What is `lim_(x->6)(x^3 - 216)/(x - 6)`

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The limit `lim_(x->6)(x^3 - 216)/(x - 6)` has to be determined.

Substituting x = 6 gives the indeterminate form `0/0` . This allows the use of l'Hopital's rule and the numerator and denominator can be replaced with their derivatives.

=> `lim_(x->6) (3x^2)/1`

Substituting x = 6 gives 108

The limit `lim_(x->6)(x^3 - 216)/(x - 6) = 108`

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tonys538 | Student, Undergraduate | (Level 1) Valedictorian

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The limit `lim_(x->6)(x^3 - 216)/(x - 6)` has to be determined.

Substituting x = 6 in the given expression gives the result `(6^3 - 216)/(6-6) = 0/0` which is indeterminate.

The factorized form of `x^3 - y^3 = (x - y)*(y^2+x*y+x^2)` .

We can rewrite the limit as:

`lim_(x->6) (x - 6)(x^2 + 6x + 6^2)/(x - 6)`

= `lim_(x->6) (x^2 + 6x + 6^2)`

Substituting x = 6 gives 36 + 36 + 36 = 108

The required limit `lim_(x->6)(x^3 - 216)/(x - 6) = 108`

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