Whats the limit: $\lim\limits_{x \to 0} \frac{1-\sqrt[3]{\cos x}}{x(1-\cos\sqrt{x})}$ I am stuck at this one, i want to find it without using L'Hôpital's Rule.
$$\lim_{x \to 0} \frac{1-\sqrt[3]{\cos x}}{x(1-\cos\sqrt{x})}$$
 A: $1-\cos \alpha=2\sin^2\frac{\alpha}2$,  $1-\cos \alpha\sim \frac {\alpha^2} 2(\alpha\to0)$, so
\begin{equation}
\begin{split}\lim_{x \to 0} \frac{1-\sqrt[3]{\cos x}}{x(1-\cos\sqrt{x})} &=\lim_{x \to 0} \frac{1-\cos x}{x(1-\cos\sqrt{x})(1+\sqrt[3]{\cos x}+\sqrt[3]{\cos^2 x})}\\&=\frac13 \lim_{x \to 0} \frac{1-\cos x}{x(1-\cos\sqrt{x})}\\
&=\frac13 \lim_{x \to 0} \frac{\frac {x^2} 2}{x\cdot\frac x2}\\
&=\frac13
\end{split}
\end{equation}
A: Use Taylor expansion on the $cos$ terms. Then Taylor expand the cube root.
Now if you take the limit, you will get the limit to be $\frac{1}{3}$.
A: Some problems are best handled by breaking them into bite-sized pieces.  To begin with, let's get rid of the cube root by writing
$$1-\sqrt[3]{\cos x}={1-\cos x\over 1+\sqrt[3]{\cos x}+\sqrt[3]{\cos^2x}}$$
and noting that the denominator just becomes a $3$ at $x=0$, so that we now have
$$\lim_{x\to0^+}\left({1-\sqrt[3]{\cos x}\over x(1-\cos\sqrt x)}\right)={1\over3}\lim_{x\to0^+}\left({1-\cos x\over x(1-\cos\sqrt x)}\right)$$
(Note, I inserted a superscript into $0^+$ for the limit, out of respect for the $\sqrt x$ inside the cosine.)
Next, write
$${1-\cos x\over x(1-\cos\sqrt x)}=\left({1-\cos x\over x^2}\right)\left({x\over1-\cos\sqrt x}\right)=\left({1-\cos x\over x^2}\right)\Big/\left({1-\cos u\over u^2}\right)$$
where $u=\sqrt x$.  Notice that as $x$ tends to $0$, so does $u$, and vice versa. It suffices, therefore, to show that
$$\lim_{x\to0}\left({1-\cos x\over x^2}\right)$$
exists and is not equal to $0$, in which case
$$\lim_{x\to0^+}\left({1-\cos x\over1-\cos\sqrt x}\right)=\lim_{x\to0}\left({1-\cos x\over x^2}\right)\Big/\lim_{u\to0}\left({1-\cos u\over u^2}\right)=1$$
and the answer to the original problem is simply $1/3$.  But
$${1-\cos x\over x^2}={1-\cos^2x\over x^2(1+\cos x)}={1\over1+\cos x}\left( \sin x\over x\right)^2$$
at which point it helps to recognize that $\lim_{x\to0}{\sin x\over x}=1$, which gives the nonzero limit $1/2$.  (If you're not allowed to recognize the limit for $\sin x\over x$, there's additional work to do, but I'm assuming you have a quiverful of standard limits at your disposal.)
