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Question Number 113341 by Khalmohmmad last updated on 12/Sep/20

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

$$\mathrm{we}\mathrm{have}\mid \sin \left(\frac{\pi }{\mathrm{x}}\right)\mid ⩽1\Rightarrow \mid \sqrt{{\mathrm{x}}^3+{\mathrm{x}}^2}\sin \left(\frac{\pi }{\mathrm{x}}\right)\mid ⩽\mid \mathrm{x}\mid \sqrt{\mathrm{x}+1}\to 0(\mathrm{x}\to 0)$$$$\Rightarrow {\lim }_{\mathrm{x}\to 0}\sqrt{{\mathrm{x}}^3+{\mathrm{x}}^2}\sin \left(\frac{\pi }{\mathrm{x}}\right)=0$$

Answered by RCRC last updated on 12/Sep/20

$$\underset{x\to 0}{lim}\sqrt{x^3+x^2}sen\left(\frac{\pi }{x}\right)=\underset{x\to 0}{lim}\left(\frac{\pi x\sqrt{x+1}}{\pi }sen\left(\frac{\pi }{x}\right)\right)$$$$=\underset{x\to 0}{lim}\left(\frac{\pi x\sqrt{x+1}}{\pi }sen\left(\frac{\pi }{x}\right)\right)$$$$=\underset{x\to 0}{lim}\left(\frac{\pi \sqrt{x+1}}{\pi x^{-1}}sen\left(\pi x^{-1}\right)\right)$$$$=\underset{x\to 0}{lim}\left(\pi \sqrt{x+1}\frac{sen\left(\pi x^{-1}\right)}{\pi x^{-1}}\right)$$$$=\underset{x\to 0}{lim}\left(\pi \sqrt{x+1}\right)\times \underset{x\to 0}{lim}\left(\frac{sen\left(\pi x^{-1}\right)}{\pi x^{-1}}\right)\measuredangle$$$$=\pi \times 0=0$$

Answered by Dwaipayan Shikari last updated on 12/Sep/20

$$\underset{x\to 0}{\lim }\sqrt{x^3+x}sin\left(\frac{\pi }{x}\right)=\underset{x\to 0}{\lim }\sqrt{x^3+x}sin\left(z\right)(z\to \mathrm{\infty })$$$$-1⩽sin\left(z\right)⩽1$$$$so\sqrt{x^3+x}sin\left(z\right)=0$$

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