What is the limit as x approaches 2 of ((x^3-4x))/(x^2-4x+4)) ?

If you look at a number very close to 2 on either side, we see:

f(x) = (x³ - 4x) / (x² - 4x + 4)

Using 1.999 and 2.001:

f(1.999) = (1.999³ - 4 * 1.999) / (1.999² - 4 * 1.999 + 4) and

f(2.001) = (2.001³ - 4 * 2.001) / (2.001² - 4 * 2.001 + 4)

f(1.999) = (7.988005999 - 7.996) / (3.996001 - 7.996 + 4) and

f(2.001) = (8.012006001 - 8.004) / (4.004001 - 8.004 + 4)

f(1.999) = -0.007994001 / 0.000001 and f(2.001) = 0.008006001 / 0.000001

f(1.999) = -7994.001 and f(2.001) = 8006.001

Since one side is negative and the other side is positive, we can make the following statements:

limit as x approaches 2 from the left is -∞

limit as x approaches 2 from the right is ∞

Since the limit is not the same on both sides, there is no defined limit as x approaches 2.

First simplify this (x^3-4x))/(x^2-4x+4)=(x(x^2-4))/(x-2)^2)= ( x(x-2)(x+2))/(x-2)^2= x (x+2)/(x-2)

Put x=2, the denominator is 0. The limit does not exist.

Factor top and bottom to get [x(x+2)(x-2)] / (x - 2)^2. At all value other than x=2, that's defined and equal to [x(x + 2)] / (x - 2). At x=2, the denominator is zero and the numerator is zero, so the limit doesn't exist.