# 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.

• Use L'Hospital's Rule – Larry Oct 21 '18 at 15:59

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.

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 $$\frac{1}{4}$$.

• $\LaTeX \text { Tip}:$ Use \sin and \lim to signify $\sin \text{ and } \lim$ – Mohammad Zuhair Khan Oct 21 '18 at 16:04
• Thanks Raptor, I'm quite new to MathStackExchange so not used yet to perfectly type in $\LaTeX$. – Piergiorgio Panero Oct 21 '18 at 16:09
• No problem, I was in your shoes not too long ago and it never hurts to help :) – Mohammad Zuhair Khan Oct 21 '18 at 16:10

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 $$\displaystyle\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f’(x)}{g’(x)}$$

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