lim-x-pi-4-4-2-cos-x-sin-x-5-cos-x-sin-x-2-

Question Number 119725 by bemath last updated on 26/Oct/20

\lim_{x \to \frac{π}{4}} \frac{4 \sqrt{2} - {(\cos x + \sin x)}^{5}}{{(\cos x - \sin x)}^{2}} = ?

Commented by bemath last updated on 26/Oct/20

c o m p a r e v i a L^{'} H o p i t a l

\lim_{x \to \frac{π}{4}} \frac{- 5 {(\cos x + \sin x)}^{4} (- \sin x + \cos x)}{2 (\cos x - \sin x) (- \cos x - \sin x)}

\lim_{x \to \frac{π}{4}} \frac{- 5 {(\cos x + \sin x)}^{4} (\cos x - \sin x)}{- 2 (\cos x - \sin x) (\cos x + \sin x)} =

\lim_{x \to \frac{π}{4}} \frac{5}{2} {(\cos x + \sin x)}^{3} =

\frac{5}{2} {(\sqrt{2})}^{3} = \frac{5}{2} \times 2 \sqrt{2} = 5 \sqrt{2}

Answered by benjo_mathlover last updated on 26/Oct/20

\lim_{x \to \frac{π}{4}} \frac{4 \sqrt{2} - 4 \sqrt{2} \sin^{5} (x + \frac{π}{4})}{1 - \sin 2 x} =

[l e t x = \frac{π}{4} + z]

\lim_{z \to 0} \frac{4 \sqrt{2} (1 - \sin^{5} (z + \frac{π}{2}))}{1 - \sin (2 z + \frac{π}{2})} =

\lim_{z \to 0} \frac{4 \sqrt{2} (1 - \cos^{5} z)}{1 - \cos 2 z} = \lim_{z \to 0} \frac{4 \sqrt{2} (1 - {(1 - \frac{z^{2}}{2})}^{5})}{1 - (1 - \frac{4 z^{2}}{2})}

= \lim_{z \to 0} \frac{4 \sqrt{2} (1 - (1 - \frac{5 z^{2}}{2}))}{2 z^{2}} = \lim_{z \to 0} \frac{2 \sqrt{2} (\frac{5 z^{2}}{2})}{z^{2}} = 5 \sqrt{2}

Answered by Dwaipayan Shikari last updated on 26/Oct/20

\lim_{x \to \frac{π}{4}} \frac{4 \sqrt{2} (1 - {c o s}^{5} (\frac{π}{4} - x))}{{(c o s x - s i n x)}^{2}}

\lim_{x \to \frac{π}{4}} \frac{4 \sqrt{2} (1 - {c o s}^{5} (\frac{π}{4} - x))}{2 {c o s}^{2} (\frac{π}{4} + x)}

\underset{x \to \frac{π}{4}}{\lim 2} \sqrt{2} \frac{1 - {c o s}^{5} (\frac{π}{4} - x)}{{s i n}^{2} (\frac{π}{4} - x)}

\underset{x \to \frac{π}{4}}{\lim 2} \sqrt{2} (1 + c o s (\frac{π}{4} - x) + {c o s}^{2} (\frac{π}{4} - x) + {c o s}^{3} (\frac{π}{4} - x) + {c o s}^{4} (\frac{π}{4} - x)) \frac{1 - c o s (\frac{π}{4} - x)}{{s i n}^{2} (\frac{π}{4} - x)}

2 \sqrt{2} (1 + 1 + 1 + 1 + 1) \frac{2 {(\frac{π}{4} - x)}^{2}}{4 . {(\frac{π}{4} - x)}^{2}} = 5 \sqrt{2}

Answered by Bird last updated on 26/Oct/20

f (x) = \frac{4 \sqrt{2} - {(c o s x + s i n x)}^{5}}{{(c o s x - s i n x)}^{2}}

\Rightarrow f (x) = \frac{4 \sqrt{2} - (\sqrt{2} c o s {(x - \frac{π}{4})}^{5}}{{(\sqrt{2} c o s (x + \frac{π}{4}))}^{2}}

= \frac{4 \sqrt{2} - 4 \sqrt{2} {c o s}^{5} (x - \frac{π}{4})}{2 {c o s}^{2} (x + \frac{π}{4})}

=_{x - \frac{π}{4} = t} 2 \sqrt{2} \times \frac{1 - {c o s}^{5} t}{{c o s}^{2} (t + \frac{π}{2})} = f (t + \frac{π}{4})

= 2 \sqrt{2} \times \frac{1 - {c o s}^{5} t}{{s i n}^{2} t}

= 2 \sqrt{2} \times \frac{(1 - c o s t) (1 + c o s t + {c o s}^{2} t + {c o s}^{3} t + {c o s}^{4} t)}{{s i n}^{2} t}

f (t + \frac{π}{4}) \sim 2 \sqrt{2} \times \frac{\frac{t^{2}}{2}}{t^{2}} (1 + c o s t + \dots + {c o s}^{4} t)

= \sqrt{2} (1 + c o s t + \dots + {c o s}^{4} t) \Rightarrow

{l i m}_{t \to 0} f (t + \frac{π}{4}) = 5 \sqrt{2} = {l i m}_{x \to \frac{π}{4}} f (x)

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