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Question Number 124273 by bramlexs22 last updated on 02/Dec/20

$$\underset{x\to \pi }{\lim }\frac{\sqrt{7+2\sin x}-\sqrt{7-2\sin x}}{\tan x}=?$$

Answered by Dwaipayan Shikari last updated on 02/Dec/20

$$\underset{x\to \pi }{\lim }\frac{\sqrt{7+2(\pi -x)}-\sqrt{7-2\pi +2x}}{\pi -x}=\sqrt{7}\left(\frac{\sqrt{1+\frac{2}{7}(\pi -x)}-\sqrt{1-\frac{2}{7}(\pi -x)}}{\pi -x}\right)$$$$=\sqrt{7}\left(\frac{1+\frac{1}{7}(\pi -x)-1+\frac{1}{7}(\pi -x)}{\pi -x}\right)cosx=-\sqrt{7}\left(\frac{2}{7}\right)=-\frac{2}{\sqrt{7}}$$

Answered by liberty last updated on 02/Dec/20

$$\underset{x\to \pi }{\lim }\frac{(7+2\sin x)-(7-2\sin x)}{(\sqrt{7+2\sin x}+\sqrt{7-2\sin x}).\tan x}=$$$$\underset{x\to \pi }{\lim }\frac{1}{\sqrt{7+2\sin x}+\sqrt{7-2\sin x}}\times \underset{x\to \pi }{\lim }\frac{4\sin x}{\tan x}$$$$=\frac{1}{2\sqrt{7}}\times \underset{x\to \pi }{\lim }4\sin x\left(\frac{\cos x}{\sin x}\right)$$$$=\frac{1}{2\sqrt{7}}\times (-4)=-\frac{2}{\sqrt{7}}.$$

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