math-lim-x-pi-sin-2-x-cos-3x-1-

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}