#### Answer

As $h\to0$, the curve $y=\frac{\sin(2x+h)-\sin2x}{h}$ gets closer and closer to the curve $y=2\cos2x$, apparently approaching to become one with $y=2\cos2x$.

#### Work Step by Step

The graphs of the function are enclosed below. The graph of $y=2\cos2x$ is portrayed by the black curve, while the other curves are variations of $y=\frac{\sin(2x+h)-\sin2x}{h}$.
From the graph, we see that as $h\to0$, the curve $y=\frac{\sin(2x+h)-\sin2x}{h}$ gets closer and closer to the curve $y=2\cos2x$, apparently approaching to become one with $y=2\cos2x$.
In other words, we can say that $$\lim_{h\to0}\frac{\sin(2x+h)-\sin2x}{h}=2\cos2x$$
From the definition of derivative, we know that $$\lim_{h\to0}\frac{\sin(2x+h)-\sin2x}{h}=(\sin2x)'$$
Therefore, $$(\sin2x)'=2\cos2x$$
So the behavior of the graphs can be summarized to show that the derivative of $\sin2x$ is $2\cos2x$.