Question Number 107930 by bemath last updated on 13/Aug/20
$$\:\:\frac{\mathbb{B}{e}\mathbb{M}{ath}}{\bullet\cap\bullet} \\ $$$$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{sin}\:{x}\right)^{\frac{\mathrm{1}}{\mathrm{ln}\:\sqrt{{x}}}} \:? \\ $$
Answered by bemath last updated on 13/Aug/20
![⇒L= lim_(x→0) (sin )^(1/(ln (√x))) = ln L= lim_(x→0) ((ln (sin x))/(ln (√x))) = 2[ lim_(x→0) ((ln (sin x))/(ln x)) ] ln L= 2[lim_(x→0) (((cos x)/(sin x))/(1/x)) ] = 2lim_(x→0) [((x cos x)/(sin x)) ] ln L = 2 ⇒L = e^2](https://www.tinkutara.com/question/Q107931.png)
$$\Rightarrow{L}=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{sin}\:\right)^{\frac{\mathrm{1}}{\mathrm{ln}\:\sqrt{{x}}}} \:= \\ $$$$\mathrm{ln}\:{L}=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left(\mathrm{sin}\:{x}\right)}{\mathrm{ln}\:\sqrt{{x}}}\:=\:\mathrm{2}\left[\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left(\mathrm{sin}\:{x}\right)}{\mathrm{ln}\:{x}}\:\right] \\ $$$$\mathrm{ln}\:{L}=\:\mathrm{2}\left[\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}}{\frac{\mathrm{1}}{{x}}}\:\right]\:=\:\mathrm{2}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\frac{{x}\:\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}\:\right] \\ $$$$\mathrm{ln}\:{L}\:=\:\mathrm{2}\:\Rightarrow{L}\:=\:{e}^{\mathrm{2}} \\ $$$$ \\ $$
Answered by Dwaipayan Shikari last updated on 13/Aug/20
$$\frac{\mathrm{1}}{{log}\sqrt{{x}}}{log}\left({sinx}\right)={logL} \\ $$$$\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}}{log}\left({x}\right)}{log}\left({x}\right)={logL}\:\:\:\:\left({sinx}\rightarrow{x}\right) \\ $$$$\mathrm{2}={logL} \\ $$$${L}={e}^{\mathrm{2}} \\ $$
Answered by mathmax by abdo last updated on 13/Aug/20
$$\mathrm{f}\left(\mathrm{x}\right)\:=\left(\mathrm{sinx}\right)^{\frac{\mathrm{1}}{\mathrm{ln}\left(\sqrt{\mathrm{x}}\right)}} \:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{ln}\left(\sqrt{\mathrm{x}}\right)}\mathrm{ln}\left(\mathrm{sinx}\right)} \\ $$$$=\mathrm{e}^{\frac{\mathrm{2}}{\mathrm{lnx}}\mathrm{ln}\left(\mathrm{sinx}\right)} \:\sim\:\mathrm{e}^{\mathrm{2}} \:\:\left(\mathrm{x}\:\sim\mathrm{0}\right)\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{e}^{\mathrm{2}} \\ $$