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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.

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  • $\begingroup$ Use L'Hospital's Rule $\endgroup$ – Larry Oct 21 '18 at 15:59
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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.

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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 $\frac{1}{4}$.

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

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

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