Finding the limit $ \lim_{x\to 0}\frac{(1-3x)^\frac{1}{3} -(1-2x)^\frac{1}{2}}{1-\cos(\pi x)}$ I cannot find this limit: 
$$
\lim_{x\to 0}\frac{(1-3x)^\frac{1}{3} -(1-2x)^\frac{1}{2}}{1-\cos(\pi x)}.
$$
Please, help me. 
Upd: I need to solve it without L'Hôpital's Rule and Taylor expansion.

Write it as $\left(\frac{\cos\pi x-\cos0}{x-0}\right)^{-1}\times\frac{\left(1-3x\right)^{\frac{1}{3}}-\left(1-2x\right)^{\frac{1}{2}}}{x-0}$.
Then $\lim_{x\rightarrow0}\frac{\cos\pi x-\cos0}{x-0}$ can be recognized
as $f'\left(0\right)$ for $f\left(x\right)=\cos\pi x$ 
and $\lim_{x\rightarrow0}\frac{\left(1-3x\right)^{\frac{1}{3}}-\left(1-2x\right)^{\frac{1}{2}}}{x-0}$
as $g'\left(0\right)$ for $g\left(x\right)=\left(1-3x\right)^{\frac{1}{3}}-\left(1-2x\right)^{\frac{1}{2}}$.
If you are blamed to use de l'Hôpital after all then just claim that
you were not aware of that. They will believe you.
Edit:
This only works if $f'\left(0\right)\neq0$ and unfortunately
that is not the case here.
So to be honest: this does not answer your question. Keep it in mind
however for next occasions.
