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 .

$$\lim_{x\to0}\;\frac{\left(x+\sin(x)\right)}{\left(x^2-\sin(x)\right)}\cdot \frac{1/x}{1/x} \quad =\quad \lim_{x\to 0}\;\frac{1+\frac{\sin(x)}x}{x-\frac{\sin(x)}x}$$
Now we evaluate, using the fact that 

$$\lim_{x \to 0} \frac {\sin(x)}x = 1$$

we can see that:
$$\lim_{x\to 0}\;\frac{1+\frac{\sin(x)}x}{x-\frac{\sin(x)}x} = \frac{1 + 1}{0 - 1} = -2$$
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