Need help finding the limit $\lim_{x\to -\infty}\frac{\ln(2-x)}{\sqrt[3]{1-6x}}$. 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.
 A: Your calculation 
$$\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}\frac{\frac{-1}{(2-x)}}{\color{red}{(-1/3)*(1-6x)^{-4/3}*6}}$$
is wrong. 
Try to calculate 
$$
\frac{d}{dx}(1-6x)^{1/3}
$$
instead of 
$$
\frac{d}{dx}(1-6x)^{-1/3}
$$
When applying the L'Hôpital's rule.
A: Perhaps easier:
$$\left(\sqrt[3]{1-6x}\right)'=\left((1-6x)^{1/3}\right)'=\frac13\cdot(-6)(1-6x)^{-2/3}=-\frac2{\sqrt[3]{(1-6x)^2}}$$
so applying l'Hospital we get
$$\lim_{x\to-\infty}\frac{-\frac1{2-x}}{-\frac2{(1-6x)^{2/3}}}=\lim_{x\to-\infty}\frac{(1-6x)^{2/3}}{2-x}=0$$
A: One can avoid the use of L'Hospital's Rule by using the standard limit $$\lim_{x \to \infty}\frac{\log x}{x^{a}} = 0\tag{1}$$ where $a > 0$.
We have
\begin{align}
L &= \lim_{x \to -\infty}\frac{\log(2 - x)}{\sqrt[3]{1 - 6x}}\notag\\
&= \lim_{y \to \infty}\frac{\log(2 + y)}{\sqrt[3]{1 + 6y}}\text{ (putting }y = -x)\notag\\
&= \lim_{y \to \infty}\frac{\log(2 + y)}{(2 + y)^{1/3}}\cdot\left(\frac{2 + y}{1 + 6y}\right)^{1/3}\notag\\
&= \lim_{t \to \infty}\frac{\log t}{t^{1/3}}\cdot\lim_{y \to \infty}\left(\frac{1 + 2/y}{6 + 1/y}\right)^{1/3}\text{ (putting }t = y + 2)\notag\\
&= 0\cdot(1/6)^{1/3}\notag\\
&= 0\notag
\end{align}
