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Question Number 113507 by Khalmohmmad last updated on 13/Sep/20

Answered by Dwaipayan Shikari last updated on 13/Sep/20

$$\underset{x\to 0^{+}}{\lim }\sqrt{x}{e}^{sin\left(\frac{\pi }{x}\right)}$$$$=\sqrt{x}{e}^{sinz}z\to +\mathrm{\infty }$$$$-1⩽sinz⩽1$$$$\frac{1}{e}⩽e^{sinz}⩽e$$$$\sqrt{x}{e}^{sinz}=0$$

Answered by mathmax by abdo last updated on 13/Sep/20

$$\mathrm{let}\mathrm{g}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}}{\mathrm{e}}^{\sin \left(\frac{\pi }{\mathrm{x}}\right)}\mathrm{changement}\frac{\pi }{\mathrm{x}}=\mathrm{t}\mathrm{give}\mathrm{x}=\frac{\pi }{\mathrm{t}}$$$$\mathrm{g}\left(\mathrm{x}\right)=\sqrt{\frac{\pi }{\mathrm{t}}}{\mathrm{e}}^{\mathrm{sint}}(\mathrm{x}\to 0^{+}\Rightarrow \mathrm{t}\to +\mathrm{\infty })\mathrm{we}\mathrm{have}$$$$-1⩽\mathrm{sint}⩽1\Rightarrow {\mathrm{e}}^{-1}⩽{\mathrm{e}}^{\mathrm{sint}}⩽\mathrm{e}\Rightarrow {\mathrm{e}}^{-1}\sqrt{\frac{\pi }{\mathrm{t}}}(\to 0)⩽\sqrt{\frac{\pi }{\mathrm{t}}}{\mathrm{e}}^{\mathrm{sint}}⩽\mathrm{e}\sqrt{\frac{\pi }{\mathrm{t}}}(\to 0)$$$$\Rightarrow {\lim }_{\mathrm{x}\to 0^{+}}\mathrm{g}\left(\mathrm{x}\right)=0$$$$$$

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