Answer
$$ \frac{3}{-5}$$
Work Step by Step
We have
$$
\lim _{x \rightarrow -\infty} \frac{3x-2}{1-5x}=-\frac{\infty}{\infty}.
$$
is an intermediate form, then we can apply L’Hôpital’s Rule as follows
$$
\lim _{x \rightarrow -\infty} \frac{3x-2}{1-5x}=\lim _{x \rightarrow -\infty} \frac{3}{-5}= \frac{3}{-5}.
$$
