Question

Limits help

Answer 1

\[\LARGE \lim_{x \rightarrow 0} \frac{\tan 6 \theta}{\sin3\theta}\]

Answer 2

\(\lim_{x\to 0} \dfrac{\sin 6x}{\cos 6x}\dfrac{1}{\sin 3x}=\lim_{x\to 0}\dfrac{2\cos 3x}{\cos 6x}=\dfrac{2(1)}{1}\)

Answer 3

How did you get that 2cos3x?

Answer 4

\[\sin 6x=2\sin 3x \cos 3x\]

so

\[\dfrac{\tan 6x}{\sin 3x}=\dfrac{\sin 6x}{\cos 6x}\frac{1}{\sin 3x}=\dfrac{2\sin3x \cos 3x}{\cos 6x\sin 3x}=\frac{2\cos 3x}{\cos 6x}\]

Answer 5

Oh! I forgot about those identities, thank you for your help :)

Answer 6

your welcome