On an evaluation of limits

Question

Can someone explain to me how the $4 \theta$ gets introduced to this solution? Also, how does the $\sin$ and $\cos$ get figured out in the final answer per the answer key? I feel there are steps that are not shown but I do not understand what those might be? Please advise.

Evalute the limit, if it exists, (without L'Hopital's rule), as $\theta \to 0$, $\frac{\tan 4 \theta}{6 \theta}$

$$\lim_{\theta \to 0}\frac{\tan 4 \theta}{6 \theta}=\lim_{\theta \to 0} \left (\frac{\sin 4 \theta}{\cos 4 \theta} \cdot \frac{1}{6 \theta} \right ) =\lim_{\theta \to 0}\left (\frac{\sin 4 \theta}{4 \theta} \cdot \frac{4 \theta}{\cos 4 \theta} \cdot \frac{1}{6 \theta} \right ) = \frac 46$$

Answer 1

The simple answer is, because it works. The reason it works is that you're trying to emulate a similar limit whose answer is already known; in this case, $\lim \frac{ \sin u}{u}$.

So you manipulate the function as needed (multiplying by $1$ in a useful form) to achieve that form (where $u = 4 \theta$, so that $\theta \to 0 \Rightarrow u \to 0$), and then hope that what's left over is more tractable. In this case, the wish is granted. Note that for $\theta \neq 0$:

$$ \frac {4 \theta}{\cos 4 \theta} \frac {1}{6 \theta} = \frac{4 \theta}{6 \theta}\frac{1}{\cos 4 \theta}= \frac 23 \frac {1}{\cos 4 \theta}.$$

Answer 2

Indeed, you are correct in stating that $$\lim_{\theta\to0}\frac{\tan(4\theta)}{6\theta}=\lim_{\theta\to0}\frac{\sin(4\theta)}{\cos(4\theta)}\cdot\frac{1}{6\theta}.$$ What you are missing is that $$\lim_{\theta\to0}\frac{\sin(4\theta)}{\cos(4\theta)}\cdot\frac{1}{6\theta}=\lim_{\theta\to0}\frac{1}{\cos(4\theta)}\cdot\frac{\sin(4\theta)}{6\theta}=\frac16\lim_{\theta\to0}\frac{1}{\cos(4\theta)}\cdot\frac{\sin(4\theta)}{\theta},$$ followed by $$\frac16\lim_{\theta\to0}\frac{1}{\cos(4\theta)}\cdot\frac{\sin(4\theta)}{\theta}=\frac16\lim_{\theta\to0}\frac{4}{\cos(4\theta)}\cdot\frac{\sin(4\theta)}{4\theta}=\frac46\lim_{\theta\to0}\frac{1}{\cos(4\theta)}\cdot\lim_{\theta\to0}\frac{\sin(4\theta)}{4\theta}.$$ The rest is easy from here. The reason $$\frac16\lim_{\theta\to0}\frac{1}{\cos(4\theta)}\cdot\frac{\sin(4\theta)}{\theta}=\frac16\lim_{\theta\to0}\frac{4}{\cos(4\theta)}\cdot\frac{\sin(4\theta)}{4\theta}$$ works is because $\frac44=1$.