Limits with Trig functions help please

[Dorian Gray]

Hello Everyone,

I am in Calculus 1, and I am having difficulties answer this following math problem.

lim (tan 6x)/(sin2x)
x--0


Thanks!

 

[soroban]

Hello, Dorian Gray!

You are expected to know this identity: .\(\displaystyle \displaystyle\lim_{\theta\to0} \frac{\sin\theta}{\theta} \:=\:1\)

\(\displaystyle \displaystyle\lim_{x\to0}\frac{\tan6x}{\sin2x}\)

We have: .\(\displaystyle \dfrac{\tan6x}{\sin2x}\;=\;\dfrac{\frac{\sin6x}{\cos6x}}{\sin2x} \;=\;\sin6x\cdot\dfrac{1}{\cos6x}\cdot\dfrac{1}{\sin2x} \)


Multiply by \(\displaystyle \frac{6x}{6x}\) and \(\displaystyle \frac{2x}{2x}\)

. . \(\displaystyle \dfrac{6x}{6x}\cdot\sin6x \cdot\dfrac{1}{\cos6x}\cdot\dfrac{1}{\sin2x}\cdot\dfrac{2x}{2x} \;=\;6x\cdot\dfrac{\sin6x}{6x}\cdot\dfrac{1}{\cos6x}\cdot\dfrac{2x}{\sin2x}\cdot\dfrac{1}{2x}\)

. . . \(\displaystyle =\;3\cdot\dfrac{\sin6x}{6x}\cdot\dfrac{1}{\cos6x}\cdot\dfrac{2x}{\sin2x} \)


Then: .\(\displaystyle \displaystyle \lim_{x\to0}\dfrac{\tan6x}{\sin2x} \;=\;3\cdot\underbrace{\lim_{x\to0}\left(\dfrac{\sin6x}{6x}\right)}_{\text{This is 1}}\cdot \underbrace{\lim_{x\to0}\left(\dfrac{1}{\cos6x}\right)}_{\text{This is 1}}\cdot \underbrace{\lim_{x\to0}\left(\dfrac{2x}{\sin2x}\right)}_{\text{This is 1}}\)

. . . . . . . . . . . . . . \(\displaystyle =\;3\cdot1\cdot1\cdot1 \;=\;3\)

 

[Dorian Gray]

update!

Okay! I just figured out the 3! I see where you get the 3 from, but I am still confused about the middle "1"...

 

[srmichael]

Okay! I just figured out the 3! I see where you get the 3 from, but I am still confused about the middle "1"...

\(\displaystyle \cos(6\cdot0)=\cos(0)=1\)

 

[Dorian Gray]

thank you everyone

Thank you everyone for your help.