← Back to learning objective

Evaluate a limit by factoring a difference of cubes

Consider the function f(x)=x3−8x−2f(x) = \dfrac{x^3 - 8}{x - 2}.

Evaluate the limit lim⁡x→2x3−8x−2.\lim_{x \to 2} \frac{x^3 - 8}{x - 2}. Note that direct substitution gives the indeterminate form 0/00/0, so you will need to factor the numerator first.

Hint

Recall the difference of cubes factorization: a3−b3=(a−b)(a2+ab+b2)a^3 - b^3 = (a-b)(a^2+ab+b^2). Apply this with a=xa = x and b=2b = 2, then cancel the common factor with the denominator before substituting.

Step-by-step solution

Since direct substitution gives 0/00/0, factor the numerator using the difference of cubes formula: x3−8=(x−2)(x2+2x+4)x^3 - 8 = (x-2)(x^2 + 2x + 4). This is valid since 8=238 = 2^3.

Substituting this into the expression gives (x−2)(x2+2x+4)x−2.\frac{(x-2)(x^2+2x+4)}{x-2}. For x≠2x \neq 2, the factor (x−2)(x-2) cancels, leaving x2+2x+4x^2 + 2x + 4. Now substituting x=2x = 2 directly gives 22+2(2)+4=4+4+4=122^2 + 2(2) + 4 = 4 + 4 + 4 = 12.

Related problems

Practice similar Calculus problems interactively.

Start Closing the Gap