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

*

*We have yet not done Taylor expansions.

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

 A: If Series Expansion is allowed,
$$\ln(1-\sin x)=-\sin x-\frac{\sin^2 x}2-O(\sin^3x)$$
Again,$$\sin x=x-O(x^3)$$
A: You might use Taylor series expansions:
$$\begin{align} \sin x &= x+O(x^3)\\
\ln(1+x)&= x-\frac12x^2+O(x^3)\\
\implies \ln(1-\sin x)+x&=\frac12x^2+O(x^3)\\
\implies \frac{\ln(1-\sin x)+x}{x^2}&=\frac12+O(x)\end{align}$$
A: You need to know the two limits (in addition to the standard limits like $\lim\limits_{x \to 0}\dfrac{\sin x}{x} = 1$): $$\lim_{x \to 0}\frac{x - \log(1 + x)}{x^{2}} = \frac{1}{2},\,\,\lim_{x \to 0}\frac{x - \sin x}{x^{2}} = 0$$ The first of these limits is bit difficult to handle without L'Hospital's Rule and has been calculated in this answer. The second limit is solved in this answer.
Note that if we put $x = -t$ in the first limit we get $$\lim_{t \to 0}\frac{\log(1 - t) + t}{t^{2}} = -\frac{1}{2}$$ and we will use this form here. We proceed as follows $$\begin{aligned}L &= \lim_{x \to 0}\frac{\log(1 - \sin x) + x}{x^{2}}\\
&= \lim_{x \to 0}\frac{\log(1 - \sin x) + \sin x + x - \sin x}{x^{2}}\\
&= \lim_{x \to 0}\frac{\log(1 - \sin x) + \sin x}{x^{2}} + \lim_{x \to 0}\frac{x - \sin x}{x^{2}}\\
&= \lim_{x \to 0}\frac{\log(1 - \sin x) + \sin x}{\sin^{2}x}\cdot\frac{\sin^{2}x}{x^{2}}\\
&= \lim_{x \to 0}\frac{\log(1 - \sin x) + \sin x}{\sin^{2}x}\\
&= \lim_{t \to 0}\frac{\log(1 - t) + t}{t^{2}}\text{ (putting }t = \sin x)\\
&= -\frac{1}{2}\end{aligned}$$
A: Recall that $\ln (1 - x) = -\sum_{i=1}^{\infty} x^n/n$, near $0$. 
So, $$\ln (1 - \sin  x) = -\sum_{i=1}^{\infty} (\sin x)^n/n = - x -x^2/2 + O(x^3).$$ 
I guess you can finish from there.
