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Question Number 130406 by john_santu last updated on 25/Jan/21

$$\underset{x\to \pi /2}{\lim }\cos (2x+\sin (\frac{3\pi }{2}+x)=?$$

Commented by Dwaipayan Shikari last updated on 25/Jan/21

$$-1$$

Answered by EDWIN88 last updated on 25/Jan/21

$$\underset{x\to \mathrm{a}}{\lim }\mathrm{f}\left(\mathrm{g}\left(\mathrm{x}\right)\right)=\mathrm{f}\left(\underset{x\to \mathrm{a}}{\lim }\mathrm{g}\left(\mathrm{x}\right)\right)$$$$\underset{x\to \pi /2}{\lim }\cos (2\mathrm{x}+\sin (\frac{3\pi }{2}+\mathrm{x}))=$$$$\cos (\underset{x\to \pi /2}{\lim }(2\mathrm{x}+\sin (\frac{3\pi }{2}+\mathrm{x}))=$$$$\cos (\pi +\sin 2\pi )=\cos \pi =-1$$

Answered by mathmax by abdo last updated on 25/Jan/21

$$\mathrm{f}\left(\mathrm{x}\right)=\cos (2\mathrm{x}+\sin (\frac{3\pi }{2}+\mathrm{x}))\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\cos (2\mathrm{x}+\sin (\pi +\frac{\pi }{2}+\mathrm{x}))$$$$=\cos (2\mathrm{x}-\mathrm{cosx})\Rightarrow {\lim }_{\mathrm{x}\to \frac{\pi }{2}}\mathrm{f}\left(\mathrm{x}\right)=\cos (\pi -0)=-1$$

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