Factor and cancel to reveal a limit

A hole in a line

The chalk curve shows \(\dfrac{x^2-x-12}{x-4}\), and its top equals \((x-4)(x+3)\). At \(x=4\) the top and the bottom are both 0, so the curve has a hole there.

Drag the orange dot toward 4. At x = 3.9 the height is 6.9, and at x = 4.1 it is 7.1, so the heights approach 7.

Press the button to draw the dashed line \(y=x+3\) over the curve. The two have the same height at every input except 4, where the curve has a hole and the line has a point.

Two expressions that agree except at one input

The chalk curve \(f\) is any quotient whose top and bottom both have the factor \(x-c\). Both are 0 at \(c\), so \(f\) has a hole there.

At every input other than \(c\), the factor \(x-c\) is not 0, so the factor divided by itself is 1. Canceling it gives an expression \(g\), the dashed orange curve, with the same values as \(f\) at those inputs.

Drag the dot across \(c\). The two dots stay together at every input except \(c\), where \(f\) has no value and \(g\) has one.

A limit as \(x\) approaches \(c\) uses only inputs different from \(c\), so \(f\) and \(g\) have the same limit. The curve \(g\) has no hole at \(c\), so you find its limit by substituting \(c\).

Notation

1. Substitute \(c\). If the top and the bottom are both 0, the quotient is indeterminate.

2. Factor the top and the bottom. Each one is a polynomial that is 0 at \(c\), so each one has the factor \(x-c\).

   \[f(x)=\frac{\cq{(x-c)}\,p(x)}{\cq{(x-c)}\,q(x)}\]

3. Cancel \(x-c\). This is allowed for every \(x\) different from \(c\), because \(x-c\) is not 0 there.

   \[f(x)=\frac{p(x)}{q(x)}\quad (x\neq c)\]

4. The two expressions agree at every input except \(c\), so they have the same limit. If \(q(c)\) is not 0, substitute \(c\).

   \[\lim_{x\to c}f(x)=\lim_{x\to c}\frac{p(x)}{q(x)}=\cy{\frac{p(c)}{q(c)}}\]

When it fails

The bottom now has the factor \(x-4\) twice. Canceling one copy leaves \(\tfrac{x+3}{x-4}\).

The new bottom still approaches 0, while the top approaches 7. The curve falls to the left of the dashed vertical line and rises to the right, so there is no limit. Check that the new bottom is not 0 at the input before you substitute.

Example 1: A common factor

Find the limit of \(\dfrac{2x^2+6x}{x^2-9}\) as \(x\) approaches \(-3\).

1. Substitute −3. The top and the bottom are both 0, so the next step is to factor.

   \[\begin{gathered}2(-3)^2+6(-3)=0\\ (-3)^2-9=0\end{gathered}\]

2. Factor the top and the bottom. The top has the common factor 2x, and the bottom is a difference of squares.

   \[\frac{2x(x+3)}{(x-3)(x+3)}\]

3. Cancel \(x+3\). This is allowed because \(x\) is not \(-3\) in a limit, so \(x+3\) is not 0.

   \[\frac{2x}{x-3}\]

4. The new expression is defined at −3, so substitute.

   \[\frac{2(-3)}{-3-3}=\frac{-6}{-6}=1\]

Example 2: Only factors cancel

Find the limit of \(\dfrac{x^2+3x+2}{x^2-1}\) as \(x\) approaches \(-1\).

1. Substitute −1. The top and the bottom are both 0.

   \[\begin{gathered}(-1)^2+3(-1)+2=0\\ (-1)^2-1=0\end{gathered}\]

2. A tempting wrong move: Cancel the x² in the top with the x² in the bottom. Then substitute.

   \[\frac{3x+2}{-1}=\frac{-1}{-1}=1\]

   Only a factor of the whole top and of the whole bottom can cancel, and x² is one term of each sum, not a factor. The curve approaches the ring at height −1/2, so the red dot at height 1 is too high.

3. Factoring shows the common factor \(x+1\). Cancel it, and substitute \(-1\) into what remains.

   \[\begin{gathered}\frac{(x+1)(x+2)}{(x-1)(x+1)}=\frac{x+2}{x-1}\\ \to\ \frac{1}{-2}=-\frac12\end{gathered}\]

Practice

Find the limit of \(\dfrac{x^2+x-12}{x^2+3x-4}\) as \(x\) approaches \(-4\).

The top and the bottom are both 0 at −4, so the curve has an open ring there and each one has the factor x + 4.

Questions

The chalk curve is \(f\). The orange dashed line is \(g\), the expression that remains after a factor of \(f\) is canceled. The open ring marks a missing point of \(f\).

1. What is the limit of \(f(x)\) as \(x\) approaches 2?

   • 8
   • 0
   • It does not exist
   Answer

   8

2. A student writes \(f(x)=2x+4\) for every input \(x\). At which input is this false?

   • x = 2
   • x = 0
   • x = −2
   Answer

   x = 2

3. What is the limit of \(\dfrac{x^2+4x+3}{x^2-9}\) as \(x\) approaches \(-3\)?

   • \(\dfrac13\)
   • 1
   • 0
   • It does not exist
   Answer

   \(\dfrac13\)

Summary

You factor the top and the bottom, cancel the common factor that is 0 at the input, and substitute into what remains.

This works because the canceled expression has the same values as the original at every input except the one that the limit ignores.

Use it when substitution gives 0 over 0 and the top and the bottom are polynomials that you can factor.

This is one of four skills in topic 1.6 of Which & Why: AP Calculus.