easy-question-if-lim-x-0-1-cos-1-cos-1-cos-x-x-8-2-a-then-a-

Question Number 141885 by mnjuly1970 last updated on 24/May/21

e a s y q u e s t i o n :

i f {l i m}_{x \to 0} \frac{1 - c o s (1 - c o s (1 - c o s (x)))}{x^{8}} = 2^{a}

t h e n a = ? ?

Answered by Dwaipayan Shikari last updated on 24/May/21

\lim_{x \to 0} \frac{1 - c o s (1 - c o s (1 - c o s (x)))}{x^{8}} = \frac{1 - c o s (1 - c o s \frac{x^{2}}{2})}{x^{8}} = \frac{1 - c o s (2 {s i n}^{2} \frac{x^{2}}{4})}{x^{8}}

= \frac{1 - c o s (\frac{x^{4}}{8})}{x^{8}} = \frac{2 {s i n}^{2} (\frac{x^{4}}{16})}{x^{8}} = \frac{1}{128} = 2^{- 7}

Commented by mnjuly1970 last updated on 24/May/21

t h a n k s a l o t . .

Answered by iloveisrael last updated on 24/May/21

\lim_{x \to 0} \frac{2 \sin^{2} (\frac{1 - \cos (1 - \cos x)}{2})}{x^{8}}

= \lim_{x \to 0} \frac{2 \sin^{2} (\frac{2 \sin^{2} (\frac{1 - \cos x}{2})}{2})}{x^{8}}

= \lim_{x \to 0} \frac{2 \sin^{2} (\frac{\cancel 2 s i n^{2} (\frac{\cancel 2 s i n^{2} (\frac{x}{2})}{\cancel 2})}{\cancel 2})}{x^{8}}

= \lim_{x \to 0} \frac{2 . \sin^{2} (\frac{x^{4}}{2^{4}})}{x^{8}}

= \lim_{x \to 0} \frac{2 . (\frac{x^{8}}{2^{8}})}{x^{8}} = \frac{1}{2^{7}}

Answered by mathmax by abdo last updated on 24/May/21

1 - cosx \sim \frac{x^{2}}{2} \Rightarrow \cos (1 - cosx) \sim \cos (\frac{x^{2}}{2}) \sim 1 - \frac{x^{4}}{8} \Rightarrow

1 - \cos (1 - cosx) \sim \frac{x^{4}}{8} \Rightarrow \cos (1 - \cos (1 - cosx)) \sim \cos (\frac{x^{4}}{8})

\sim 1 - \frac{x^{8}}{2.64} \Rightarrow 1 - \cos (1 - \cos (1 - cosx)) \sim \frac{x^{8}}{2^{7}} \Rightarrow

\lim_{x \to 0} \frac{1 - \cos (1 - \cos (1 - cosx))}{x^{8}} = \frac{1}{128} = 2^{- 7} \Rightarrow a = - 7