2
$\begingroup$

Find the limit when $x$ approaches zero of $\lim\limits_{x \to 0}{\frac{1-\cos(1-\cos x)}{x^4}}$

My teacher already told us that the result is $1/8$

  • 4
    Got to applaud the honest approach.2012-11-10

2 Answers 2

6

I use that $\lim\limits_{x\to 0}\frac{1-\cos x}{x^2}=\frac 1 2$

Now, consider the following manipulation $\lim\limits_{x\to 0}\frac{1-\cos(1-\cos x )}{x^4}=\\ \lim\limits_{x\to 0}\frac{1-\cos(1-\cos x )}{(1-\cos x )^2}\frac {(1-\cos x )^2}{x^4}=\\ \lim\limits_{x\to 0}\frac{1-\cos(1-\cos x )}{(1-\cos x )^2}\left(\frac {1-\cos x }{x^2}\right)^2=$

When $x\to 0$, $1-\cos x \to 0$, so

$\lim\limits_{u\to 0}\frac{1-\cos u}{u^2}\lim\limits_{x\to 0}\left(\frac {1-\cos x }{x^2}\right)^2=\frac 1 2 \frac 1 4=\frac 1 8$

  • 0
    Nice solution..2012-11-11
5

What about the L'Hospital rule? $\frac{1-\cos(1-\cos x)}{x^4} $ Differentiate both the nominator and the denominator: $\big(1-\cos(1-\cos x)\big)' = \sin(1-\cos x)\cdot(1-\cos x)' = \sin(1-\cos x)\cdot\sin x $ and so on..

  • 0
    @Daryl, Cameron was reacting to an incorrect version of the expression that was visible for about ten minutes, until I fixed it to agree with the actual question asked. You are doing fine.2012-11-10