Question

find the one - sided limit (if it exists). (if the limit is infinite, enter
lim_{x
ightarrow - 8^{-}}\frac{x + 8}{x^{2}+3x - 40}

Explanation:

Step1: Factor denominator

$x^2 + 3x - 40 = (x + 8)(x - 5)$

Step2: Simplify function

$\frac{x + 8}{(x + 8)(x - 5)} = \frac{1}{x - 5}$ (for $x
eq -8$)

Step3: Evaluate limit

As $x \to -8^-$, $x - 5 \to -8 - 5 = -13$, so $\frac{1}{x - 5} \to \frac{1}{-13}$

Answer:

$-\frac{1}{13}$