AP Calculus AB and BC

Limit of (x^2 - 9)/(x - 3) at 3 Is 6

Q: Why can you cancel x - 3 when it is zero at x = 3?

Because the limit only uses inputs near 3 and not equal to 3. For all of those, x-3 is a nonzero number and the cancellation is standard algebra.

The limit of x squared minus 9, over x minus 3, as x approaches 3 is 6. The numerator factors as x minus 3 times x plus 3, the shared factor cancels, and substituting 3 into x plus 3 gives 6.

\lim_{x \to 3} \frac{x^{2}-9}{x-3} = 6

Settled by factoring the difference of squares.

Factor, cancel, substitute

x^{2}-9 = x^{2}-3^{2} = (x-3)(x+3)

\lim_{x \to 3}\frac{(x-3)(x+3)}{x-3} = \lim_{x \to 3}(x+3) = 6

Direct substitution gives $\frac{0}{0}$ , which is not an answer. It is a report that the numerator and the denominator share the factor $x-3$ .

Why cancelling a zero factor is allowed

The original quotient and the line $x+3$ are not the same function. At $x = 3$ the quotient has no value at all, while $x+3$ has the value $6$ . Everywhere else the two agree exactly.

A limit at $3$ inspects inputs near $3$ and never $3$ itself. For every such input the factor $x-3$ is a nonzero number, so cancelling it is ordinary arithmetic. The single point of disagreement is invisible to the limit, which is why the limit exists at a place where the function does not.

What the graph looks like

The graph is the line y = x + 3 with one point punched out at (3, 6). The limit is 6 because the hole sits at height 6. That is a removable discontinuity, and defining the value at 3 to be 6 would repair it.

The mistakes students make

Two of these stop short of an answer and one lands on a specific wrong number.

Writing $\frac{0}{0}$ as the final answer. That form is a signal to factor, not a value.
Cancelling the $9$ against the $3$ to leave $\frac{x^{2}}{x}$ and answering $3$ . Only whole factors cancel, never single terms.
Saying the limit does not exist because $f(3)$ is undefined. A limit never asks for the value at the point.

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 $\frac{x^{2}-9}{x-3}$ as x approaches 3?

It is $6$ .

Why can you cancel x - 3 when it is zero at x = 3?

Because the limit only uses inputs near $3$ and not equal to $3$ . For all of those, $x-3$ is a nonzero number and the cancellation is standard algebra.

Is the function continuous at x = 3?

No. It is undefined there, so it has a removable discontinuity at $x = 3$ . The limit still exists and equals $6$ .