Question

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

Answer 1

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

Explanation:

Since `x -> 7^-` we should consider only values of `x < 7`. In that case we have `8x < 56`, and therefore `8x - 56 < 0`. Hence:

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

`(-8x+56)/ (x-7)= - 8 (x -7)/(x-7)`. We can simplify `(x-7)` since `x != 7`, so we have:

`lim_(x-> 7^-)(-8x+56)/ (x-7)= lim_(x-> 7^-)- 8 (x -7)/(x-7)= -8`