# Evaluate `lim_(x->0)((x+3)^3 - 27)/x`

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You can use the formula of difference of cubes:

a^3-b^3=(a-b)(a^2+ab+b^2)

(x+3)^3-27=(x+3-3)[(x+3)^2+3(x+3)+9]

(x+3)^3-27=(x)[(x+3)^2+3(x+3)+9]

(x+3)^3-27=x(x^2+6x+9+3x+9+9)

(x+3)^3-27=x(x^2+9x+27)

Calculate the  limit

lim [(x+3)^3-27]/x = lim x(x^2+9x+27)/x

Reduce x and substitute x by 0

lim (x^2+9x+27)= 0^2+9*0+27=27

ANSWER: When x tends to 0, the limit of the function is 27.

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to determine `lim_(x->0)((x + 3)^3 - 27)/x`

Substituting x = 0 gives 0/0 which is indeterminate.

We can use the L'Hopital's rule and substitute the numerator and denominator by their derivatives.

=> `lim_(x->0)3*(x+3)^2`

substitute x = 0

=> 3*(0+3)^2

=> 3*9

=> 27

The required limit is 27

Wiggin42 | Student, Undergraduate | (Level 2) Valedictorian

Posted on

Substitute in 0 in ( (x + 3)^3 - 27 ) / x

and you will get 0/0. This is an indeterminate form where you can use L' Hopital's Rule.

Take the derivative of the numerator and the denominator:

3(x + 3)^2 / 1

Plug in 0

= 27