# How do I evaluate the limit $\;\lim\limits_{x\to 0}\dfrac{\sin3x}{x\cos2x}$?

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

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?

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

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 $$\lim_{p\to 0}\dfrac{\sin p}{p}=1.$$

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,$$\lim_{x\to0}\frac{\sin(3x)}{x\cos(2x)}=3\cdot 1=3.$$

• I would have done the same steps as you 😊 +1 Mar 2, 2021 at 22:54
• @Sebastiano Hi amigo from Italy of zoff. Mar 2, 2021 at 22:58
• I think we made the exact same steps. I just put another extra explanation why $L \sin(3x)/(3x)$ is really equal to $\sin x/x$. Mar 2, 2021 at 22:58
• @hamam_Abdallah Ahahahah 😊😊😊😊😊 I remember the final in Spain from Italy-Germany 3-1 in 1982. Mar 2, 2021 at 23:00
• @Sebastiano The third goal was by Altobelli. the best player was Conti. Mar 2, 2021 at 23:01

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 :)