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Question Number 19271 by Joel577 last updated on 08/Aug/17

$$\underset{x\to \pi }{\lim }{(2-{\cos }^{2}x)}^{\frac{2\sqrt{2(1+\cos x)}}{{(x-\pi )}^3}}$$

Answered by ajfour last updated on 09/Aug/17

$$=\underset{x\to \pi }{\lim }{\left\{{[1+\sin {}^2(\pi -\mathrm{x})]}^{\frac{1}{\sin {}^2(\pi -\mathrm{x})}}\right\}}^{4\cos \left(\frac{\mathrm{x}}{2}\right)\times \frac{\sin {}^2(\pi -\mathrm{x})}{{(\pi -\mathrm{x})}^2}\times \frac{-1}{(\pi -\mathrm{x})}}$$$$={\mathrm{e}}^{\underset{x\to \pi }{\lim }\mid 4\sin (\frac{\pi }{2}-\frac{\mathrm{x}}{2})\mid \times \frac{\sin {}^2(\pi -\mathrm{x})}{{(\pi -\mathrm{x})}^2}\times \frac{-1}{(\pi -\mathrm{x})}}$$$$={\mathrm{e}}^{-\underset{x\to \pi }{\lim }\frac{\mid 4\sin (\frac{\pi }{2}-\frac{\mathrm{x}}{2})\mid }{2(\frac{\pi }{2}-\frac{\mathrm{x}}{2})}}$$$$\mathrm{R}.\mathrm{H}.\mathrm{L}.={\mathbf{e}}^2;\mathrm{L}.\mathrm{H}.\mathrm{L}.={\mathbf{e}}^{-2}=\frac{1}{{\mathbf{e}}^2}.$$$$\mathrm{so}\mathrm{the}\mathrm{limit}\mathrm{dont}\mathrm{exist}\mathrm{at}\mathrm{x}=\pi .$$

Commented by Joel577 last updated on 09/Aug/17

$$\mathrm{Can}\mathrm{u}\mathrm{explain}\mathrm{with}\mathrm{more}\mathrm{detail}\mathrm{Sir}?$$$$\mathrm{I}\mathrm{dont}\mathrm{understand}\mathrm{from}\mathrm{the}\mathrm{second}\mathrm{row}$$

Commented by ajfour last updated on 09/Aug/17

$$\mathrm{since}\underset{x\to 0}{\lim }{(1+\mathrm{x})}^{1/\mathrm{x}}=\mathrm{e}$$$$\mathrm{here}\mathrm{we}\mathrm{have}\underset{x\to \pi }{\lim }{[1+\sin {}^2(\pi -\mathrm{x})]}^{\frac{1}{\sin {}^2(\pi -\mathrm{x})}}=\mathrm{e}$$$$..$$

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