Question

Find limit x->2[x3-4x2+4x/x2-4]

Answer 1

Answer:

0

Step-by-step explanation:

$\frac{ {x}^{3} - 4 {x}^{2} + 4x }{ {x}^{2} - 4} \\ = \frac{x( {x}^{2} - 4x + 4) }{ {x}^{2} - {2}^{2} } \\ = \frac{x( {x}^{2} - 2x - 2x + 4) }{(x - 2)(x + 2)} \\ = \frac{x {(x - 2)}^{2} }{(x - 2)(x + 2)} \\ = \frac{x(x - 2)}{(x + 2)} \\ substituting \: x = 2 \\ = \frac{2(2 - 2)}{4} \\ = 0$

so the limit exist and it is 0

Hope it helps :D