What is the limit of the fraction (x^3-2x^2)/(x-2) for x-->2 ?

Topic:

Calculus

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justaguide's profile pic

Posted on

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

giorgiana1976's profile pic

Posted on

We notice that substituting x by 2, we'll get a indetermination case:

lim  (x^3-2x^2)/(x-2) = (8 - 8)/(2-2) = 0/0

We'll factorize the numerator by x^2:

lim  (x^3-2x^2)/(x-2) = lim x^2(x - 2)/(x-2)

We'll simplify and we'll get:

lim x^2(x - 2)/(x-2) = lim x^2

We'll substitute x by 2:

lim x^2 = 2^2

lim  (x^3-2x^2)/(x-2) = 4

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