You need to factor (x^2 - x - 6)
(x^2 - x - 6) = (x - 3)(x + 2)
Now you have lim (x - 3)(x + 2) / (x - 3)
(x - 3) cancels out, so you are left with lim (x + 2)
The limit as x approaches 3 of (x + 2) is 5.
Notice, though that there is a hole in the equation because of the (x - 3) in the denominator. Although the limit exists, at x = 3 the value is undefined.
The limit exists because the limit as x approaches 3 from the positive and negative directions both exist. That does not mean the function exists at that point.
how would would u find f(3) and the
lim f(x)
x->3
of f(x)=(x^2-x-6) / (x-3)
1 answer