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Question Number 125790 by mohammad17 last updated on 13/Dec/20

Answered by liberty last updated on 13/Dec/20

$$\underset{x\to 2}{\lim }\frac{(x+2)(x-2)}{\sin \pi x}=4\times \underset{x\to 2}{\lim }\left[\frac{x-2}{\sin \pi x}\right]$$$$[letx=2+t\wedge t\to 0]$$$$=4\times \underset{t\to 0}{\lim }\frac{t}{\sin \pi (t+2)}=4\times \underset{t\to 0}{\lim }\frac{t}{\sin (2\pi +\pi t)}$$$$=4\times \underset{t\to 0}{\lim }\frac{t}{\sin \pi t}=4\times \underset{t\to 0}{\lim }\frac{\pi t}{\pi \sin \pi t}$$$$=\frac{4}{\pi }\times \underset{t\to 0}{\lim }\frac{\pi t}{\sin \pi t}=\frac{4}{\pi }\times 1=\frac{4}{\pi }$$

Commented by mohammad17 last updated on 13/Dec/20

$$nisesirthankyou$$

Answered by mathmax by abdo last updated on 14/Dec/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^2-4}{\sin \left(\pi \mathrm{x}\right)}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{(\mathrm{x}-2)(\mathrm{x}+2)}{\sin \left(\pi \mathrm{x}\right)}{=}_{\mathrm{x}-2=\mathrm{t}}\frac{\mathrm{t}(\mathrm{t}+4)}{\sin \left(\pi (\mathrm{t}+2)\right)}$$$$=\frac{\mathrm{t}(\mathrm{t}+4)}{\mathrm{sn}\left(\pi \mathrm{t}\right)}\mathrm{we}\mathrm{have}\mathrm{x}\to 2\iff \mathrm{t}\to 0\Rightarrow \frac{\mathrm{t}(\mathrm{t}+4)}{\sin \left(\pi \mathrm{t}\right)}=\sim \frac{\mathrm{t}(\mathrm{t}+4)}{\pi \mathrm{t}}\sim \frac{\mathrm{t}+4}{\pi }\to \frac{4}{\pi }$$$$\Rightarrow {\lim }_{\mathrm{x}\to 2}\mathrm{f}\left(\mathrm{x}\right)=\frac{4}{\pi }$$

Answered by Dwaipayan Shikari last updated on 14/Dec/20

$$\underset{x\to 2}{\lim }\frac{x^2-4}{sin\pi x}=\underset{x\to 2}{\lim }-\frac{x^2-4}{sin(2\pi -\pi x)}=\underset{x\to 2}{\lim }-\frac{(x-2)(x+2)}{\pi (2-x)}=\frac{x+2}{\pi }=\frac{4}{\pi }$$

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