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Question Number 116576 by bemath last updated on 05/Oct/20

$$\underset{x\to 0}{\lim }{\left(\frac{\sin \mathrm{x}}{\mathrm{x}}\right)}^{\frac{1}{{\mathrm{x}}^2}}=?$$

Commented by mathmax by abdo last updated on 05/Oct/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)={\left(\frac{\mathrm{sinx}}{\mathrm{x}}\right)}^{\frac{1}{{\mathrm{x}}^2}}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{\frac{1}{{\mathrm{x}}^2}\ln \left(\frac{\mathrm{sinx}}{\mathrm{x}}\right)}\mathrm{we}\mathrm{have}$$$$\mathrm{sinx}=\mathrm{x}-\frac{{\mathrm{x}}^3}{6}+\mathrm{o}\left({\mathrm{x}}^3\right)\Rightarrow \frac{\mathrm{sinx}}{\mathrm{x}}=1-\frac{{\mathrm{x}}^2}{6}+\mathrm{o}\left({\mathrm{x}}^2\right)\Rightarrow$$$$\ln \left(\frac{\mathrm{sinx}}{\mathrm{x}}\right)\sim \ln (1-\frac{{\mathrm{x}}^2}{6})\sim -\frac{{\mathrm{x}}^2}{6}\Rightarrow \frac{1}{{\mathrm{x}}^2}\ln \left(\frac{\mathrm{sinx}}{\mathrm{x}}\right)\sim -\frac{1}{6}\Rightarrow$$$${\lim }_{\mathrm{x}\to 0}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{-\frac{1}{6}}=\frac{1}{\left({}^6\sqrt{\mathrm{e}}\right)}$$

Answered by Olaf last updated on 05/Oct/20

$$\ln {\left(\frac{\sin x}{x}\right)}^{\frac{1}{x^2}}=\frac{1}{x^2}\ln \left(\frac{\sin x}{x}\right)$$$$\ln {\left(\frac{\sin x}{x}\right)}^{\frac{1}{x^2}}\underset{0}{\sim }\frac{1}{x^2}\ln \left(\frac{x-\frac{x^3}{6}}{x}\right)$$$$\ln {\left(\frac{\sin x}{x}\right)}^{\frac{1}{x^2}}\underset{0}{\sim }\frac{1}{x^2}\ln (1-\frac{x^2}{6})$$$$\ln {\left(\frac{\sin x}{x}\right)}^{\frac{1}{x^2}}\underset{0}{\sim }\frac{1}{x^2}\times (-\frac{x^2}{6})=-\frac{1}{6}$$$$\Rightarrow \underset{x\to 0}{\lim }{\left(\frac{\sin x}{x}\right)}^{\frac{1}{x^2}}=e^{-\frac{1}{6}}$$$$$$$$$$

Answered by bobhans last updated on 05/Oct/20

$$\underset{x\to 0}{\lim }{\left(\frac{\sin \mathrm{x}}{\mathrm{x}}\right)}^{\frac{1}{{\mathrm{x}}^2}}=?$$$$\ln \mathrm{L}=\underset{x\to 0}{\lim }\frac{\ln \left(\frac{\sin \mathrm{x}}{\mathrm{x}}\right)}{{\mathrm{x}}^2}=\underset{x\to 0}{\lim }\frac{\ln \left(\frac{\mathrm{x}-\frac{{\mathrm{x}}^3}{6}}{\mathrm{x}}\right)}{{\mathrm{x}}^2}$$$$\ln \mathrm{L}=\underset{x\to 0}{\lim }\frac{\ln (1-\frac{{\mathrm{x}}^2}{6})}{{\mathrm{x}}^2}=\underset{x\to 0}{\lim }\frac{-\frac{{\mathrm{x}}^2}{6}+\mathrm{o}\left({\mathrm{x}}^2\right)}{{\mathrm{x}}^2}$$$$\ln \mathrm{L}=-\frac{1}{6}\Rightarrow \mathrm{L}={\mathrm{e}}^{-\frac{1}{6}}=\frac{1}{\sqrt[6]{\mathrm{e}}}\therefore$$

Answered by 1549442205PVT last updated on 05/Oct/20

$$\mathrm{Put}\mathrm{A}=\lim {\left(\frac{\sin \mathrm{x}}{\mathrm{x}}\right)}^{\frac{1}{{\mathrm{x}}^2}}$$$$\Rightarrow \mathrm{lnI}=\underset{\mathrm{x}\to 0}{\lim }\left[\frac{1}{{\mathrm{x}}^2}\ln \left(\frac{\mathrm{sinx}}{\mathrm{x}}\right)\right]=\underset{\mathrm{x}\to 0}{\lim }\frac{\ln \left(\frac{\mathrm{sinx}}{\mathrm{x}}\right)}{{\mathrm{x}}^2}$$$$\mathrm{This}\mathrm{is}\mathrm{the}\mathrm{form}\frac{0}{0}\mathrm{hence}\mathrm{using}{\mathrm{L}}^{\prime }\mathrm{Hopital}\mathrm{we}\mathrm{get}$$$$\underset{\mathrm{x}\to 0}{\lim }\frac{\frac{\mathrm{x}}{\mathrm{sinx}}\times \frac{\mathrm{xcosx}-\mathrm{sinx}}{{\mathrm{x}}^2}}{2\mathrm{x}}=\underset{\mathrm{x}\to 0}{\lim }\frac{\mathrm{xcosx}-\mathrm{sinx}}{{2\mathrm{x}}^2\mathrm{sinx}}$$$$\underset{{\mathrm{L}}^{\prime }\mathrm{Hopital}}{=}\underset{\mathrm{x}\to 0}{\lim }\frac{\mathrm{cosx}-\mathrm{xsinx}-\mathrm{cosx}}{4\mathrm{xsinx}+{2\mathrm{x}}^2\mathrm{cosx}}$$$$\underset{{\mathrm{L}}^{\prime }\mathrm{Hopital}}{=}\underset{\mathrm{x}\to 0}{\lim }\frac{-\mathrm{sinx}}{4\mathrm{sinx}+2\mathrm{xcosx}}=\underset{\mathrm{x}\to 0}{\lim }\frac{-\mathrm{cosx}}{4\mathrm{cosx}+2\mathrm{cosx}-2\mathrm{xsinx}}$$$$=\frac{-1}{4+2-0}=\frac{-1}{6}$$$$\Rightarrow \mathrm{I}={\mathrm{e}}^{\frac{-1}{6}}=\frac{1}{{}^6\sqrt{\mathrm{e}}}$$

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