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