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Find a limit, $$\lim_{x\to 0} \frac{1-\cos{(1-\cos{(1-\cos x)})}}{x^8}$$ without using Maclaurin series. My attempt was to use L'hopital's rule but that's just too much, and chances of not making a mistake trough repetitive differentiation are very low.

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Hint: Use repetitively "$1-\cos x=2\sin^2 x/2$" $$\newcommand{\b}[1]{\left(#1\right)} \newcommand{\f}{\frac} \newcommand{\t}{\text} \newcommand{\u}{\underbrace} $$ $$\lim_{x\to0}\frac{1-\cos{(1-\cos{(1-\cos x)})}}{x^8} =\lim_{x\to0}\f{2\sin^2\b{\sin^2\b{\sin^2\b{\f x2}}}}{x^8}\\ =\lim_{x\to0}\f{2\sin^2\b{\color{red}{\sin^2\b{\color{blue}{\sin^2\b{\color{green}{\f x2}}}}}}}{\b{\color{red}{\sin^2\b{\color{blue}{\sin^2\b{\color{green}{\f x2}}}}}}^2}.\f{\b{\color{red}{\sin^2\b{\color{blue}{\sin^2\b{\color{green}{\f x2}}}}}}^2}{\b{\color{blue}{\sin^2\b{\color{green}{\f x2}}}}^4}.\f{\b{\color{blue}{\sin^2\b{\color{green}{\f x2}}}}^4}{\b{\color{green}{\f x2}}^8}.\b{\f 12}^8 \\=\lim_{x\to0}\f1{128}\b{\f{\sin\b{\color{fuchsia}{\sin^2\b{\sin^2\b{\f x2}}}}}{\color{fuchsia}{\sin^2\b{\sin^2\b{\f x2}}}}}^2.\b{\f{\sin\b{\color{purple}{\sin^2\b{\f x2}}}}{\color{purple}{\sin^2\b{\f x2}}}}^4.\b{\f{\sin\b{\color{crimson}{\f x2}}}{\color{crimson}{\f x2}}}^8 \\={\large\f1{128}}\quad\b{\because \f{\sin x}x=1}$$

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    $\begingroup$ I wanted to give the same hint! $\endgroup$ – Mhenni Benghorbal Jan 24 '15 at 9:55
  • $\begingroup$ @Transcedental edited $\endgroup$ – RE60K Jan 24 '15 at 10:17
  • $\begingroup$ @sciona did you see this and this $\endgroup$ – RE60K Jan 24 '15 at 10:21
  • $\begingroup$ @ADG =P I see .. nice formatting. $\endgroup$ – sciona Jan 24 '15 at 10:23
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Hint: Use $1-\cos t = 2\sin^2 \frac{t}{2}$ and the chain rule for limit. (ask if you need further hint)

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  • $\begingroup$ Oh i'm an idiot. Thank you. $\endgroup$ – Transcendental Jan 24 '15 at 10:05
  • $\begingroup$ @Transcendental wc :) $\endgroup$ – sciona Jan 24 '15 at 10:11

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