Answer
(a) See graph, $0.6666$, (b) See table $0.6667$, (c) $\frac{2}{3}$
Work Step by Step
(a) See graph, we can estimated the limit as $$\lim_{x\to 0}\frac{x}{\sqrt {1+3x}-1}=0.6666$$
(b) See table for function values when $x$ approaches zero, and we can estimate the limit as $0.6667$
(c) Based on the Limit Laws, we have: $$\lim_{x\to 0}\frac{x}{\sqrt {1+3x}-1}=\lim_{x\to 0}\frac{x(\sqrt {1+3x}+1)}{(\sqrt {1+3x}-1)(\sqrt {1+3x}+1)}=\lim_{x\to 0}\frac{x(\sqrt {1+3x}+1)}{(1+3x-1)}=\lim_{x\to 0}\frac{\sqrt {1+3x}+1}{3}=\frac{\sqrt {1+3\times0}+1}{3}=\frac{2}{3}$$
