If f(x)= (x^2+4x+4)/(x+2), why does f(-2) not exist? Show algebraically that lim┬(x→ -2)⁡〖f(x)〗 does exist. Support your answer graphically and through table of values.

Question

If f(x)= (x^2+4x+4)/(x+2), why does f(-2) not exist? Show algebraically that lim┬(x→ -2)⁡〖f(x)〗 does exist. Support your answer graphically and through table of values.

Answer 1

division by zero is impossible. It's undefined. It has no real quotient. It's either infinity or negative infinity, which are not real numbers. It's even worse, it's 0/0 as ((-2)^2 +4(-2) +4)/(-2+2) = 0/0. 0/0 is very undefined, it can be any real number or a surreal number (infinity, negative infinity or an infinitely small number. 0/0 = y where 0(y) = 0. y can be any number. 0 times any number = 0.

(x^2 +4x +4)/(x+2) = (x+2)(x+2)/x+2) = x+2. Cancel the (x+2) in the numerator and denominator, leaving just x+2. then substitute x=-2 to get -2+2 = 0

Graph the equation f(x) = (x^2+4x+4)/(x+2) and you get a straight line f(x)=x+2, except for a missing point, a discontinuity at x=-2. It's a line with a "hole" in it.

you can create a table of values for f(x)= (x^2+4x+4)/(x+2) which is

identical to f(x)=x+2, except for x=-2, y=0, the point (-2,0)

as you approach x=-2 from left or right, y approaches 0.

it's the limit from left or right, but the point is never reached

when f(x) = (x^2+4x+4)/(x+2)