Question

How do you find the limit of `abs(x-3)/ (x-3)` as x approaches `3^-`?

Answer 1

`lim_(x rarr 3^-) |x-3|/(x-3) = -1`

Explanation:

`\ \ \ \ \ \ lim_(x rarr 3^-) |x-3|/(x-3) = lim_(x rarr 3^-) -(x-3)/(x-3)` (as `x<3`)
`:. lim_(x rarr 3^-) |x-3|/(x-3) = lim_(x rarr 3^-) -1`
`:. lim_(x rarr 3^-) |x-3|/(x-3) = -1`

NB `{ lim_(x rarr 3^+) |x-3|/(x-3) = 1 }`