AP Calculus AB and BC

Limit of (x^2-16)/(x-4) as x Approaches 4

Q: What does the graph look like?

The line y = x + 4 with a single point removed at (4, 8). Filling that hole would make the function continuous, which is what removable means.

The limit of x squared minus sixteen over x minus four as x approaches four is eight. Factoring the numerator as a difference of squares cancels the offending factor, and direct substitution then finishes the job.

\lim_{x \to 4} \frac{x^{2}-16}{x-4} = 8

Settled by factoring a difference of squares.

Factor and cancel

\frac{x^{2}-16}{x-4} = \frac{(x-4)(x+4)}{x-4} = x+4 \quad (x \ne 4)

Substituting $x = 4$ into the simplified form gives 8. The cancellation is valid because a limit never asks about the value AT the point, only about values near it, and $x \ne 4$ throughout.

The original function has a hole at $x = 4$ : it is undefined there, while everywhere else it agrees with the line $y = x + 4$ . That is a removable discontinuity.

Two other ways to see it

The expression is the difference quotient for $f(x) = x^{2}$ at $x = 4$ , so its limit is $f'(4) = 2 \cdot 4 = 8$ . Recognising a derivative in disguise is often the fastest route on a multiple-choice question.

L'Hopital's rule also applies, giving $\frac{2x}{1} \to 8$ . All three routes agree, and factoring is the one that also reveals the hole.

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

Why can I cancel a factor that is zero at the limit point?

Because the limit only concerns values near the point, never the value at it. For every $x \ne 4$ the cancellation is ordinary algebra.

What does the graph look like?

The line $y = x + 4$ with a single point removed at $(4, 8)$ . Filling that hole would make the function continuous, which is what removable means.