Answer
12
Work Step by Step
1. f is a closed function (we know by Th.10.1 that it is continuous), that is, L= $\displaystyle \lim_{x\rightarrow a}f(x)$ = $f(a)$,
for all a from the domain of f.
2. evaluating: $f(2)$, (plugging $x=2$) results in an indeterminate form 0/0.
So, we see that $x=2$ is NOT in the domain of f.
We apply case 2 (of step 2). Simplifying $f(x)$,
$\displaystyle \frac{x^{3}-8}{x -2}=\frac{x^{3}-2^3}{x -2}=$... difference of cubes...
$=\displaystyle \frac{(x-2)(x^{2}+2x+4)}{(x-2)}=$...reduce...$=x^{2}+2x+4=g(x)$
$g(2)=2^{2}+2(2)+4=12$
($2$ is in the domain of g, g is closed-form), so
$\displaystyle \lim_{x\rightarrow 2}f(x)=\lim_{x\rightarrow 2}g(x)=g(2)=12$
