Limit of $\lim_{x\to 0} \frac{\log(1+\sqrt[3]{x})(\sin x -. x)}{(\cos x-1)^3 x^2}$ without hopital I am struggling to make this limit:
$$\lim_{x\to 0} \frac{\log(1+\sqrt[3]{x})(\sin x - x)}{(\cos x-1)^3x^2}$$
I know it doesn't exist, and I am really near to prove it but something is blocking me. Skipping some step, I have that $$\frac{\sqrt[3]x(\sin x-x)}{-\frac{1}{2}x^8}$$
From this, I see that $x^8$ could be $x^7$, by making the top term (1-1), and from this I see that this goes to +-infinity, so it doesn't exist. But this isn't a formal method, nor probably a method at all. Would love to see your methods to solve this without hopital!
 A: Just rewrite it as
$$\frac{\ln (1+\sqrt[3]{x})}{\sqrt[3]{x}}\sqrt[3]{x}\frac{\sin x -x}{x^3}x^3\frac{x^6}{(\cos x-1)^3}\frac{1}{x^6}\frac{1}{x^2}$$ you get a limit of $\infty$.
A: Use Taylor expansion of $\sin(x)$ about $0$ which is
$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} ...$
with which you'll get
$\frac{x^{\frac{3}{2}}(- \frac{x^3}{3!} + \frac{x^5}{5!} ...)}{\frac{-1}{2}x^8}$. You can follow from here.
A: For all $\varepsilon>0$ and for all $x$ sufficiently close to $0$ we have:
$$
\begin{align*}
(1-\varepsilon)\sqrt[3]{x}&\le\ln(1+\sqrt[3]{x})\\
-\varepsilon x&\le\sin(x)-x\\
\varepsilon x^2&\ge\cos(x)-1.
\end{align*}
$$
Thus for $c=\frac{\varepsilon^2-\varepsilon}{\varepsilon^3}$ we have
$$
\lim_{x\to0}\frac{c}{x^{20/3}}=\lim_{x\to0}\frac{((1-\varepsilon)\sqrt[3]{x})(-\varepsilon x)}{(\varepsilon x^2)^3x^2}\le\lim_{x\to0}\frac{\ln(1+\sqrt[3]{x})(\sin(x)-x))}{(\cos(x)-1)^3x^2}.
$$
It is easily seen that the LHS diverges as $x\to 0$ which forces the RHS to diverge as well. 
A: A general comment... 
Even though you are not allowed to use l'H$\hat{\operatorname{o}}$pital's, I do believe you may use things like the definition of derivative, product and quotient rule for limits .   
For example
$$\lim_{x\to0}\frac{\log(1+x)}{\sin x} =\lim_{x\to0}\frac{\frac{\log(1+x)-\log(0)}{x}}{\frac{\sin x-\sin 0}{x}} =\frac{\frac{d}{dx}\log(1+x)\big|_{x=0}}{\frac{d}{dx}\sin x\big|_{x=0}}=\frac{1/(1+0)}{\cos 0}=1$$
