Bobbie Comstock

2021-12-31

How do I evaluate the limit $\underset{x\to 0}{lim}\frac{\mathrm{sin}3x}{x\mathrm{cos}2x}$?
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 $\underset{x\to 0}{lim}\frac{\mathrm{sin}3x}{1}$ and then somehow introducing 3x so it turns into a special limit.
Otherwise, I'm just stuck.

Marcus Herman

We want to evaluate
$\underset{x\to 0}{lim}\frac{\mathrm{sin}3x}{x\mathrm{cos}2x}$
Set $p=3x$, so $x=\frac{p}{3}$. Note that as $x\to 0,p\to 0$
$3\underset{p\to 0}{lim}\frac{\mathrm{sin}p}{p\mathrm{cos}\frac{2p}{3}}=3\underset{p\to 0}{lim}\frac{\mathrm{sin}p}{p}\underset{p\to 0}{lim}\frac{1}{\mathrm{cos}\frac{2p}{3}}=3$
Here is a nice geometric proof for
$\underset{p\to 0}{lim}\frac{\mathrm{sin}p}{p}=1$

Rita Miller

Note that $\underset{x\to 0}{lim}\frac{1}{\mathrm{cos}\left(2x\right)}=1$. On the other hand
$\underset{x\to 0}{lim}\frac{\mathrm{sin}\left(3x\right)}{x}=3\underset{x\to 0}{lim}\frac{\mathrm{sin}3x}{3x}=3$
Therefore,
$\underset{x\to 0}{lim}\frac{\mathrm{sin}3x}{x\mathrm{cos}2x}=3\cdot 1=3$

Vasquez

We can use also use power series for this limit. Using the Maclaurin series expansion for sin and cos we see that
$\underset{x\to 0}{lim}\frac{\mathrm{sin}3x}{x\mathrm{cos}2x}=\underset{x\to 0}{lim}\frac{3x-\frac{\left(3x{\right)}^{3}}{3!}+\dots }{x\left(1-\frac{\left(2x{\right)}^{2}}{2!}\right)+\dots }=\underset{x\to 0}{lim}\frac{3-\frac{{3}^{3}{x}^{2}}{3!}+\dots }{1-\frac{\left(2x{\right)}^{2}}{2!}+\dots }=3$

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