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Question Number 119074 by benjo_mathlover last updated on 22/Oct/20

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

Answered by Dwaipayan Shikari last updated on 22/Oct/20

$$\underset{x\to 1}{\lim }(x-1)cot(\frac{\pi }{2}-\frac{\pi }{2}x)$$$$=\underset{x\to 1}{\lim }-(1-x)\frac{cos(1-x)\frac{\pi }{2}}{sin(1-x)\frac{\pi }{2}}$$$$=\underset{x\to 1}{\lim }-(1-x)\frac{1}{(1-x)\frac{\pi }{2}}=-\frac{2}{\pi }\underset{x\to 0}{\lim }sinx=x$$

Answered by bemath last updated on 22/Oct/20

$$letx=1+m\wedge m\to 0$$$$\underset{m\to 0}{\lim }\frac{m}{\cot \left(\frac{(1+m)\pi }{2}\right)}=\underset{m\to 0}{\lim }\left[\frac{m}{-\tan \left(\frac{m\pi }{2}\right)}\right]$$$$=-\frac{2}{\pi }$$

Answered by 1549442205PVT last updated on 22/Oct/20

$$\mathrm{Put}\mathrm{x}-1=\mathrm{t}\Rightarrow (\mathrm{x}\to 1\sim \mathrm{t}\to 0\sim \pi \mathrm{t}/2\to 0)$$$$\mathrm{I}=\lim (x-1)\tan \left(\frac{\pi x}{2}\right)=\underset{\mathrm{t}\to 0}{\lim }\left[\mathrm{ttan}\frac{\pi (1+\mathrm{t})}{2}\right]$$$$=\underset{\mathrm{t}\to 0}{\lim }(-\mathrm{tcot}\frac{\pi \mathrm{t}}{2})=\underset{\mathrm{t}\to 0}{\lim }(-\frac{1}{\pi })\left(\frac{\pi \mathrm{t}}{\sin \frac{\pi \mathrm{t}}{2}}\times \cos \frac{\pi \mathrm{t}}{2}\right)$$$$=\underset{\mathrm{t}\to 0}{\lim }(-\frac{1}{\pi })\times \frac{2}{\frac{\sin \left(\pi \mathrm{t}/2\right)}{\left(\pi \mathrm{t}/2\right)}}\times \cos \left(\pi \mathrm{t}/2\right)=$$$$-\frac{2}{\pi }\times 1=-\frac{2}{\pi }$$

Answered by malwan last updated on 22/Oct/20

$$\underset{x\to 1}{lim}(x-1)cot(\frac{\pi }{2}-\frac{\pi x}{2})$$$$=\underset{x\to 1}{lim}(x-1)cot\frac{\pi }{2}(1-x)$$$$=\underset{x\to 1}{lim}\frac{\frac{\pi }{2}(1-x)\times (-1)}{\frac{\pi }{2}tan\frac{\pi }{2}(1-x)}=-\frac{2}{\pi }$$

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