Compute the following limits$,$ showing all work. If using L'Hospital's Rule, state where it is used:

$($a$)\underset{x{1}^{-}}{lim}\frac{x^3+2}{4x^3-4}$

$\underset{x{1}^{-}}{lim}\frac{3x^2}{8x^2}=\underset{x{1}^{-}}{lim}\frac{6x}{16x}$

$($b$)\underset{x}{lim}\frac{x^3+2}{4x^3-4}$

$($c$)\underset{x{0}^{+}}{lim}{x}^x$