# What is the value for lim x-->2 [ (x^3 -8) / (x^2 -4)]

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

We see that if x is replced by 2 in (x^3 -8) / (x^2 -4) we get 0/0 which cannot be determined.

Instead we first factorize the numerator and denominator and cancel the common terms, this gives

[ (x^3 -8) / (x^2 -4)] =>

[ (x-2) ( x^2 + 2x +4) ] / [(x-2) ( x+2)

=> (x^2 +2x +4) / (x+2)

Now lim x--> 2 [(x^2 +2x +4) / (x+2)]

=> (2^2 +2*2 + 4) / (2+2)

=> (4+4+4) / 4 => 3

Therefore lim x-->2 [ (x^3 -8) / (x^2 -4)] = 3

tonys538 | Student, Undergraduate | (Level 1) Valedictorian

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The limit `lim_(x->2) (x^3 -8) / (x^2 -4)` has to be determined.

If we directly substitute x = 2 in `(x^3 -8) / (x^2 -4)` , the result is `(8-8)/(4-4) = 0/0` .

`0/0` is an indeterminate form and for such expressions in limits it is possible to use l'Hospital's rule and replace the numerator and denominator with their derivatives.

`(x^3 - 8)' = 3x^2`

`(x^2 - 4)' = 2x`

The limit is now :

`lim_(x->2) (3x^2)/(2x)`

= `lim_(x->2) (3x)/(2)`

Substituting x = 2 gives (3*2)/2 = 3

The limit `lim_(x->2) (x^3 -8) / (x^2 -4) = 3`

neela | High School Teacher | (Level 3) Valedictorian

Posted on

lt x-->2{x^3-8}/(x^2-4).

We see that if we put  x = 3,  both numerator and denomonator  becomes zer. So this is a 0/0 form of indeterminates.

Therefore we can diffrentiate both numerator and denominator and then take limit. Or we can divide by the common factor (x-2) and then take limt.

(x^3-8)/(x^2-4) = (x-2)(x^2+2x+4)/(x-2)(x+2) = (x^2+2x+4)/(x+2)

Therefore lt (x^3-8)/(x^2-4) = Lt(x^2+2x+4)/(x+2) = (2^2+2*2+4)/(2+2) =(12/4) = 3