lim-x-pi-4-pi-4x-1-sin-2x-

Question Number 143995 by liberty last updated on 20/Jun/21

\lim_{x \to π / 4} \frac{π - 4 x}{\sqrt{1 - \sqrt{\sin 2 x}}} = ?

Answered by mathmax by abdo last updated on 20/Jun/21

f (x) = \frac{π - 4 x}{\sqrt{1 - \sqrt{\sin 2 x}}} \Rightarrow f (x) =_{\frac{π}{4} - x = t} \frac{π - 4 (\frac{π}{4} - t)}{\sqrt{1 - \sqrt{\sin (2 (\frac{π}{4} - t)}}}

= \frac{4 t}{\sqrt{1 - \sqrt{\sin (\frac{π}{2} - 2 t)}}} = \frac{4 t}{\sqrt{1 - \sqrt{\cos (2 t)}}} = g (t) (t \to 0)

\cos (2 t) \sim 1 - \frac{{4 t}^{2}}{2} = 1 - {2 t}^{2} \Rightarrow \sqrt{\cos (2 t)} \sim \sqrt{1 - {2 t}^{2}} \sim 1 - \frac{1}{2} ({2 t}^{2}) = 1 - t^{2}

- \sqrt{\cos (2 t)} \sim t^{2} - 1 \Rightarrow 1 - \sqrt{\cos (2 t)} \sim t^{2} \Rightarrow \sqrt{1 - \sqrt{\cos (2 t)}} \sim t \Rightarrow

g (t) \sim \frac{4 t}{t} \Rightarrow \lim_{t \to 0} g (t) = 4

Commented by liberty last updated on 21/Jun/21

in my book the limit doesnot exist

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