Determine the limit below (if it exists)
Problem:
lim 8x^2+40x+32/x+4
x→-4
I got 0/0, is that right?
8x^2+40x+32 = 8(x^2+5x+4) = 8(x+4)(x+1)
So, (8x^2+40x+32)/(x+4) = 8(x+1) at every value of x except x = -4.
So, the limit as you approach x = -4 is just 8(-4+1) = -24.
The graph has a hole at x = -4, since 0/0 is not defined. But everywhere else, no matter how close, f(x) = 8(x+1).
0/0 is not defined because, treating it as a normal fraction, if it has a value n, then
0/0 = n
0 = 0*n
But that is true for any value of n! In this case, it holds for n = -24.