AP Calculus AB and BC

Limit of (x^2 - x - 6)/(x - 3) at 3 Is 5

Q: How do I factor x^2 - x - 6 quickly?

Find two numbers multiplying to -6 and adding to -1: they are -3 and 2, so x^{2}-x-6 = (x-3)(x+2).

Q: Is the function the same as x + 2?

Everywhere except x = 3, yes. At x = 3 the original quotient is undefined while x+2 equals 5, and that single difference is invisible to the limit.

The limit of x squared minus x minus 6, over x minus 3, as x approaches 3 is 5. The numerator factors as x minus 3 times x plus 2, the shared factor cancels, and substituting 3 into x plus 2 gives 5.

\lim_{x \to 3} \frac{x^{2}-x-6}{x-3} = 5

Settled by factoring the quadratic.

You already know one of the factors

Both the numerator and the denominator are zero at $x = 3$ , an indeterminate $\frac{0}{0}$ form that is also a piece of information. A polynomial vanishing at $3$ must have $x-3$ among its factors, so the search narrows to finding that factor's partner.

The partner has to multiply with $-3$ to give the constant $-6$ , so it is $x+2$ . Expanding confirms it: $(x-3)(x+2) = x^{2}+2x-3x-6 = x^{2}-x-6$ .

\lim_{x \to 3}\frac{(x-3)(x+2)}{x-3} = \lim_{x \to 3}(x+2) = 5

Factoring with a head start

General factoring means hunting for two numbers at once. Here the zero over zero form has already handed you one factor, so you only have to find what it is multiplied by, and the constant term settles that in a line.

What the graph looks like

After cancelling, the function agrees with the line $y = x+2$ at every $x$ except $3$ , where it is undefined. The graph is that line with a single point missing at $(3,5)$ .

This is a removable discontinuity. Defining the value at $x = 3$ to be $5$ would patch the hole and make the function continuous, which is another way of saying the limit is $5$ .

The mistakes students make

Cancelling something that is not a factor, swapping the signs inside the factors, and refusing the limit because the point is missing from the domain account for nearly every wrong answer here.

Cancelling the $x$ terms to get $\frac{-x-6}{-3}$ and then $3$ . Cancellation applies to factors, never to individual terms.
Factoring as $(x+3)(x-2)$ . That expands to $x^{2}+x-6$ , nothing cancels, and the problem looks unsolvable.
Reporting no limit because $x = 3$ is outside the domain. A hole in the graph does not stop a limit, since 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 (x^2 - x - 6)/(x - 3) as x approaches 3?

It is $5$ .

How do I factor x^2 - x - 6 quickly?

Find two numbers multiplying to $-6$ and adding to $-1$ : they are $-3$ and $2$ , so $x^{2}-x-6 = (x-3)(x+2)$ .

Is the function the same as x + 2?

Everywhere except $x = 3$ , yes. At $x = 3$ the original quotient is undefined while $x+2$ equals $5$ , and that single difference is invisible to the limit.