## Calculus, 10th Edition (Anton)

$\lim\limits_{x \to 4^{-}}\frac{3-x}{x²-2x-8}$ = +$\infty$
x² - 2x - 8 = (x - 4)(x + 2), so: $\lim\limits_{x \to 4^{-}}\frac{3-x}{x²-2x-8}$ = $\lim\limits_{x \to 4^{-}}\frac{3-x}{(x - 4)(x + 2)}$ = $\frac{3-4^{4}}{(4^{-}-4)(4^{-}+2)}$ = $\frac{-1}{0^{-}\times6}$ = $\frac{-1}{0^{-}}$ = +$\infty$