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Question Number 18627 by Arnab Maiti last updated on 26/Jul/17

lim_(x→0) (((sin x)/x))^((sin x)/(x−sin x))

$$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{x}}\right)^{\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{x}−\mathrm{sin}\:\mathrm{x}}} \\ $$

Commented by Arnab Maiti last updated on 26/Jul/17

Please help.

$$\mathrm{Please}\:\mathrm{help}. \\ $$

Answered by 433 last updated on 26/Jul/17

((sin x)/x)=y x→0 ⇒ y→1 ((sin x)/(x−sin x))=(((sin x)/x)/((x/x)−((sin x)/x)))=(y/(1−y))=−1+(1/(1−y)) lim_(y→1) (y^(1/(1−y)) /y)=lim_(y→1) (e^((1/(1−y))ln (y)) /y)=e^(−1) ((ln y)/(1−y))=^(DLH) ((1/y)/(−1))→^(y→1) −1

$$ \\ $$$$ \\ $$$$\frac{\mathrm{sin}\:{x}}{{x}}={y} \\ $$$${x}\rightarrow\mathrm{0}\:\Rightarrow\:{y}\rightarrow\mathrm{1} \\ $$$$\frac{\mathrm{sin}\:{x}}{{x}−\mathrm{sin}\:{x}}=\frac{\frac{\mathrm{sin}\:{x}}{{x}}}{\frac{{x}}{{x}}−\frac{\mathrm{sin}\:{x}}{{x}}}=\frac{{y}}{\mathrm{1}−{y}}=−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}−{y}} \\ $$$$\underset{{y}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{{y}^{\frac{\mathrm{1}}{\mathrm{1}−{y}}} }{{y}}=\underset{{y}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{e}^{\frac{\mathrm{1}}{\mathrm{1}−{y}}\mathrm{ln}\:\left({y}\right)} }{{y}}={e}^{−\mathrm{1}} \\ $$$$\frac{\mathrm{ln}\:{y}}{\mathrm{1}−{y}}\overset{{DLH}} {=}\frac{\frac{\mathrm{1}}{{y}}}{−\mathrm{1}}\overset{{y}\rightarrow\mathrm{1}} {\rightarrow}−\mathrm{1} \\ $$

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