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Question Number 122550 by liberty last updated on 18/Nov/20

$$\underset{x\to 1/2}{\lim }\frac{{2\mathrm{x}}^2+\mathrm{x}-1}{\tan \left(\pi \mathrm{x}\right)}?$$

Answered by bramlexs22 last updated on 18/Nov/20

$$\underset{x\to 1/2}{\lim }\frac{(2x-1)(x+1)}{\tan \left(\pi x\right)}=$$$$\frac{3}{2}\times \underset{x\to 1/2}{\lim }\frac{2x-1}{\tan \left(\pi x\right)}=$$$$\frac{3}{2}\times \underset{x\to 1/2}{\lim }\frac{2}{\pi \sec {}^2\left(\pi x\right)}=$$$$3\times \underset{x\to 1/2}{\lim }\frac{\cos {}^2\left(\pi x\right)}{\pi }=0.$$

Answered by Dwaipayan Shikari last updated on 18/Nov/20

$$\underset{x\to \frac{1}{2}}{\lim }\frac{2x^2+x-1}{tan\pi x}=\frac{(2x-1)(x+1)}{sin\left(\pi x\right)}sin(\frac{\pi }{2}-\pi x)$$$$=\frac{3}{2}.(2x-1)(\frac{\pi }{2}-\pi x)=0$$

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