Answer
$$\lim_{x\to\sqrt2}g(x)=2\sqrt2$$
Work Step by Step
$$g(x)=\frac{x^2-2}{x-\sqrt2}$$
(a) The table is shown below.
As we can see from the, as $x$ gets more decimals and approaches the exact value of $\sqrt2$, the value of $g(x)$ also gets more decimals and apparently approaches $2\sqrt2\approx2.8284271247462...$. So I would estimate here that $\lim_{x\to\sqrt2}g(x)=2\sqrt2$.
(b) The graph is shown below.
Again, looking at the graph, the closer $x$ approaches $\sqrt2$, the closer $g(x)$ apparently approaches $2\sqrt2$. So my estimate again is $\lim_{x\to\sqrt2}g(x)=2\sqrt2$.
(c) $$\lim_{x\to\sqrt2}g(x)=\lim_{x\to\sqrt2}\frac{x^2-2}{x-\sqrt2}=\lim_{x\to\sqrt2}\frac{(x-\sqrt2)(x+\sqrt2)}{x-\sqrt2}$$ $$=\lim_{x\to\sqrt2}(x+\sqrt2)=\sqrt2+\sqrt2=2\sqrt2$$
Therefore, $$\lim_{x\to\sqrt2}g(x)=2\sqrt2$$
