AP Calculus AB and BC

Limit of (x^4 - 16)/(x - 2) at 2 Is 32

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

Yes. The limit as x \to a is na^{n-1}, because the quotient is the alternate form of the derivative of x^{n} at a. It holds for any positive integer n, and for any real exponent n provided a > 0. Here n = 4 and a = 2, giving 4 \cdot 2^{3} = 32.

The limit of x to the fourth minus 16, over x minus 2, as x approaches 2 is 32. Factoring twice gives x minus 2 times x plus 2 times x squared plus 4, and the first factor cancels. The same quotient is the derivative of x to the fourth at 2.

\lim_{x \to 2} \frac{x^{4}-16}{x-2} = 32

Settled by factoring twice, or recognising a derivative at a point.

Route one: factor twice

x^{4}-16 = \left(x^{2}-4\right)\left(x^{2}+4\right) = (x-2)(x+2)\left(x^{2}+4\right)

Only the first bracket splits again. $x^{2}+4$ is a sum of squares and has no real factors, so the factoring stops there.

\lim_{x \to 2}(x+2)\left(x^{2}+4\right) = 4 \cdot 8 = 32

Route two: it is already a derivative

Put $f(x) = x^{4}$ . Then $f(2) = 16$ , so the quotient is $\frac{f(x)-f(2)}{x-2}$ , which is the alternate form of $f'(2)$ .

f'(x) = 4x^{3}, \qquad f'(2) = 4 \cdot 2^{3} = 32

No algebra at all, once the shape is recognised. The general statement is $\lim_{x \to a}\frac{x^{n}-a^{n}}{x-a} = na^{n-1}$ for any positive integer $n$ , and for any real exponent $n$ provided $a > 0$ . Carry it into Unit 2 with those conditions attached, since root and reciprocal powers need the restriction on $a$ .

The mistakes students make

The first comes from over eager factoring, the second from stopping too early, the third from the power rule.

Factoring $x^{2}+4$ as $(x+2)(x-2)$ , cancelling, and answering $0$ . A sum of squares does not factor over the reals.
Stopping at $\left(x^{2}-4\right)\left(x^{2}+4\right)$ , finding nothing to cancel, and reporting no limit.
Evaluating the derivative as $4 \cdot 2^{4}$ and answering $64$ . The power rule drops the exponent by one, so it is $4 \cdot 2^{3}$ .

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 limit of $\frac{x^{4}-16}{x-2}$ at 2?

It is $32$ .

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

Yes. The limit as $x \to a$ is $na^{n-1}$ , because the quotient is the alternate form of the derivative of $x^{n}$ at $a$ . It holds for any positive integer $n$ , and for any real exponent $n$ provided $a > 0$ . Here $n = 4$ and $a = 2$ , giving $4 \cdot 2^{3} = 32$ .

Does $x^{2}+4$ factor?

Not over the real numbers. Differences of squares factor, sums of squares do not, so $x^{2}-4$ splits and $x^{2}+4$ stays whole.