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Question Number 72836 by aliesam last updated on 03/Nov/19

Commented by mathmax by abdo last updated on 04/Nov/19
$changement x=(π/3)+t give lim_(x→(π/3)) (((√3)cosx−sinx)/(1−2cosx)) =lim_(t→0) (((√3)cos((π/3)+t)−sin((π/3)+t))/(1−2cos((π/3)+t))) =lim_(t→0) (((√3)(cos((π/3))cost −sin((π/3))sint)−(sin((π/3))cost +cos((π/3))sint))/(1−2{cos((π/3))cost −sin((π/3))sint})) =lim_(t→0) (((√3)((1/2)cost−((√3)/2)sint)−((√3)/2)cost−(1/2)sint)/(1−2{(1/2)cost−((√3)/2)sint})) =lim_(t→0) ((−2sint)/(1−cost+(√3)sint)) =lim_(t→0) ((−2((sint)/t))/(((1−cost)/t)+(√3)((sint)/t))) =((−2)/(0+(√3))) =−(2/(√3)) because lim_(t→0) ((sint)/t) =1 and lim_(t→0) ((1−cost)/t)=0$
$c h a n g e m e n t x = \frac{π}{3} + t g i v e$ ${l i m}_{x \to \frac{π}{3}} \frac{\sqrt{3} c o s x - s i n x}{1 - 2 c o s x} = {l i m}_{t \to 0} \frac{\sqrt{3} c o s (\frac{π}{3} + t) - s i n (\frac{π}{3} + t)}{1 - 2 c o s (\frac{π}{3} + t)}$ $= {l i m}_{t \to 0} \frac{\sqrt{3} (c o s (\frac{π}{3}) c o s t - s i n (\frac{π}{3}) s i n t) - (s i n (\frac{π}{3}) c o s t + c o s (\frac{π}{3}) s i n t)}{1 - 2 {c o s (\frac{π}{3}) c o s t - s i n (\frac{π}{3}) s i n t}}$ $= {l i m}_{t \to 0} \frac{\sqrt{3} (\frac{1}{2} c o s t - \frac{\sqrt{3}}{2} s i n t) - \frac{\sqrt{3}}{2} c o s t - \frac{1}{2} s i n t}{1 - 2 {\frac{1}{2} c o s t - \frac{\sqrt{3}}{2} s i n t}}$ $= {l i m}_{t \to 0} \frac{- 2 s i n t}{1 - c o s t + \sqrt{3} s i n t} = {l i m}_{t \to 0} \frac{- 2 \frac{s i n t}{t}}{\frac{1 - c o s t}{t} + \sqrt{3} \frac{s i n t}{t}}$ $= \frac{- 2}{0 + \sqrt{3}} = - \frac{2}{\sqrt{3}} b e c a u s e {l i m}_{t \to 0} \frac{s i n t}{t} = 1 a n d {l i m}_{t \to 0} \frac{1 - c o s t}{t} = 0$
	Answered by $@ty@m123 last updated on 04/Nov/19

	$= \lim_{x \to \frac{π}{3}} \frac{2 (\frac{\sqrt{3}}{2} \cos x - \frac{1}{2} \sin x)}{2 (\frac{1}{2} - \cos x)}$ $= \lim_{x \to \frac{π}{3}} \frac{\sin (\frac{π}{3} - x)}{\cos \frac{π}{3} - \cos x}$ $= \lim_{x \to \frac{π}{3}} \frac{2 \sin (\frac{π}{6} - \frac{x}{2}) \cos (\frac{π}{6} - \frac{x}{2})}{- 2 \sin (\frac{π}{6} + \frac{x}{2}) \sin (\frac{π}{6} - \frac{x}{2})}$ $= \lim_{x \to \frac{π}{3}} \frac{\cos (\frac{π}{6} - \frac{x}{2})}{- \sin (\frac{π}{6} + \frac{x}{2})}$ $= \frac{\cos 0}{- \sin \frac{π}{3}}$ $= - \frac{1}{\frac{\sqrt{3}}{2}}$ $= - \frac{2}{\sqrt{3}}$
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