1-lim-x-pi-2-cos-2x-tan-x-2-lim-x-pi-2-cos-x-1-pi-x-2-

Question Number 115604 by bemath last updated on 27/Sep/20

(1) \lim_{x \to \frac{π}{2}} \frac{\cos 2 x}{\tan x} = ?

(2) \lim_{x \to π} \frac{\sqrt{2 + \cos x} - 1}{{(π - x)}^{2}} = ?

Commented by Dwaipayan Shikari last updated on 27/Sep/20

\lim_{x \to π} \frac{2 + cosx - 1}{{(π - x)}^{2}} . \frac{1}{\sqrt{2 + cosx} + 1} = \frac{1}{4} \frac{{2 \cos}^{2} \frac{x}{2}}{{(\frac{π}{2} - \frac{x}{2})}^{2}} . \frac{1}{2}

= \frac{1}{4} . \frac{\sin^{2} (\frac{π}{2} - \frac{x}{2})}{{(\frac{π}{2} - \frac{x}{2})}^{2}} = \frac{1}{4}

Answered by bobhans last updated on 27/Sep/20

(2) \lim_{x \to π} \frac{\frac{- \sin x}{2 \sqrt{2 + \cos x}}}{- 2 (π - x)} = \lim_{x \to π} \frac{1}{4 \sqrt{2 + \cos x}} \times \lim_{x \to π} \frac{\sin x}{π - x}

= \frac{1}{4} \times \lim_{x \to π} \frac{\cos x}{- 1} = - \frac{1}{4} \times (- 1) = \frac{1}{4}

Answered by Olaf last updated on 27/Sep/20

(2)

x = π - u

\lim_{u \to 0} \frac{\sqrt{2 - \cos u} - 1}{u^{2}}

\lim_{u \to 0} \frac{\sqrt{2 - (1 - \frac{u^{2}}{2})} - 1}{u^{2}}

\lim_{u \to 0} \frac{\sqrt{1 + \frac{u^{2}}{2}} - 1}{u^{2}}

\lim_{u \to 0} \frac{1 + \frac{u^{2}}{4} - 1}{u^{2}}

\lim_{u \to 0} \frac{\frac{u^{2}}{4}}{u^{2}} = \frac{1}{4}

Answered by mathmax by abdo last updated on 27/Sep/20

let f (x) = \frac{\cos (2 x)}{tanx} changement t = x - \frac{π}{2} give

f (x) = \frac{\cos (2 (t + \frac{π}{2}))}{\tan (t + \frac{π}{2})} = - \frac{\cos (2 t)}{- \frac{1}{tant}} = tant . \cos (2 t)

(x \to \frac{π}{2} \Rightarrow t \to 0) \Rightarrow \tan (t) \sim t and \cos (2 t) \sim 1 - {2 t}^{2} \Rightarrow

tant . \cos (2 t) \sim t (1 - {2 t}^{2}) \to 0 \Rightarrow \lim_{x \to \frac{π}{2}} f (x) = 0

Answered by mathmax by abdo last updated on 27/Sep/20

2) let g (x) = \frac{\sqrt{2 + cosx} - 1}{{(π - x)}^{2}} we do the changement π - x = t \Rightarrow

g (x) = g (π - t) = \frac{\sqrt{2 + \cos (π - t)} - 1}{t^{2}} = \frac{\sqrt{2 - cost} - 1}{t^{2}}

we have cost \sim 1 - \frac{t^{2}}{2} \Rightarrow - cost \sim - 1 + \frac{t^{2}}{2} \Rightarrow 2 - cost \sim 1 + \frac{t^{2}}{2}

\Rightarrow \sqrt{2 - cost} \sim \sqrt{1 + \frac{t^{2}}{2}} \sim 1 + \frac{t^{2}}{4} \Rightarrow g (π - t) \sim \frac{t^{2}}{{4 t}^{2}} = \frac{1}{4} \Rightarrow

\lim_{t \to 0} g (π - t) = \frac{1}{4} = \lim_{x \to π} g (x)

Answered by bobhans last updated on 27/Sep/20

(1) s e t x = z + \frac{π}{2}

\lim_{z \to 0} \frac{\cos (2 z + π)}{\tan (z + \frac{π}{2})} = \lim_{z \to 0} \frac{- \cos 2 z}{- \cot z}

= \lim_{z \to 0} \tan z . \cos 2 z = 0 .

Answered by Olaf last updated on 27/Sep/20

\lim_{x \to \frac{π}{2}} \cos 2 x = - 1

\lim_{x \to \frac{π}{2}} \tan x = \pm \infty

\Rightarrow \lim_{x \to \frac{π}{2}} \frac{\cos 2 x}{\tan x} = 0

Facebook Tweet Pin