Let
\[ g(x) = \begin{cases} \frac {x^3 - 8}{x-2} &\text {if $x<1$,} \\ x^3+1 &\text {if $x>1$.} \end{cases} \]
Does \(\lim _{x \to 2} g(x)\) exist? If it does, give its value. Otherwise write DNE. \[ \lim _{x \to 2} g(x) = \answer {9} \]
Note that,
close to \(x=2\), the rule for \(g(x)\) is \(x^3+1\).