We have to find the value of lim x-->2 [(x^3-2x^2)/(x-2)]

IF we substitute x=2, we get the indeterminate form 0/0. We can use the l'Hopital's Rule and replace the numerator and denominator with their derivatives.

We get :

lim x-->2 [(3x^2 - 4x)/1]

substitute x = 2

=> 3*2^2 - 4*2

=> 3*4 - 8

=> 4

**The required value of lim x-->2 [(x^3-2x^2)/(x-2)] = 4**