How do I evaluate the limit limx→0sin⁡3xxcos⁡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 limx→0sin⁡3x1 and then somehow introducing 3x so it turns into a special limit.Otherwise, I'm just stuck.

Question

How do I evaluate the limit limx→0sin⁡3xxcos⁡2x?I&#039;m having trouble doing this problem: the farthest I&#039;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 limx→0sin⁡3x1 and then somehow introducing 3x so it turns into a special limit.Otherwise, I&#039;m just stuck.

Marcus Herman · Accepted Answer

We want to evaluate  limx→0sin⁡3xxcos⁡2x  Set p=3x, so x=p3. Note that as x→0,p→0  3limp→0sin⁡ppcos⁡2p3=3limp→0sin⁡pplimp→01cos⁡2p3=3  Here is a nice geometric proof for  limp→0sin⁡pp=1

Rita Miller · Accepted Answer

Note that limx→01cos⁡(2x)=1. On the other hand  limx→0sin⁡(3x)x=3limx→0sin⁡3x3x=3  Therefore,  limx→0sin⁡3xxcos⁡2x=3⋅1=3

Vasquez · Accepted Answer

We can use also use power series for this limit. Using the Maclaurin series expansion for sin and cos we see that limx→0sin⁡3xxcos⁡2x=limx→03x−(3x)33!+…x(1−(2x)22!)+…=limx→03−33x23!+…1−(2x)22!+…=3 I hope that was helpful.

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

Answered question

Answer & Explanation

New Questions in Trigonometry