lim_(x→0) ((x tan x)/( (√3) cos x−sin^2 x−(√3))) ?


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Question Number 114405 by bemath last updated on 19/Sep/20

$\lim_{x \to 0} \frac{x \tan x}{\sqrt{3} \cos x - \sin^{2} x - \sqrt{3}} ?$
	Answered by bobhans last updated on 19/Sep/20

	$\lim_{x \to 0} \frac{x \tan x}{\sqrt{3} (\cos x - 1) - \sin^{2} x} =$ $\lim_{x \to 0} \frac{x^{2}}{\sqrt{3} (1 - \frac{x^{2}}{2} - 1) - x^{2}} =$ $\lim_{x \to 0} \frac{x^{2}}{x^{2} (- \frac{\sqrt{3}}{2} - 1)} = \frac{- 2}{\sqrt{3} + 2} = \frac{- 2 (2 - \sqrt{3})}{4 - 3}$ $= 2 \sqrt{3} - 2$
	Commented by bemath last updated on 19/Sep/20

	$g a v e k u d o s ✓ ≎$
	Answered by Dwaipayan Shikari last updated on 19/Sep/20

	$\lim_{x \to 0} \frac{x t a n x}{\sqrt{3} (c o s x - 1) - {s i n}^{2} x} = \frac{x^{2}}{- 2 \sqrt{3} {s i n}^{2} \frac{x}{2} - 4 {s i n}^{2} \frac{x}{4} {c o s}^{2} \frac{x}{2}}$ $= \frac{x^{2}}{\frac{- \sqrt{3}}{2} x^{2} - x^{2}} = - \frac{2}{\sqrt{3} + 2} = 2 (\sqrt{3} - 2)$
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