How would you "remove the discontinuity" of f? in other words, how would you define
f(5) in order to make f continuous at 5?
f(x) = [(x^2 -2x -15)/x-5]

The numerator of this function given may be factored out as follows,
                     x² - 2x - 15 = (x - 5)(x + 3)
Given that the expression x - 5 appears in both the numerator and the denominator, we can cancel them and the new function becomes,
                              f(x) = x + 3
which is continuous at 5. 

A.5 is your answer to this problem