Answer
(a) See graph, $0.29$, (b) See table, $0.29$, (c) $\frac{\sqrt 3}{6}$
Work Step by Step
(a) See graph, the limit can be estimated at the value of the function when $x\to 0$ as $0.29$
(b) See table, the values of the function when $x\to 0$ are listed, and the limit can be estimated as $0.29$
(c) Based on the Limit Laws, we have: $$\lim_{x\to 0}\frac{\sqrt {3+x}-\sqrt 3}{x}=\lim_{x\to 0}\frac{(\sqrt {3+x}-\sqrt 3)(\sqrt {3+x}+\sqrt 3)}{x(\sqrt {3+x}+\sqrt 3)}=\lim_{x\to 0}\frac{3+x-3}{x(\sqrt {3+x}+\sqrt 3)}=\lim_{x\to 0}\frac{1}{\sqrt {3+x}+\sqrt 3}=\frac{1}{\sqrt {3+0}+\sqrt 3}=\frac{1}{2\sqrt 3}=\frac{\sqrt 3}{6}$$
