\left(\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{3}}+{o}\left({x}^{\mathrm{4}} \right)\right)\:\sim\frac{{x}^{\mathrm{2}} }{\mathrm{3}}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{ln}\left(\frac{{tanx}}{{x}}\right)\:\sim\:\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} {f}\left({x}\right)\:={e}^{\frac{\mathrm{1}}{\mathrm{3}}} \:=^{\mathrm{3}} \sqrt{{e}} \\ $$

Commented by jagoll last updated on 22/Feb/20

lim_(x→0) (1+((tan x)/x)−1)^(1/x^2 ) = lim_(x→0) (1+((tan x−x)/x))^(1/x^2 ) = e^(lim_(x→0) (((tan x−x)/x^3 ))) = e^(lim_(x→0) (((sec^2 x−1)/(3x^2 )))) = e^(lim_(x→0) (((1−cos^2 x)/(3x^2 cos^2 x)))) = e^(1/3)

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{1}+\frac{\mathrm{tan}\:{x}}{{x}}−\mathrm{1}\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \:=\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{1}+\frac{\mathrm{tan}\:{x}−{x}}{{x}}\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \:= \\ $$$${e}\:^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{tan}\:{x}−{x}}{{x}^{\mathrm{3}} }\right)} =\:{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{sec}\:^{\mathrm{2}} {x}−\mathrm{1}}{\mathrm{3}{x}^{\mathrm{2}} }\right)} \\ $$$$=\:{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {x}}{\mathrm{3}{x}^{\mathrm{2}} \:\mathrm{cos}\:^{\mathrm{2}} {x}}\right)} \:=\:{e}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$

Commented by john santu last updated on 22/Feb/20

nice

$${nice} \\ $$

Commented by niroj last updated on 22/Feb/20

great.

$$\:{great}. \\ $$

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