Given #(x - 4) / (x^2 - 2x - 8)# how do you find the limit as x approaches -2?

1 Answer
Jan 24, 2017

It does't have a limit.

Explanation:

The numerator tends to -6 and the denominator tends to zero at #x=-2# and #x=+4#. It doesn't help if you cancel the #x-4# as the numerator becomes one and the denominator still tends to zero. However, original function tends to #1/6# as #x# tends to #+4#. The graph of the original function has a single vertical asymptote at #x=-2# and has a hole at x=+4 as the function is undefined there. If you cancel the common factor the new function is the same as the old one but is now defined (as #1/6#) at #x=4#, and the hole in the graph is filled.