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Question Number 70886 by Omer Alattas last updated on 09/Oct/19

Commented by mathmax by abdo last updated on 10/Oct/19

$1 - c o s u \sim \frac{u^{2}}{2} a n d s i n u \sim u i f u \in V (0) \Rightarrow$ $1 - c o s (\frac{π t}{3}) \sim \frac{{(\frac{π t}{3})}^{2}}{2} = π^{2} \frac{t^{2}}{18} a n d s i n (\frac{π t}{3}) \sim \frac{π t}{3} . . . .$
Commented by mathmax by abdo last updated on 09/Oct/19
$let f(x)=((1−4sin^2 (((πx)/6)))/(1−x^2 )) ⇒f(x)=((1−4((1−cos(((πx)/3)))/2))/(1−x^2 )) =((1−2(1−cos(((πx)/3))))/(1−x^2 )) =((2cos(((πx)/3))−1)/(1−x^2 )) =((1−2cos(((πx)/3)))/(x^2 −1)) changement x−1=t give f(x)=g(t)=((1−2cos((π/3)(1+t)))/(t(t+2))) =((1−2cos((π/3)+((πt)/3)))/(t(t+2))) =((1−2{(1/2)cos(((πt)/3))−((√3)/2)sin(((πt)/3))})/(t(t+2))) =((1−cos(((πt)/3))+(√3)sin(((πt)/3)))/(t(t+2))) x→1 ⇔t→0 so 1−co(((πt)/3))∼(((((πt)/3))^2 )/2) and sin(((πt)/3))∼((πt)/3) ⇒ g(t)∼ ((((π^2 t^2 )/(18)) +(√3)×((πt)/3))/(t(t+2))) ∼ ((((π^2 t)/(18))+((π(√3))/3))/(t+2)) →((π(√3))/6) ⇒ lim_(x→1) f(x)=((π(√3))/6)$
$l e t f (x) = \frac{1 - 4 {s i n}^{2} (\frac{π x}{6})}{1 - x^{2}} \Rightarrow f (x) = \frac{1 - 4 \frac{1 - c o s (\frac{π x}{3})}{2}}{1 - x^{2}}$ $= \frac{1 - 2 (1 - c o s (\frac{π x}{3}))}{1 - x^{2}} = \frac{2 c o s (\frac{π x}{3}) - 1}{1 - x^{2}} = \frac{1 - 2 c o s (\frac{π x}{3})}{x^{2} - 1}$ $c h a n g e m e n t x - 1 = t g i v e f (x) = g (t) = \frac{1 - 2 c o s (\frac{π}{3} (1 + t))}{t (t + 2)}$ $= \frac{1 - 2 c o s (\frac{π}{3} + \frac{π t}{3})}{t (t + 2)} = \frac{1 - 2 {\frac{1}{2} c o s (\frac{π t}{3}) - \frac{\sqrt{3}}{2} s i n (\frac{π t}{3})}}{t (t + 2)}$ $= \frac{1 - c o s (\frac{π t}{3}) + \sqrt{3} s i n (\frac{π t}{3})}{t (t + 2)}$ $x \to 1 \Leftrightarrow t \to 0 s o 1 - c o (\frac{π t}{3}) \sim \frac{{(\frac{π t}{3})}^{2}}{2} a n d s i n (\frac{π t}{3}) \sim \frac{π t}{3} \Rightarrow$ $g (t) \sim \frac{\frac{π^{2} t^{2}}{18} + \sqrt{3} \times \frac{π t}{3}}{t (t + 2)} \sim \frac{\frac{π^{2} t}{18} + \frac{π \sqrt{3}}{3}}{t + 2} \to \frac{π \sqrt{3}}{6} \Rightarrow$ ${l i m}_{x \to 1} f (x) = \frac{π \sqrt{3}}{6}$
Commented by kaivan.ahmadi last updated on 09/Oct/19

Commented by Omer Alattas last updated on 09/Oct/19

$t h a n k s$
Commented by Omer Alattas last updated on 09/Oct/19

$I d o n t u n d e r s t s n d t h e l a s t s t a p c a n y o u$ $e x p l a n t h e l a s t s t a p$
Commented by Omer Alattas last updated on 10/Oct/19

$I d o n t k n o w w h a t i s t h i s 1 - c o s u \sim u^{2} / 2$ $c a n y o u t i l l m e h o w y o u g e t t h i s ? a n d w h a t s m e a n t h i s \sim$
Commented by mathmax by abdo last updated on 10/Oct/19

$i t s a f o r m u l a 1 - c o s u = 2 {s i n}^{2} (\frac{x}{2}) f o r x \in V (0)$ $s i n (\frac{x}{2}) \sim \frac{x}{2} \Rightarrow {s i n}^{2} (\frac{x}{2}) \sim \frac{x^{2}}{4} \Rightarrow 2 {s i n}^{2} (\frac{x}{2}) \sim \frac{x^{2}}{2}$ $\Rightarrow 1 - c o s u \sim \frac{x^{2}}{2} f o r x \in V (0)$
Commented by mathmax by abdo last updated on 11/Oct/19

$\sim i s t h e s y m b o l o f e q u i v a l e n c e . .$
Commented by Omer Alattas last updated on 11/Oct/19

$t h a n k y o u v a r y m u c h$
Commented by Omer Alattas last updated on 11/Oct/19

$t h i s f o r m u l a f o r x \in 0 o n l y r i g h t$
Commented by mathmax by abdo last updated on 11/Oct/19

$y e s .$
	Answered by Henri Boucatchou last updated on 09/Oct/19

	$\underset{x \to 1}{l i m} \frac{1 - 4 {s i n}^{2} (\frac{π}{6} x)}{1 - x^{2}} = \underset{x \to 1}{l i m} \frac{\frac{4 {s i n}^{2} (\frac{π}{6} 1) - 4 {s i n}^{2} (\frac{π}{6} x)}{1 - x}}{\frac{1 - x^{2}}{1 - x}} = \frac{{4 {s i n}^{2} (\frac{π}{6} x)]}_{(x = 1)}^{'}}{{x^{2}]}_{(x = 1)}^{'}} = \frac{{4 \times 2 \frac{π}{6} s i n (\frac{π}{6} x) c o s (\frac{π}{6} x)]}_{(x = 1)}}{2}$ $= \frac{\sqrt{3}}{6}$
	Answered by Henri Boucatchou last updated on 09/Oct/19

	$S o r r y, t h e a n s w e r i s \frac{\sqrt{3} π}{6} i n s t e d o f \frac{\sqrt{3}}{6}$
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