lim-x-pi-2-4sin-x-6-sin-x-10-pi-2-x-

Question Number 100134 by bemath last updated on 25/Jun/20

\lim_{x \to \frac{π}{2}} \frac{4 \sin x - \sqrt{6 \sqrt{\sin x} + 10}}{\frac{π}{2} - x} ?

Commented by bobhans last updated on 25/Jun/20

set \frac{π}{2} - x = t, x = \frac{π}{2} - t

\lim_{t \to 0} \frac{4 \cos t - \sqrt{6 \sqrt{\cos t} + 10}}{t} =

\lim_{t \to 0} \frac{16 \cos^{2} t - (6 \sqrt{\cos t} + 10)}{8 t} =

\lim_{t \to 0} \frac{- 16 \sin 2 t - \frac{6 (- \sin t)}{2 \sqrt{\cos t}}}{8} = 0

Commented by Dwaipayan Shikari last updated on 25/Jun/20

\lim_{x \to \frac{π}{2}} \frac{- 4 cosx - \frac{\frac{1}{2} . \frac{3}{\sqrt{sinx}} cosx}{\sqrt{6 \sqrt{sinx} + 10}}}{- 1} = 0

Answered by mathmax by abdo last updated on 25/Jun/20

changement \frac{π}{2} - x = t give f (x) = \frac{4 sinx - \sqrt{6 \sqrt{sinx} + 10}}{\frac{π}{2} - x}

= \frac{4 cost - \sqrt{6 \sqrt{cost} + 10}}{t} = g (t) we have cost \sim 1 - \frac{t^{2}}{2} \Rightarrow \sqrt{cost} \sim 1 - \frac{t^{2}}{4} \Rightarrow

6 \sqrt{cost} \sim 6 - \frac{3}{2} t^{2} \Rightarrow \sqrt{6 \sqrt{cost} + 10} \sim \sqrt{16 - \frac{3}{2} t^{2}} = 4 \sqrt{1 - \frac{{3 t}^{2}}{32}} \sim 4 (1 - \frac{{3 t}^{2}}{64}) \Rightarrow

g (t) \sim \frac{4 - {2 t}^{2} - 4 + \frac{{3 t}^{2}}{16}}{t} = (- 2 + \frac{3}{16}) t \to 0 \Rightarrow \lim_{x \to \frac{π}{2}} f (x) = 0