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Question Number 109091 by bemath last updated on 21/Aug/20

$$\frac{△\mathrm{♭}e\mathcal{M}ath▽}{\equiv ⊸\equiv ⊸\equiv }$$$$\underset{x\to \pi }{\lim }\frac{\sin x}{\sqrt{\pi +\tan x}-\sqrt{\pi -\tan x}}?$$

Answered by bobhans last updated on 21/Aug/20

Commented by bobhans last updated on 21/Aug/20

$$typothelastline$$$$\underset{x\to \pi }{\lim }\frac{\sin x.\cos x}{2\sin x}\times 2\sqrt{\pi }=-\sqrt{\pi }$$

Answered by bemath last updated on 21/Aug/20

$$setx=\pi +q$$$$\underset{q\to 0}{\lim }\frac{-\sin q}{\sqrt{\pi +\tan q}-\sqrt{\pi -\tan q}}=$$$$\underset{q\to 0}{\lim }\frac{-\sin q}{\sqrt{\pi }\{\sqrt{1+\frac{\tan q}{\pi }}-\sqrt{1-\frac{\tan q}{\pi }}\}}=$$$$\frac{1}{\sqrt{\pi }}.\underset{q\to 0}{\lim }\frac{-q}{(1+\frac{q}{2\pi })-(1-\frac{q}{2\pi })}=$$$$\frac{1}{\sqrt{\pi }}.\underset{q\to 0}{\lim }\frac{-q}{\left(\frac{q}{\pi }\right)}=\frac{-\pi }{\sqrt{\pi }}=-\sqrt{\pi }.$$

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