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Question Number 169004 by mathlove last updated on 23/Apr/22

$$\underset{x\to \frac{\pi }{2}}{\lim }\left(\frac{sinx-1}{x-\frac{\pi }{2}}\right)=?$$

Commented by infinityaction last updated on 23/Apr/22

$$\left(let\right)p=\underset{x\to \frac{\pi }{2}}{\lim }\left(\frac{\sin x-\sin \frac{\pi }{2}}{x-\frac{\pi }{2}}\right)$$$$p=\underset{x\to \frac{\pi }{2}}{\lim }\left\{\frac{2\cos \left(\frac{x+\pi /2}{2}\right)\sin \left(\frac{x-\pi /2}{2}\right)}{2\left(\frac{x-\pi /2}{2}\right)}\right\}$$$$p=\cos \frac{\pi }{2}$$$$p=0$$$$$$

Commented by mathlove last updated on 23/Apr/22

$$sir$$$$sinp+sinq=2cos\frac{p+q}{2}\cdot sin\frac{p-q}{2}$$

Commented by mr W last updated on 23/Apr/22

$$\underset{x\to \frac{\pi }{2}}{\lim }\left(\frac{sinx-1}{x-\frac{\pi }{2}}\right)$$$$=\underset{x\to \frac{\pi }{2}}{\lim }\left(\frac{sinx-\sin \frac{\pi }{2}}{x-\frac{\pi }{2}}\right)$$$$={\left(\sin x\right)}^{\prime }{\mid }_{x=\frac{\pi }{2}}$$$$=\left(\cos x\right){\mid }_{x=\frac{\pi }{2}}$$$$=0$$

Answered by qaz last updated on 23/Apr/22

$$\underset{\mathrm{x}\to \frac{\pi }{2}}{\lim }\frac{\sin \mathrm{x}-1}{\mathrm{x}-\frac{\pi }{2}}=\underset{\mathrm{x}\to \frac{\pi }{2}-\frac{\pi }{2}}{\lim }\frac{\sin (\mathrm{x}+\frac{\pi }{2})-1}{(\mathrm{x}+\frac{\pi }{2})-\frac{\pi }{2}}=\underset{\mathrm{x}\to 0}{\lim }\frac{\cos \mathrm{x}-1}{\mathrm{x}}=\underset{\mathrm{x}\to 0}{\lim }\frac{-\frac{1}{2}{\mathrm{x}}^2}{\mathrm{x}}=0$$

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