Question Number 141964 by gsk2684 last updated on 25/May/21
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{sin}\:{x}\right)^{\frac{\mathrm{1}}{{x}}} =? \\ $$
Commented by gsk2684 last updated on 25/May/21
$${solution}\:{please} \\ $$
Answered by iloveisrael last updated on 25/May/21
$$\:{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{sin}\:{x}\right)^{\frac{\mathrm{1}}{{x}}} \\ $$$$\mathrm{ln}\:{L}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left(\mathrm{sin}\:{x}\right)}{{x}}\: \\ $$$$\mathrm{ln}\:{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left(\mathrm{1}+\left(\mathrm{sin}\:{x}−\mathrm{1}\right)\right)}{{x}} \\ $$$$\mathrm{ln}\:{L}=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:{x}−\mathrm{1}}{{x}}\:=−\infty \\ $$$$\:{L}\:=\:{e}^{−\infty} \:=\:\mathrm{0} \\ $$
Commented by gsk2684 last updated on 25/May/21
$${thank}\:{you} \\ $$
Answered by mathmax by abdo last updated on 25/May/21
$$\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{sinx}\right)^{\frac{\mathrm{1}}{\mathrm{x}}} \:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{x}}\mathrm{log}\left(\mathrm{sinx}\right)} \:\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{sinx}\sim\mathrm{x}\:\:\Rightarrow\mathrm{log}\left(\mathrm{sinx}\right)\sim\mathrm{logx}\:\Rightarrow\frac{\mathrm{log}\left(\mathrm{sinx}\right)}{\mathrm{x}}\sim\frac{\mathrm{logx}}{\mathrm{x}} \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}^{+} } \:\:\frac{\mathrm{logx}}{\mathrm{x}}\:=_{\mathrm{x}=\frac{\mathrm{1}}{\mathrm{t}}} \:\:\mathrm{lim}_{\mathrm{t}\rightarrow+\infty} \:\:−\mathrm{tlogt}\:=−\infty\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{f}\left(\mathrm{x}\right)=\mathrm{e}^{−\infty} \:=\mathrm{0} \\ $$$$ \\ $$
Commented by gsk2684 last updated on 25/May/21
$${thank}\:{you} \\ $$