We have indeterminateness of type
oo/-oo,
i.e. limit for the numerator is
$$\lim_{x \to -\infty}\left(\log{\left(x \right)} + 1\right) = \infty$$
and limit for the denominator is
$$\lim_{x \to -\infty} x = -\infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to -\infty}\left(\frac{\log{\left(x \right)} + 1}{x}\right)$$
=
$$\lim_{x \to -\infty}\left(\frac{\frac{d}{d x} \left(\log{\left(x \right)} + 1\right)}{\frac{d}{d x} x}\right)$$
=
$$\lim_{x \to -\infty} \frac{1}{x}$$
=
$$\lim_{x \to -\infty} \frac{1}{x}$$
=
$$0$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)