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.

Solving this without Taylor expansion is quite artificial, if you ask me (any justification, pedagogical or otherwise?), hence here we go for artificiality. 
If $u\to1$ and $v\to1$,
$$
u^{1/6}-v^{1/6}=\frac{u-v}{u^{5/6}+u^{4/6}v^{1/6}+u^{3/6}v^{2/6}+u^{2/6}v^{3/6}+u^{1/6}v^{4/6}+v^{5/6}},
$$
and each term in the denominator has limit $1$ hence
$$
u^{1/6}-v^{1/6}\sim\frac{u-v}6.
$$
Using this for $u=(1-3x)^2$ and $v=(1-2x)^3$ yields $u-v=-3x^2+8x^3$ hence 
$$u-v\sim-3x^2.
$$
On the other hand, $1-\cos(\pi x)=2\sin^2\left(\frac12\pi x\right)$ and $\sin t\sim t$ (but are we allowed to use this? this is Taylor, after all...) hence 
$$1-\cos(\pi x)\sim2\left(\frac12\pi x\right)^2=\frac12\pi^2x^2.$$
Thus, the limit of the ratio is
$$
\frac{-3\cdot\frac16}{\frac12\pi^2}=-\frac1{\pi^2}.
$$
