Limit $\lim_\limits{x\to0}{2x-\sin{x}\over3x+\sin{x}}$ 
What is the limit of:
  $$\lim_\limits{x\to0}{2x-\sin{x}\over3x+\sin{x}}$$

What I've tried:
$$\lim_\limits{x\to0}{2x-2\sin{x\over2}\cos{x\over2}\over3x+2\sin{x\over2}\cos{x\over2}}=\lim_\limits{x\to0}{2(x-\sin{x\over2}\cos{\frac{x}{2}})\over3x+2\sin{x\over2}\cos{x\over2}}$$
I'm lost at this point, no idea what to do, or if I should've done something else.
 A: Divide by $x$!
$$\lim_{x\to0}\frac{2x-\sin x}{3x+\sin x}=\lim_{x\to0}\frac{2-\frac{\sin x}x}{3+\frac{\sin x}x}=\frac{2-1}{3+1}=\frac14$$
where the well-known limit $\lim_{x\to0}\frac{\sin x}x=1$ was used.
A: Since $\sin x = x + O(x)$ for $x \rightarrow 0$, your limit becomes: $\large{\lim_{x\to 0} \frac{2x - x + O(x^3)}{3x + x - O(x^3)}}$, thence for $x \rightarrow 0$, the result is $\frac{1}{4}$.
A: You may do this with L Hospital rule. Differentiate numerator and denominator with respect to x.
$$\lim_{x\to0}\frac{2x-\sin x}{3x+\sin x}=\frac{2-cosx}{3+cosx}=\frac{2-1}{3+1}=\frac14$$

L’Hospital’s rule says that if $\displaystyle \lim_{x \to a}f(x) = 0$ and $\displaystyle \lim_{x \to a}g(x) = 0$ (or $\displaystyle \lim_{x \to a}f(x) = \infty$ and $\displaystyle \lim_{x \to a}g(x) = \infty$) then
  $\displaystyle\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f’(x)}{g’(x)}$

A: As Larry mentioned in comment, use L'hospital, as $2x-\sin{x}\to 0 $ and aslo $3x+\sin x\to 0$ as $x\to 0$. $$\lim_{x\to 0}\frac{2x-\sin x}{3x+\sin x}=\lim_{x\to 0}\frac{2-\cos x}{3+\cos x}=\frac{2-1}{3+1}=\frac14.$$
