How do I evaluate the limit \lim_{x \to 0} \frac{\sin 3x}

Bobbie Comstock

Bobbie Comstock

Answered question

2021-12-31

How do I evaluate the limit limx0sin3xxcos2x?
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 limx0sin3x1 and then somehow introducing 3x so it turns into a special limit.
Otherwise, I'm just stuck.

Answer & Explanation

Marcus Herman

Marcus Herman

Beginner2022-01-01Added 41 answers

We want to evaluate
limx0sin3xxcos2x
Set p=3x, so x=p3. Note that as x0,p0
3limp0sinppcos2p3=3limp0sinpplimp01cos2p3=3
Here is a nice geometric proof for
limp0sinpp=1
Rita Miller

Rita Miller

Beginner2022-01-02Added 28 answers

Note that limx01cos(2x)=1. On the other hand
limx0sin(3x)x=3limx0sin3x3x=3
Therefore,
limx0sin3xxcos2x=31=3
Vasquez

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
limx0sin3xxcos2x=limx03x(3x)33!+x(1(2x)22!)+=limx0333x23!+1(2x)22!+=3
I hope that was helpful.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?