((♥JS♥)/(≤°≡°≤)) lim_(x→π/6) ((2−(√3) cos x−sin x)/((6x−π)^2 )) ?


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Question Number 108217 by john santu last updated on 15/Aug/20

$\frac{♡ J S ♡}{⩽ ° \equiv ° ⩽}$ $\lim_{x \to π / 6} \frac{2 - \sqrt{3} \cos x - \sin x}{{(6 x - π)}^{2}} ?$
Commented by john santu last updated on 15/Aug/20

$g o o d a l l s i r . . .$
	Answered by bemath last updated on 15/Aug/20
	$((BeMath)/(★°•°★)) lim_(x→π/6) ((2−(√3) cos x−sin x)/((6x−π)^2 )) ? set x = (π/6)+w ⇒6x=π+6w lim_(w→0) ((2−{(√3) cos (w+(π/6))+sin (w+(π/6))})/((6w)^2 ))= lim_(w→0) ((2−2cos (w+(π/6)−(π/6)))/(36w^2 ))= lim_(w→0) ((2(1−cos w))/(36w^2 )) = lim_(w→0) ((2(2sin^2 ((w/2))))/(36w^2 )) = 4×(1/4)×(1/(36)) = (1/(36))$
	$\frac{B e M a t h}{★ ° ∙ ° ★}$ $\lim_{x \to π / 6} \frac{2 - \sqrt{3} \cos x - \sin x}{{(6 x - π)}^{2}} ?$ $s e t x = \frac{π}{6} + w \Rightarrow 6 x = π + 6 w$ $\lim_{w \to 0} \frac{2 - {\sqrt{3} \cos (w + \frac{π}{6}) + \sin (w + \frac{π}{6})}}{{(6 w)}^{2}} =$ $\lim_{w \to 0} \frac{2 - 2 \cos (w + \frac{π}{6} - \frac{π}{6})}{36 w^{2}} =$ $\lim_{w \to 0} \frac{2 (1 - \cos w)}{36 w^{2}} = \lim_{w \to 0} \frac{2 (2 \sin^{2} (\frac{w}{2}))}{36 w^{2}}$ $= 4 \times \frac{1}{4} \times \frac{1}{36} = \frac{1}{36}$
	Answered by 1549442205PVT last updated on 15/Aug/20

	$Put x - \frac{π}{6} = t we have 6 x = 6 t + π$ $\lim_{x \to \frac{π}{6}} \frac{2 - \sqrt{3} \cos x - \sin x}{{(6 x - π)}^{2}} = \lim_{t \to 0} \frac{2 - \sqrt{3} \cos (t + \frac{π}{6}) - \sin (t + \frac{π}{6})}{{36 t}^{2}}$ $= \lim_{t \to 0} \frac{2 - \sqrt{3} [\frac{\sqrt{3}}{2} cost - \frac{1}{2} sint) - \frac{\sqrt{3}}{2} sint - \frac{1}{2} cost}{{36 t}^{2}}$ $Extra \left or missing \right$ $= \frac{1}{36}$
	Answered by Dwaipayan Shikari last updated on 15/Aug/20

	$\lim_{x \to \frac{π}{6}} \frac{\sqrt{3} s i n x - c o s x}{2.6 . (6 x - π)} = \frac{\sqrt{3} c o s x + s i n x}{2.6.6} = \frac{\frac{3}{2} + \frac{1}{2}}{2.6.6} = \frac{1}{36}$
	Answered by mathmax by abdo last updated on 15/Aug/20
	$let f(x) =((2−(√3)cosx−sinx)/(36(x−(π/6))^2 )) changement x−(π/6)=t give f(x)=((2−(√3)cos(t+(π/6))−sin(t+(π/6)))/(36t^2 )) (x→(π/6)⇒t→0) =((2−(√3){((√3)/2)cost−(1/2)sint}−(((√3)/2)sint +(1/2)cost))/(36t^2 )) =((2−2cost )/(36t^2 ))=((1−cost)/(18t^2 )) ∼((t^2 /2)/(18t^2 )) =(1/(36)) (t→0) ⇒lim_(x→(π/6)) f(x)=(1/(36))$
	$let f (x) = \frac{2 - \sqrt{3} cosx - sinx}{36 {(x - \frac{π}{6})}^{2}} changement x - \frac{π}{6} = t give$ $f (x) = \frac{2 - \sqrt{3} \cos (t + \frac{π}{6}) - \sin (t + \frac{π}{6})}{{36 t}^{2}} (x \to \frac{π}{6} \Rightarrow t \to 0)$ $= \frac{2 - \sqrt{3} {\frac{\sqrt{3}}{2} cost - \frac{1}{2} sint} - (\frac{\sqrt{3}}{2} sint + \frac{1}{2} cost)}{{36 t}^{2}}$ $= \frac{2 - 2 cost}{{36 t}^{2}} = \frac{1 - cost}{{18 t}^{2}} \sim \frac{\frac{t^{2}}{2}}{{18 t}^{2}} = \frac{1}{36} (t \to 0) \Rightarrow \lim_{x \to \frac{π}{6}} f (x) = \frac{1}{36}$
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