△((♭ε)/(math))▽ lim_(x→π) ((sin^2 x)/(cos 3x+1))


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Question Number 109728 by bemath last updated on 25/Aug/20

$△ \frac{♭ ϵ}{m a t h} ▽$ $\lim_{x \to π} \frac{\sin^{2} x}{\cos 3 x + 1}$
Commented by bemath last updated on 25/Aug/20

$t h a n k y o u a l l m a s t e r$
	Answered by Her_Majesty last updated on 25/Aug/20

	$= - {l i m}_{x \to π} \frac{2 s i n x c o s x}{3 s i n 3 x}$ $= {l i m}_{x \to π} \frac{2 - 4 {c o s}^{2} x}{9 c o s 3 x}$ $= \frac{2}{9}$
	Answered by bobhans last updated on 25/Aug/20

	Answered by 1549442205PVT last updated on 25/Aug/20

	$Put π - t = x we have sinx = sint$ $\cos 3 x = \cos (3 π - 3 t) = \cos (π - 3 t) = - \cos 3 t . Hence$ $\lim_{x \to π} \frac{\sin^{2} x}{\cos 3 x + 1} = \lim_{t \to 0} \frac{\sin^{2} t}{- \cos 3 t + 1} =$ $\underset{L^{'} Hopital}{=} \lim_{t \to 0} \frac{2 sintcost}{3 \sin 3 t} = \lim_{t \to 0} \frac{2 sintcost}{3 (3 sint - {4 \sin}^{3} t)}$ $= \lim_{t \to 0} \frac{2 cost}{3 (3 - 4 \sin^{2} t)} = \frac{2.1}{3 (3 - 4.0)} = \frac{2}{9}$
	Answered by mathmax by abdo last updated on 25/Aug/20

	$let f (x) = \frac{\sin^{2} x}{1 + \cos (3 x)} changement x = t + π give$ $f (x) = \frac{\sin^{2} t}{1 + \cos (3 t + π)} = \frac{\sin^{2} t}{1 - \cos (3 t)} = g (t) (t \to 0)$ $\cos (3 t) \sim 1 - \frac{{9 t}^{2}}{2} \Rightarrow 1 - \cos (3 t) \sim \frac{{9 t}^{2}}{2} and \sin^{2} t \sim t^{2} \Rightarrow$ $g (t) \sim \frac{2}{9} \Rightarrow \lim_{x \to π} f (x) = \frac{2}{9}$
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