BeMath-lim-t-0-1-cos-2t-sin-pi-2-t-cos-2t-

Question Number 107516 by bemath last updated on 11/Aug/20

⊚ B e M a t h ⊚

\lim_{t \to 0} \frac{1 - \sqrt{\cos 2 t}}{\sin (\frac{π}{2} - t) - \cos 2 t} ?

Answered by bemath last updated on 11/Aug/20

\lim_{t \to 0} \frac{1 - \cos 2 t}{\cos t - \cos 2 t} \times \lim_{t \to 0} \frac{1}{1 + \sqrt{\cos 2 t}}

= \frac{1}{2} \times \lim_{t \to 0} \frac{2 \sin^{2} t}{- 2 \sin (\frac{3}{2} t) (\sin - \frac{t}{2})}

= \frac{1}{2} \times \lim_{t \to 0} \frac{2 \sin^{2} t}{2 \sin (\frac{3 t}{2}) \sin (\frac{t}{2})}

= \frac{1}{2} \times \frac{4}{3} = \frac{2}{3}

Answered by Dwaipayan Shikari last updated on 11/Aug/20

\lim_{t \to 0} \frac{1 - \sqrt{1 - \frac{4 t^{2}}{2}}}{c o s t - c o s 2 t} = \lim_{t \to 0} \frac{1 - 1 + t^{2}}{1 - \frac{t^{2}}{2} - 1 + \frac{4 t^{2}}{2}} = \lim_{x \to 0} \frac{t^{2}}{\frac{3 t^{2}}{2}} = \frac{2}{3}

Answered by mathmax by abdo last updated on 11/Aug/20

let f (t) = \frac{1 - \sqrt{\cos (2 t)}}{\sin (\frac{π}{2} - t) - \cos (2 t)} we have \cos (2 t) \sim 1 - {2 t}^{2}

\Rightarrow \sqrt{\cos (2 t)} \sim 1 - t^{2} \Rightarrow 1 - \sqrt{\cos (2 t)} \sim t^{2} also \sin (\frac{π}{2} - t) = cost \sim 1 - \frac{t^{2}}{2} \Rightarrow

\sin (\frac{π}{2} - t) - \cos (2 t) \sim 1 - \frac{t^{2}}{2} - 1 + {2 t}^{2} = \frac{3}{2} t^{2} \Rightarrow f (t) \sim \frac{3}{2} \Rightarrow

\lim_{t \to 0} f (t) = \frac{3}{2}

Commented by mathmax by abdo last updated on 11/Aug/20

sorry f (x) \sim \frac{2}{3} \Rightarrow \lim_{t \to 0} f (x) = \frac{2}{3}