For #f(x)# to be continuous at #x = 3#, the following must be true:

- #lim_(x->3)f(x)# exists.
- #f(3)# exists

Let's investigate the first postulate. We know that for a limit to exist, **the left hand and right hand limits must be equal**. Mathematically:

#lim_(x->3^-)f(x) = lim_(x->3^+)f(x)#

As defined above, we can see that to the left of #x = 3#, #f(x) = x^3 - 5x^2 + 2x + 8#.

Also, to the right of (and including) #x = 3#, #f(x) = -2x + k#

Hence, the limits of these functions as they approach #3# must be equal:

#=> lim_(x->3) x^3 - 5x^2 + 2x + 8 = lim_(x->3) -2x + k#

Evaluating the limits:

#(3)^3 - 5(3)^2 + 2(3) + 8 = -2(3) + k#

#=>26 - 45 + 6 + 8 = -6 + k#

#=> -5 = -6 + k#

#=> k = 1#

#k# being 1 guarantees that the limit exists, and we already know that #f(3)# exists (we actually just evaluated that above), and hence we can say that the function is continuous at #x = 3# IFF #k = 1#.