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Question Number 172872 by mnjuly1970 last updated on 02/Jul/22

lim_( x→ 0^+ ) (x^( (1/(ln(xsin(x^3 ))))) )=?

$$ \\ $$$$\:\:\:\:\:\mathrm{lim}_{\:{x}\rightarrow\:\mathrm{0}^{+} } \left({x}^{\:\frac{\mathrm{1}}{\mathrm{l}{n}\left({xsin}\left({x}^{\mathrm{3}} \right)\right)}} \right)=? \\ $$$$ \\ $$$$ \\ $$

Answered by mnjuly1970 last updated on 02/Jul/22

e^( lim _(x→0^+ ) ln x^( (1/((ln(xsin(x^( 3) ))))) ) =e^( lim_( x→0^+ ) ((ln(x))/(ln(x)+ln(sin(x^( 3) ))))) = e^( lim_( x→0^( +) ) ((1/x)/((1/x) + ((3x^( 2) cos(x^( 3) ))/(sin(x^( 3) )))))) = = e^( lim_( x→0^( +) ) ((1/x)/((1/x) + (3/x))) =lim((1/4))) =(e)^(1/4)

$$\:\:\:\:\:\mathrm{e}^{\:\mathrm{lim}\:_{\mathrm{x}\rightarrow\mathrm{0}^{+} } \mathrm{ln}\:\mathrm{x}^{\:\frac{\mathrm{1}}{\left(\mathrm{ln}\left(\mathrm{xsin}\left(\mathrm{x}^{\:\mathrm{3}} \right)\right)\right.}} } =\mathrm{e}^{\:\mathrm{lim}_{\:\mathrm{x}\rightarrow\mathrm{0}^{+} } \:\frac{\mathrm{ln}\left(\mathrm{x}\right)}{\mathrm{ln}\left(\mathrm{x}\right)+\mathrm{ln}\left(\mathrm{sin}\left(\mathrm{x}^{\:\mathrm{3}} \right)\right)}} \\ $$$$\:\:\:=\:\mathrm{e}^{\:\mathrm{lim}_{\:\mathrm{x}\rightarrow\mathrm{0}^{\:+} } \frac{\frac{\mathrm{1}}{\mathrm{x}}}{\frac{\mathrm{1}}{\mathrm{x}}\:+\:\frac{\mathrm{3x}^{\:\mathrm{2}} \mathrm{cos}\left(\mathrm{x}^{\:\mathrm{3}} \right)}{\mathrm{sin}\left(\mathrm{x}^{\:\mathrm{3}} \right)}}} = \\ $$$$\:\:\:\:=\:\mathrm{e}^{\:\mathrm{lim}_{\:\mathrm{x}\rightarrow\mathrm{0}^{\:+} } \frac{\frac{\mathrm{1}}{\mathrm{x}}}{\frac{\mathrm{1}}{\mathrm{x}}\:+\:\frac{\mathrm{3}}{\mathrm{x}}}\:=\mathrm{lim}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)} =\sqrt[{\mathrm{4}}]{\mathrm{e}} \\ $$$$ \\ $$

Answered by Mathspace last updated on 02/Jul/22

=lim_(x→0^+ ) e^((lnx)/(xln(sinx))) lim_(x→0^+ ) xln(sinx)=lim_(x→0^+ ) xlnx=o^− lim_(x→0^+ ) lnx=−∞ ⇒ lim_(x→0^+ ) e^((lnx)/(xln(sinx))) =e^(+∞) =+∞

$$={lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:{e}^{\frac{{lnx}}{{xln}\left({sinx}\right)}} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\:{xln}\left({sinx}\right)={lim}_{{x}\rightarrow\mathrm{0}^{+} } {xlnx}={o}^{−} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:{lnx}=−\infty\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:{e}^{\frac{{lnx}}{{xln}\left({sinx}\right)}} ={e}^{+\infty} =+\infty \\ $$

Commented by mnjuly1970 last updated on 02/Jul/22

pls recheck the answer sir

$$\mathrm{pls}\:\:\mathrm{recheck}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{sir} \\ $$

Answered by floor(10²Eta[1]) last updated on 02/Jul/22

L=lim_(x→0^+ ) x^(1/(ln(xsin(x^3 )))) lnL=lim_(x→0^+ ) ((lnx)/(ln(xsin(x^3 ))))=lim_(x→0^+ ) ((1/x)/((sin(x^3 )+3x^3 cos(x^3 ))/(xsin(x^3 )))) =lim_(x→0^+ ) ((sin(x^3 ))/(sin(x^3 )+3x^3 cos(x^3 ))) =lim_(x→0^+ ) ((3x^2 cos(x^3 ))/(12x^2 cos(x^3 )−9x^4 sin(x^3 ))) =lim_(x→0^+ ) ((3cos(x^3 ))/(12cos(x^3 )−9x^2 sin(x^3 )))=(3/(12))=(1/4) ⇒lnL=(1/4)⇒L=(e)^(1/4)

$$\mathrm{L}=\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}x}^{\frac{\mathrm{1}}{\mathrm{ln}\left(\mathrm{xsin}\left(\mathrm{x}^{\mathrm{3}} \right)\right)}} \\ $$$$\mathrm{lnL}=\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{\mathrm{lnx}}{\mathrm{ln}\left(\mathrm{xsin}\left(\mathrm{x}^{\mathrm{3}} \right)\right)}=\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{\frac{\mathrm{1}}{\mathrm{x}}}{\frac{\mathrm{sin}\left(\mathrm{x}^{\mathrm{3}} \right)+\mathrm{3x}^{\mathrm{3}} \mathrm{cos}\left(\mathrm{x}^{\mathrm{3}} \right)}{\mathrm{xsin}\left(\mathrm{x}^{\mathrm{3}} \right)}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{\mathrm{sin}\left(\mathrm{x}^{\mathrm{3}} \right)}{\mathrm{sin}\left(\mathrm{x}^{\mathrm{3}} \right)+\mathrm{3x}^{\mathrm{3}} \mathrm{cos}\left(\mathrm{x}^{\mathrm{3}} \right)} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{\mathrm{3x}^{\mathrm{2}} \mathrm{cos}\left(\mathrm{x}^{\mathrm{3}} \right)}{\mathrm{12x}^{\mathrm{2}} \mathrm{cos}\left(\mathrm{x}^{\mathrm{3}} \right)−\mathrm{9x}^{\mathrm{4}} \mathrm{sin}\left(\mathrm{x}^{\mathrm{3}} \right)} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{\mathrm{3cos}\left(\mathrm{x}^{\mathrm{3}} \right)}{\mathrm{12cos}\left(\mathrm{x}^{\mathrm{3}} \right)−\mathrm{9x}^{\mathrm{2}} \mathrm{sin}\left(\mathrm{x}^{\mathrm{3}} \right)}=\frac{\mathrm{3}}{\mathrm{12}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{lnL}=\frac{\mathrm{1}}{\mathrm{4}}\Rightarrow\mathrm{L}=\sqrt[{\mathrm{4}}]{\mathrm{e}} \\ $$

Commented by mnjuly1970 last updated on 03/Jul/22

thanks alot...very nice solution

$$\:\:\:\:\:\mathrm{thanks}\:\mathrm{alot}...\mathrm{very}\:\mathrm{nice}\:\mathrm{solution} \\ $$

Answered by a.lgnaoui last updated on 04/Jul/22

lnL=ln[x^(1/(ln(xsin (x^3 )))) ]=(1/(ln(xsin (x^3 ))))×lnx =((lnx)/(ln(xsin (x^3 ))))=((lnx)/(ln[x^4 ×(((sin (x^3 ))/x^3 ) )])) lim_(x→0) (((sin x)/x))=lim_(x→0) ((sin (x^3 ))/x^3 )=1 lnL=((lnx)/(ln(x^4 )))=(1/4)⇒L=e^(1/4) =^4 (√e).

$${lnL}={ln}\left[{x}^{\frac{\mathrm{1}}{{ln}\left({x}\mathrm{sin}\:\left({x}^{\mathrm{3}} \right)\right)}} \right]=\frac{\mathrm{1}}{{ln}\left({x}\mathrm{sin}\:\left({x}^{\mathrm{3}} \right)\right)}×{lnx} \\ $$$$=\frac{{lnx}}{{ln}\left({x}\mathrm{sin}\:\left({x}^{\mathrm{3}} \right)\right)}=\frac{{lnx}}{{ln}\left[{x}^{\mathrm{4}} ×\left(\frac{\mathrm{sin}\:\left({x}^{\mathrm{3}} \right)}{{x}^{\mathrm{3}} }\:\right)\right]} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \left(\frac{\mathrm{sin}\:{x}}{{x}}\right)={lim}_{{x}\rightarrow\mathrm{0}} \frac{\mathrm{sin}\:\left({x}^{\mathrm{3}} \right)}{{x}^{\mathrm{3}} }=\mathrm{1} \\ $$$${lnL}=\frac{{lnx}}{{ln}\left({x}^{\mathrm{4}} \right)}=\frac{\mathrm{1}}{\mathrm{4}}\Rightarrow{L}={e}^{\frac{\mathrm{1}}{\mathrm{4}}} =^{\mathrm{4}} \sqrt{{e}}. \\ $$

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