Evaluating $\lim_{x\to0}\frac{x+\sin(x)}{x^2-\sin(x)}$ I did the math, and my calculations were:
$$\lim_{x\to0}\frac{x+\sin(x)}{x^2-\sin(x)}= \lim_{x\to0}\frac{x}{x^2-\sin(x)}+\frac{\sin(x)}{x^2-\sin(x)}$$ But I can not get out of it. I would like do it without using derivation or L'Hôpital's rule .

Another way of solving it is expanding in Maclaurin series: $\sin x =x + O(x^3)$, hence you get after cancelling out $x$ in numerator and denominator 
$$
\lim_{x \to 0}\frac{2+O(x^2)}{x-1+O(x^2)}=-2
$$
