$$\lim_{x\to 0}\dfrac{\sin3x}{x\cos2x}$$
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?
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,$$\lim_{x\to0}\frac{\sin(3x)}{x\cos(2x)}=3\cdot 1=3.$$
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 :)
