Question

Use formula 2 and trigonometric identities to evaluate the limit: lim t->0 (tan6t)/(sin2t)

Answer 1

Formula 2 is: lim theta->0 sin(theta)/theta=1

Answer 2

There's a variant of the formula that says the same about the reciprocal:
\[\lim_{\theta\to0}\frac{\sin a\theta}{a\theta}=\lim_{\theta\to0}\frac{a\theta}{\sin a\theta}=1~~\text{for}~~a\not=0\]
For your given limit, the identity you should be concerned with is
\[\tan6t=\frac{\sin6t}{\cos6t}\]
So with this info, you can rewrite the limit:
\[\large\begin{align*}
\lim_{t\to0}\frac{\tan6t}{\sin2t}&=\lim_{t\to0}\frac{\frac{\sin6t}{\cos6t}}{\sin2t}\\\\
&=\lim_{t\to0}\frac{\sin6t}{1}\cdot\frac{1}{\sin2t}\cdot\frac{1}{\cos6t}\\\\
&=\lim_{t\to0}\frac{\color{red}{6t}}{\color{blue}{2t}}\cdot\frac{\sin6t}{\color{red}{6t}}\cdot\frac{\color{blue}{2t}}{\sin2t}\cdot\frac{1}{\cos6t}
\end{align*}\]

Answer 3

Ok, i followed until the last step...

Answer 4

Ohhhhh just kidding I got it

Answer 5

Good. So you see that the sine and cosine limits are 1, right? This leaves you with \(\dfrac{6t}{2t}\). Since \(t\to0\), you never actually have \(t=0\), which means you can divide through the \(t\) factors:
\[\frac{6t}{2t}=\frac{6}{2}\]

Answer 6

Ok that all makes sense
What happened to the 1/cos6t though?

Answer 7

Direct substitution gives \(\dfrac{1}{\cos0}=1\).

Answer 8

Right ok got it! Thanks a bunch!

Answer 9

You're welcome!