Other ways to evaluate $\lim_{\theta\to 0} \frac{\sin2\theta}{2\theta}$? What steps should be taken to find the limit:
$$\lim_{\theta\to 0} \frac{\sin2\theta}{2\theta}$$?
I went about evaluating the limit using the fundamental rules of limits. I noticed that $\lim_{\theta\to 0}$ $\sin\theta\over\theta$ $=1$ and that the $\lim_{x\to a}$ $cx=ca$
Therefore, this meant that the limit of the function evaluated will go as followed, $\sin2\theta\over2 \theta$ $\to$ $2\times1\over2$ $\to$ $1$
I was wondering if I evaluated the function correctly, or did I get to this by luck? Is there another way to evaluate this equation?

Well, since very fortunately L'Hopital rule is not prohibited here, why don't we use it:
$$
\lim_{x\to 0 }\frac{\sin (2x)}{2x}=\lim_{x\to 0}\frac{2\cos 2x}{2}=1.
$$
