limits - How to evaluate $lim_{x to 0} ( ln(1 - sin x) + x)/x^2$ without using l'Hôpital?

How to evaluate
$$\lim_{x \to 0} \frac{\ln(1 - \sin x) + x} {x^2}$$

without using l'Hôpital?
I am not able to substitute the right infinitesimal. Is there a substitute?

Background

1. We have yet not done Taylor expansions.


2. I know that $\ln$ around 1 tends to 0 and $\sin$ around 0 tends to 0.