AP Calculus AB and BC

Limit of (x^3-27)/(x-3) as x Approaches 3

Q: What is the general formula for this shape?

\lim_{x \to a}\frac{x^{n}-a^{n}}{x-a} = n a^{n-1}, because the expression is the difference quotient for x^{n} at a.

Q: How do I factor a difference of cubes?

a^{3}-b^{3} = (a-b)(a^{2}+ab+b^{2}). The sum of cubes is the same with the outer signs flipped: (a+b)(a^{2}-ab+b^{2}).

The limit of x cubed minus twenty-seven over x minus three as x approaches three is twenty-seven. Factoring the difference of cubes cancels the vanishing factor and leaves a quadratic that can be evaluated directly.

\lim_{x \to 3} \frac{x^{3}-27}{x-3} = 27

Settled by factoring the difference of cubes.

The difference of cubes

a^{3}-b^{3} = (a-b)(a^{2}+ab+b^{2}) \quad\Longrightarrow\quad x^{3}-27 = (x-3)(x^{2}+3x+9)

Cancelling $x - 3$ leaves $x^{2}+3x+9$ , and substituting $x = 3$ gives $9 + 9 + 9 = 27$ .

The middle term is $ab$ , not $2ab$ . Confusing this with the difference of squares expansion is the usual slip, and it produces a wrong quadratic that still looks plausible.

The derivative shortcut

This is the difference quotient for $f(x) = x^{3}$ at $x = 3$ , so the limit is $f'(3) = 3 \cdot 3^{2} = 27$ . The general pattern $\frac{x^{n}-a^{n}}{x-a} \to n a^{n-1}$ handles every case of this shape instantly.

Worth checking against the factored form: $x^{2}+3x+9$ at $x = 3$ has three equal terms of 9, which is exactly $3 \cdot 3^{2}$ .

Not sure which technique a limit wants?

The Limit Method Chooser walks the decision from direct substitution through factoring, the conjugate, and L'Hopital, and says why each one applies or fails.

Frequently asked questions

What is the general formula for this shape?

$\lim_{x \to a}\frac{x^{n}-a^{n}}{x-a} = n a^{n-1}$ , because the expression is the difference quotient for $x^{n}$ at $a$ .

How do I factor a difference of cubes?

$a^{3}-b^{3} = (a-b)(a^{2}+ab+b^{2})$ . The sum of cubes is the same with the outer signs flipped: $(a+b)(a^{2}-ab+b^{2})$ .