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$$\lim_{x\to 0}\dfrac{\sin3x}{x\cos2x}$$

I'm having trouble doing this problem: the farthest I've gotten is just using a limit law for division and then moving the constant $x$ in the denominator in front. I also thought about leaving $\lim_{x\to 0}\sin3x/1$ and then somehow introducing $3x$ so it turns into a special limit.

Otherwise, I'm just stuck. Hmm, the original problem does resemble tangent, not sure if that has to do with anything?

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    $\begingroup$ $\sin(3x)/(3x) $ goes to one. $\endgroup$ Mar 2, 2021 at 22:31

3 Answers 3

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We want to evaluate $$\lim\limits_{x\to 0}\dfrac{\sin3x}{x\cos2x}.$$ Set $p=3x$, so $x=p/3$. Note that as $x\to0, p\to0.$ $$3\lim_{p\to 0}\dfrac{\sin p}{p\cos\tfrac{2p}{3}}=3\lim_{p\to 0}\dfrac{\sin p}{p}\lim_{p\to 0}\frac{1}{\cos\tfrac{2p}{3}}=3.$$ Here is a nice geometric proof for $$\lim_{p\to 0}\dfrac{\sin p}{p}=1.$$

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Note that $\lim_{x\to0}\frac1{\cos(2x)}=1$. On the other hand$$\lim_{x\to0}\frac{\sin(3x)}x=3\lim_{x\to0}\frac{\sin(3x)}{3x}=3.$$Therefore,$$\lim_{x\to0}\frac{\sin(3x)}{x\cos(2x)}=3\cdot 1=3.$$

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    $\begingroup$ I would have done the same steps as you 😊 +1 $\endgroup$
    – Sebastiano
    Mar 2, 2021 at 22:54
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    $\begingroup$ @Sebastiano Hi amigo from Italy of zoff. $\endgroup$ Mar 2, 2021 at 22:58
  • $\begingroup$ I think we made the exact same steps. I just put another extra explanation why $L \sin(3x)/(3x)$ is really equal to $\sin x/x$. $\endgroup$
    – vitamin d
    Mar 2, 2021 at 22:58
  • $\begingroup$ @hamam_Abdallah Ahahahah 😊😊😊😊😊 I remember the final in Spain from Italy-Germany 3-1 in 1982. $\endgroup$
    – Sebastiano
    Mar 2, 2021 at 23:00
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    $\begingroup$ @Sebastiano The third goal was by Altobelli. the best player was Conti. $\endgroup$ Mar 2, 2021 at 23:01
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We can use also use power series for this limit. Using the Maclaurin series expansion for $\sin$ and $\cos$ we see that $$\lim_{x\to 0}\frac{\sin3x}{x\cos2x}=\lim_{x\to 0}\frac{3x-\frac{(3x)^3}{3!}+\cdots}{x(1-\frac{(2x)^2}{2!}+\cdots)}=\lim_{x\to 0}\frac{3-\frac{3^3x^2}{3!}+\cdots}{1-\frac{(2x)^2}{2!}+\cdots}=3$$


I hope that was helpful. If you have any questions please don't hesitate to ask :)

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