# How do you find the limit of abs(8x-56)/ (x-7) as x->7^-?

Mar 12, 2018

You need to consider exclusively values of $x < 7$

#### Explanation:

Since $x \to {7}^{-}$ we should consider only values of $x < 7$. In that case we have $8 x < 56$, and therefore $8 x - 56 < 0$. Hence:

$| 8 x - 56 | = - 8 x + 56$, and so the expression becomes:

$\frac{- 8 x + 56}{x - 7} = - 8 \frac{x - 7}{x - 7}$. We can simplify $\left(x - 7\right)$ since $x \ne 7$, so we have:

${\lim}_{x \to {7}^{-}} \frac{- 8 x + 56}{x - 7} = {\lim}_{x \to {7}^{-}} - 8 \frac{x - 7}{x - 7} = - 8$