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Question Number 203247 by mathlove last updated on 13/Jan/24

$$\underset{x\to 0}{\lim }xtan\frac{\pi }{2}(1+x)=?$$

Answered by MM42 last updated on 13/Jan/24

$$={lim}_{x\to 0}-xcot\frac{\pi }{2}x={lim}_{x\to 0}-\frac{\frac{\pi }{2}xcos\frac{\pi }{2}x}{sin\frac{\pi }{2}x}\times \frac{2}{\pi }$$$$=-\frac{2}{\pi }✓$$$$$$

Commented by mathlove last updated on 13/Jan/24

$$waytan\frac{\pi }{2}(1+x)=-cotx$$

Commented by MM42 last updated on 13/Jan/24

$$ok$$$$-cot\frac{\pi }{2}x$$

Answered by mr W last updated on 13/Jan/24

$$\underset{x\to 0}{\lim }x\tan \frac{\pi }{2}(1+x)$$$$=\underset{x\to 0}{\lim }\frac{x\sin (\frac{\pi }{2}+\frac{\pi x}{2})}{\cos (\frac{\pi }{2}+\frac{\pi x}{2})}$$$$=\underset{x\to 0}{\lim }\frac{x\sin (\frac{\pi }{2}-\frac{\pi x}{2})}{-\cos (\frac{\pi }{2}-\frac{\pi x}{2})}$$$$=\underset{x\to 0}{\lim }\frac{x\cos \left(\frac{\pi x}{2}\right)}{-\sin \left(\frac{\pi x}{2}\right)}$$$$=-\frac{2}{\pi }\underset{x\to 0}{\lim }\frac{\left(\frac{\pi x}{2}\right)}{\sin \left(\frac{\pi x}{2}\right)}\times \cos \left(\frac{\pi x}{2}\right)$$$$=-\frac{2}{\pi }$$

Commented by mathlove last updated on 13/Jan/24

$$thatsright$$$$thankssir$$

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