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.

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

Otherwise, I'm just stuck.

Marcus Herman

Beginner2022-01-01Added 41 answers

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

$\underset{p\to 0}{lim}\frac{\mathrm{sin}p}{p}=1$

Set

Here is a nice geometric proof for

Rita Miller

Beginner2022-01-02Added 28 answers

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,

$\underset{x\to 0}{lim}\frac{\mathrm{sin}3x}{x\mathrm{cos}2x}=3\cdot 1=3$

Therefore,

Vasquez

Expert2022-01-08Added 669 answers

We can use also use power series for this limit. Using the Maclaurin series expansion for sin and cos we see that

I hope that was helpful.

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