evaluation of a limit with trigonometric and expontential components I'm trying to evaluate the limit $$\lim_{x\to 0} \frac{x^3-\sin^3x}{(e^x-x-1)\sin^3x}$$ 
I see that $\large\frac{1}{e^x-x-1}\to +\infty$, since $e^x-x-1>0,\, \forall x\in \mathbb{R^*}$. Also $\large\frac{x^3-\sin^3x}{\sin^3x} \to 0$. But this of course leads nowhere. 
On the other hand, $$\left| \frac{x^3-\sin^3x}{(e^x-x-1)\sin^3x} \right|=\frac{1}{e^x-x-1}\cdot\left| \frac{x^3-\sin^3x}{\sin^3x} \right|= \frac{1}{e^x-x-1}\cdot\left| \left(\frac{x}{\sin x} \right)^3-1\right| $$
Could this lead anywhere by bounding? Any hints would be appreciated.
 A: HINT:
$$\lim_{x\to 0} \frac{x^3-\sin^3x}{(e^x-x-1)\sin^3x}$$ 
$$=\lim_{x\to 0} \frac{(x-\sin x)(x^2+x\sin x+\sin^2x)}{(e^x-x-1)\sin^3x}$$ 
$$=\lim_{x\to 0}\frac{1+\dfrac{\sin x}x+\left(\dfrac{\sin x}x\right)^2}{\left(\dfrac{\sin x}x\right)^3}\cdot\lim_{x\to 0}\frac{x-\sin x}{x(e^x-x-1)} $$
Using Taylor's Expansion $\displaystyle\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$
$\displaystyle\implies x-\sin x=\frac{x^3}{3!}+O(x^5)$
and $\displaystyle e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$
$\displaystyle\implies e^x-1-x=\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$
Can you take it home from here?
Alternatively,
$$\lim_{x\to 0}\frac{x-\sin x}{x(e^x-x-1)} $$ can be managed by L'Hospital's Rule as well
A: $$\lim_{x\to 0} \frac{x^3-\sin^3x}{(e^x-x-1)\sin^3x}$$
$$=\lim_{x\to 0} \frac{1-\frac{\sin^3x}{x^3}}{(e^x-x-1)\frac{\sin^3x}{x^3}}$$
$$=\lim_{x\to 0} \frac{(1-\frac{\sin x}{x})\left(1+\frac{\sin x}{x}+\frac{\sin^2 x}{x^2}\right)}{(e^x-x-1)\frac{\sin^3x}{x^3}}$$
$$=\lim_{x\to 0} \frac{(1-\frac{\sin x}{x})}{(e^x-x-1)}\lim_{x\to 0}\frac{\left(1+\frac{\sin x}{x}+\frac{\sin^2 x}{x^2}\right)}{\frac{\sin^3x}{x^3}}$$
$$=\lim_{x\to 0} \frac{(x-\sin x)}{(xe^x-x^2-x)}3$$
$$=\lim_{x\to 0} \frac{(1-\cos x)}{(e^x(x+1)-2x-1)}3$$
$$=\lim_{x\to 0} \frac{(\sin x)}{(e^x(x+2)-2)}3$$
$$=\lim_{x\to 0} \frac{(\cos x)}{(e^x(x+3))}3=1$$
