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..