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Question Number 128439 by Study last updated on 07/Jan/21

$$li\underset{x\to \frac{\pi }{2}}{m}\frac{cosx}{1-sinx}=???$$

Answered by benjo_mathlover last updated on 07/Jan/21

$$\mathrm{let}\mathrm{x}=\frac{\pi }{2}+\mathrm{z}\wedge \mathrm{z}\to 0$$$$\underset{\mathrm{z}\to 0}{\lim }\frac{\cos (\frac{\pi }{2}+\mathrm{z})}{1-\sin (\mathrm{z}+\frac{\pi }{2})}=\underset{\mathrm{z}\to 0}{\lim }\frac{-\sin \mathrm{z}}{1-\cos \mathrm{z}}$$$$=\underset{\mathrm{z}\to 0}{\lim }\frac{-\sin \mathrm{z}}{2\sin {}^2\left(\mathrm{z}/2\right)}=-\mathrm{\infty }$$

Answered by liberty last updated on 08/Jan/21

$$\underset{x\to \pi /2}{\lim }\frac{\cos {}^2\left(\mathrm{x}/2\right)-\sin {}^2\left(\mathrm{x}/2\right)}{\sin {}^2\left(\mathrm{x}/2\right)-2\sin \left(\mathrm{x}/2\right)\cos \left(\mathrm{x}/2\right)+\cos {}^2\left(\mathrm{x}/2\right)}$$$$=\underset{x\to \pi /2}{\lim }\frac{(\cos \frac{\mathrm{x}}{2}-\sin \frac{\mathrm{x}}{2})(\cos \frac{\mathrm{x}}{2}+\sin \frac{\mathrm{x}}{2})}{{(\cos \frac{\mathrm{x}}{2}-\sin \frac{\mathrm{x}}{2})}^2}$$$$=\underset{x\to \pi /2}{\lim }\frac{\cos \frac{\mathrm{x}}{2}+\sin \frac{\mathrm{x}}{2}}{\mid \cos \frac{\mathrm{x}}{2}-\sin \frac{\mathrm{x}}{2}\mid }=\frac{\sqrt{2}}{0}=\mathrm{\infty }$$$$\mathrm{or}\underset{x\to \pi /2}{\lim }\frac{\cos \left(\frac{\mathrm{x}}{2}\right)+\sin \left(\frac{\mathrm{x}}{2}\right)}{\mid \cos \frac{\mathrm{x}}{2}-\sin \frac{\mathrm{x}}{2}\mid }=-\mathrm{\infty }$$

Answered by MJS_new last updated on 08/Jan/21

$$\mathrm{limit}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{exist}\mathrm{because}$$$$\underset{x\to \frac{\pi }{2}{}^{-}}{\lim }\frac{\cos x}{1-\sin x}=+\mathrm{\infty }\mathrm{but}\underset{x\to \frac{\pi }{2}{}^{+}}{\lim }\frac{\cos x}{1-\sin x}=-\mathrm{\infty }$$

Commented by liberty last updated on 08/Jan/21

$$\mathrm{yes}....\mathrm{agree}$$

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