# 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 .

• Hint: divide everything by $x$, and use $\lim \frac{\sin x}{x} = 1$. – Secret Math Jun 16 '13 at 1:12
• If L'Hôpital's rule won't work, it doesn't mean that the original problem doesn't convergent. – eccstartup Jun 16 '13 at 1:17

$$\lim_{x\to0}\;\frac{\left(x+\sin(x)\right)}{\left(x^2-\sin(x)\right)}\cdot \frac{1/x}{1/x} \quad =\quad \lim_{x\to 0}\;\frac{1+\frac{\sin(x)}x}{x-\frac{\sin(x)}x}$$

Now we evaluate, using the fact that

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

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



• Nice Amy. :-=)${}$ – mrs Jun 16 '13 at 4:29
• @amWhy: I have a lot of those - and they won't last long ;-)! – Amzoti Jun 17 '13 at 0:08

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 $$\lim_{x \to 0}\frac{2+O(x^2)}{x-1+O(x^2)}=-2$$