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

1 Answer
Mar 12, 2018

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#