When solving a limit, is the variable (theta) the same as any variable ( x or y, for example)? I'm trying to analytically evaluate the limit
$$\lim_{\theta \to 0} \frac{\tan(12\theta)}{5\theta}$$
I'm having some trouble because the theta is throwing me off a little. Does $\theta$ function the same as any variable would? Whenever $\pi$ and radians appear I get confused.
So far I've moved the $12$ and $5$ out of that part of the equation to leave me with $\tan\theta/\theta$, which becomes $\frac{\sin\theta}{2\cos(\theta)}$, and as $\cos\theta$ gets closer to $0$ the result becomes closer to $2.4$.
Does this mean that $(0, 2.4)$ is the limit (in other words, limit of $f(\theta)$ as $\theta$ goes to $0$ is $2.4$)?
 A: It seems to me you are a bit confused. The variable we choose does not interfere with the limit. The important thing is that whenever the same variable is encountered, we know that it represents the same thing, whatever that may be.
We also could replace $\theta$ with a star ($\star$) and the limit would remain the same, because where there was a theta, now there is a star. I also have to point out that your solution is not correct, since you have manipulated the limit in some ways that are not allowed. For example, $\tan(12\theta) \neq 12\tan\theta$. Here is how I would evaluate the limit:
$$\begin{align}
\lim_{\omega \to 0}\frac{\tan(12\omega)}{5\omega} &= \lim_{\omega \to 0}\frac{\sin(12\omega)}{5\omega\cos(12\omega)} =\\
&=\lim_{\omega \to 0}\frac{\sin(12\omega)}{12\omega}\cdot\frac{12\omega}{5\omega\cos(12\omega)} =\\
&= \lim_{\omega \to 0}\frac{\sin(12\omega)}{12\omega}\cdot\lim_{\omega \to 0}\frac{12\omega}{5\omega\cos(12\omega)} =\\
&= 1 \times \frac{12}{5} = \frac{12}{5} = 2.4
\end{align}$$
I used another variable on purpose, so that you can get yourself accustomed to seeing different symbols.
A: Yes $\theta$ is just a letter and works exactly like any other variable letter. There's nothing about the letter $\theta$ that means that "$\pi$ and radians appear", nor about other letters that would mean they don't.
In mathematics, the trigonometric functions generally work in radians, but that is independent of which variable letters we use. So your limit is exactly the same as
$$\lim_{x\to 0} \frac{\tan 12x}{5x} \qquad\text{or}\qquad \lim_{t\to 0} \frac{\tan 12t}{5t}$$
As others have pointed out your solution method is not quite right -- if you want to move the factor of 12 out, the universal tool you should be looking for is L'Hospital's rule. Since both $\tan 12x$ and $5x$ go to zero as $x\to 0$, we can replace each side of the fraction by its derivative:
$$ \lim_{x\to 0} \frac{\tan 12x}{5x} = \lim_{x\to 0}{12\tan'(12x)}{5} $$
and then in the numerator, the chain rule has lifted 12 outside for us. Then we can just set $x=0$ (since everything is now continuous at $0$) and know that $tan'(0)=1$, which will give the limit $12/5$.
