Answer
(a) $0.5$ (b) See graph.
Work Step by Step
(a) See table, a few points of the function values in the region of $[-1,1]$ are listed.. We can estimate the limit as : $\lim_{x\to0}\frac{x}{sin2x}=0.5$
(b) See graph, we can confirm that the results.
Extra: prove algebraically, $\lim_{x\to0}\frac{x}{sin2x}=\lim_{x\to0}\frac{1}{2}\frac{(2x)}{sin(2x)}=\frac{1}{2}$. Here we used a property: $\lim_{h\to0}\frac{sin(h)}{h}=1$
