AP Calculus AB and BC

Limit of (x^3-8)/(x-2) as x Approaches 2 Is 12

Q: Is there a general pattern for \frac{x^n-a^n}{x-a}?

Yes. The limit as x \to a is n a^{n-1}, because the fraction is the difference quotient for x^n at x = a. Here n = 3 and a = 2, giving 3 \cdot 2^2 = 12. The same rule turns \lim_{x \to 3}\frac{x^4-81}{x-3} into 4 \cdot 3^3 = 108.

The limit of (x^3 - 8)/(x - 2) as x approaches 2 is 12. Direct substitution gives 0/0. Factor the difference of cubes: x^3 - 8 = (x - 2)(x^2 + 2x + 4). Cancel the (x - 2), then substitute into the quadratic to get 4 + 4 + 4 = 12.

\lim_{x \to 2} \frac{x^3-8}{x-2} = 12

Settled by factoring the difference of cubes.

Peeling off the matching factor

Since $8 = 2^3$ , the numerator is $x^3 - 2^3$ . The difference of cubes pattern always splits off the matching linear factor and leaves a quadratic behind.

a^3-b^3 = (a-b)\left(a^2+ab+b^2\right) \quad \Longrightarrow \quad x^3-8 = (x-2)\left(x^2+2x+4\right)

The $(x-2)$ now matches the denominator and cancels for every $x \neq 2$ , which is every input a limit at $2$ actually looks at.

\lim_{x \to 2} \frac{x^3-8}{x-2} = \lim_{x \to 2} \left(x^2+2x+4\right) = 4 + 4 + 4 = 12

The leftover quadratic is continuous everywhere, so substituting $x = 2$ is the last step rather than a guess.

What direct substitution gives

At $x = 2$ the numerator is $8 - 8 = 0$ and the denominator is $2 - 2 = 0$ .

\frac{2^3-8}{2-2} = \frac{0}{0}

That form says a factor of $(x-2)$ is hiding in both places, nothing more. If the cubes pattern does not come to mind, dividing $x^3 - 8$ by $x - 2$ with long or synthetic division produces the same quotient $x^2 + 2x + 4$ with remainder $0$ .

A second route to 12

The fraction is also the difference quotient for $g(x) = x^3$ based at $x = 2$ , since $g(2) = 8$ . So the limit has to be $g'(2) = 3 \cdot 2^2 = 12$ . Two independent methods landing on the same number is a good habit for checking work.

The mistakes students make

Writing $x^3 - 8 = (x-2)\left(x^2-2x+4\right)$ . The middle sign is a plus for a difference of cubes; the minus version belongs to the sum $a^3+b^3 = (a+b)\left(a^2-ab+b^2\right)$ .
Assuming $x^3 - 8 = (x-2)^3$ . Expanding $(x-2)^3$ gives $x^3 - 6x^2 + 12x - 8$ , which is a different polynomial.
Evaluating $x^2 + 2x + 4$ at $x = 2$ as $4 + 2 + 4 = 10$ . The middle term is $2x$ , and $2 \cdot 2 = 4$ , so the total is $12$ .
Deciding the limit does not exist because the denominator hits zero. A zero denominator only forces an infinite limit when the numerator does not vanish with it.

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

Is there a general pattern for $\frac{x^n-a^n}{x-a}$ ?

Yes. The limit as $x \to a$ is $n a^{n-1}$ , because the fraction is the difference quotient for $x^n$ at $x = a$ . Here $n = 3$ and $a = 2$ , giving $3 \cdot 2^2 = 12$ . The same rule turns $\lim_{x \to 3}\frac{x^4-81}{x-3}$ into $4 \cdot 3^3 = 108$ .

Do I have to memorize the difference of cubes?

It saves time, but synthetic division reaches the same place. Dividing $x^3 + 0x^2 + 0x - 8$ by $x - 2$ gives coefficients $1, 2, 4$ and a remainder of $0$ , so the quotient is $x^2 + 2x + 4$ . The zero remainder confirms that $x - 2$ was a genuine factor.

Why is there no vertical asymptote at $x = 2$ ?

An asymptote needs the denominator to go to zero while the numerator does not. Here both go to zero and the factor cancels completely, so the graph has a removable hole at $(2, 12)$ instead of blowing up.