Evaluate and simplify: $\lim_{x\to 0}\frac{\sin 2x}{3x}$ 
Evaluate and simplify:
  $$\lim_{x\to 0}\frac{\sin 2x}{3x}$$

So, by direct substitution, it is indeterminate. I got the derivative of this functio,n and now I'm going around in circles with the trig identities. 
Is there a more efficient way of solving this?
 A: $$\lim_{x\to0}\frac{\sin 2x}{3x}=\lim_{x\to0}\frac{2}{3}\cdot\frac{\sin 2x}{2x}=\frac{2}{3}\cdot\lim_{x\to0}\frac{\sin 2x}{2x}=\frac{2}{3}$$
A: Hint: Note that the function $$\frac{\sin y}{y}$$ has a limiting value of $1$ as $y$ vanishes.
A: Another way is to use the identity
$$
\sin 2x=2\sin x\cos x
$$
whence
$$
\lim_{x\to 0}\frac{\sin 2x}{3x}=\lim_{x\to 0}\frac{2\sin x\cos x}{3x}=\frac{2}{3}
\left(\lim_{x\to0}\frac{\sin x}{x}\right)\left(\lim_{x\to 0}\cos x\right)=\frac{2}{3}\times 1\times 1=\frac{2}{3}$$
where we used the limit laws in conjunction with the well-known fact that
$$
\lim_{x\to0}\frac{\sin x}{x}=1
$$
as well as the fact that cosine is a continuous function.
A: Apply L'Hôpital's rule (https://en.wikipedia.org/wiki/L%27Hôpital%27s_rule). 
We have $lim_{x\rightarrow 0}\frac{sin2x}{3x} =lim_{x\rightarrow 0}\frac{2cos2x}{3}=\frac{2}{3}.$
One more approach that you may apply using the sine series($Sin(x)=x-\frac{x^3}{3}+\ldots$) and then apply the limit $x\rightarrow 0$, you will get the same result. 
A: Use L'Hospital's Rule:
$$\lim_{x \to 0}\frac{\sin(2x)}{3x}$$
$$\lim_{x \to 0}\frac{2\cos(2x)}{3}$$
$$\lim_{x \to 0}\frac{2*1}{3}$$
$$\frac{2}{3}$$
A: Use L'Hôpital's rule
Differentiate both sides
$$\lim_{x\to 0}  \frac {2\cos {2x}}{3}$$ 
Put $x=0$ 
$\cos {(2*0)}=1$
$$\lim_{x\to 0}  \frac {2\cos {2x}}{3}=   \frac {2}{3}$$ 
