lim-x-pi-4-pi-8-2-x-tan-x-sin-x-cos-x-

Question Number 159142 by tounghoungko last updated on 13/Nov/21

\lim_{x \to - \frac{π}{4}} \frac{\frac{π}{\sqrt{8}} - \sqrt{2} x . \tan x}{\sin x + \cos x} = ?

Commented by cortano last updated on 14/Nov/21

\lim_{x \to - \frac{π}{4}} \frac{\frac{π \cos x - 4 x \sin x}{\sqrt{8}}}{\sqrt{2} \sin (x + \frac{π}{4}) \cos x}

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

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

= \frac{1}{2 \sqrt{2}} (\frac{\frac{π}{\sqrt{2}} + 2 \sqrt{2} + \frac{π}{\sqrt{2}}}{1})

\underset{―}{= \frac{1}{2 \sqrt{2}} (2 \sqrt{2} + π \sqrt{2}) = \frac{2 + π}{2}}

Answered by FongXD last updated on 13/Nov/21

L = \lim_{x \to - \frac{π}{4}} \frac{\frac{π}{\sqrt{8}} (tanx + 1) - (\frac{π}{\sqrt{8}} + \sqrt{2} x) tanx}{\sqrt{2} (\sin \frac{π}{4} sinx + \cos \frac{π}{4} cosx)}

\Leftrightarrow L = \frac{1}{4} \lim_{x \to - \frac{π}{4}} \frac{\frac{π \sin (x + \frac{π}{4})}{cosxcos \frac{π}{4}} - (π + 4 x) tanx}{\cos (x - \frac{π}{4})}

\Leftrightarrow L = \frac{1}{4} \lim_{x \to - \frac{π}{4}} \frac{π \sin (x + \frac{π}{4}) - (π + 4 x) tanxcosxcos \frac{π}{4}}{\sin (x + \frac{π}{4})} \times \frac{1}{cosxcos \frac{π}{4}}

\Leftrightarrow L = \frac{π}{2} - \frac{1}{2} \lim_{x \to - \frac{π}{4}} \frac{π + 4 x}{\sin (x + \frac{π}{4})} \times tanxcosxcos \frac{π}{4}

\Leftrightarrow L = \frac{π}{2} + \frac{1}{4} \lim_{x \to - \frac{π}{4}} \frac{π + 4 x}{\sin (x + \frac{π}{4})}

put t = x + \frac{π}{4}, if x \to - \frac{π}{4}, \Rightarrow t \to 0

we get : L = \frac{π}{2} + \frac{1}{4} \lim_{t \to 0} \frac{4 t}{sint} = \frac{π}{2} + \frac{1}{4} (4 \times 1) = \frac{π}{2} + 1

so . \begin{matrix} L = \lim_{x \to - \frac{π}{4}} \frac{\frac{π}{\sqrt{8}} - \sqrt{2} xtanx}{sinx + cosx} = \frac{π}{2} + 1 \end{matrix}

Answered by tounghoungko last updated on 14/Nov/21

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

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

= \frac{1}{2 \sqrt{2}} \lim_{x \to 0} \frac{(π \sqrt{2} - 2 \sqrt{2} x) \sin x + 2 \sqrt{2} x \cos x}{\sin x}

= \frac{1}{2 \sqrt{2}} \lim_{x \to 0} (π \sqrt{2} - 2 \sqrt{2} x + 2 \sqrt{2} \cos x)

= \frac{π \sqrt{2} + 2 \sqrt{2}}{2 \sqrt{2}} = \frac{π}{2} + 1