real analysis - Need help finding the limit $lim_{xto -infty}frac{ln(2-x)}{sqrt[3]{1-6x}}$.

This is my problem:

Find the limit if it exists:
$$\lim_{x\to -\infty}\frac{\ln(2-x)}{\sqrt[3]{1-6x}}$$

I've tried solving it, but I can't seem to figure it out.

Using hospital's rule:

I know that
$$\frac{1}{\sqrt[3]{1-6x}} = (1-6x)^{-1/3}$$

so
$$\lim_{x\to-\infty}f(x)=\frac{\frac{d}{dx}ln(2-x)}{\frac{d}{dx}\sqrt[3]{1-6x}}$$
$$\lim_{x\to-\infty}f(x)=\frac{\frac{-1}{(2-x)}}{(-1/3)*(1-6x)^{-4/3}*6}$$

After this I get really confused... I've tried stuff for hours now. I think it becomes:

$$\lim_{x\to-\infty}f(x)=\frac{\frac{-1}{(2-x)}}{\frac{-6}{3(1-6x)^{4/3}}}$$

$$\lim_{x\to-\infty}f(x)=\frac{\frac{-1}{(2-x)}}{\frac{-2}{(1-6x)^{4/3}}}$$

$$\lim_{x\to-\infty}f(x)=\frac{-(1-6x)^{4/3}}{-2(2-x)}$$

I don't dare go any further because I really don't know if I'm doing it right. I'm pretty desperate, so I appreciate the help.

Thank you guys.