Match a formula to nearby graph behavior

A fraction with no value at x = 2

The curve shows \(f(x)=\dfrac{x^2-4}{x-2}\). The orange dot is on the curve at the input \(x\) that you choose, and the readout gives its height \(f(x)\).

If you move x toward 2 from the left, the height rises toward 4. From the right, the height falls toward 4.

At \(x=2\) the bottom of the fraction is 0, so \(f\) has no value there, and the open ring marks the missing point.

A shared factor

This graph has no numbers, so it stands for any fraction with the bottom \(x-c\). The faint vertical line marks the input \(c\).

Suppose the top is also 0 at \(c\). Then the top has the factor \(x-c\), which cancels with the bottom at every input except \(c\).

After the cancellation, the fraction has the same values as a simpler formula \(g\) at every input except \(c\). The orange dashed curve shows \(g\). Move the sliders. The ring stays on \(g\).

A limit uses only the values near \(c\), and not the value at \(c\). The height of \(g\) at \(c\) is therefore the height that the curve approaches.

Notation

1. Substitute \(c\) into the top. If the result is 0, the top has the factor \(x-c\), so factor it out.

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

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

   \[f(x)=\cy{g(x)}\quad (x\neq c)\]

3. Near \(c\) the curve is the graph of \(g\), so it approaches \(g(c)\).

   \[\lim_{x\to c} f(x)=\cy{g(c)}\]

4. The graph of \(f\) is the graph of \(g\) with an open ring at \(c\), so the limit is the height of the ring.

When it fails

The top is now \(x^2-3\). At \(x=2\) it equals 1, and not 0, so nothing cancels.

Near 2 the top is close to 1 and the bottom is close to 0, so the fraction is large. It is positive on the right of 2 and negative on the left, so it has no limit at 2.

Example 1: the graph approaches a height

Describe the graph of \(f(x)=\dfrac{x^2-x-12}{x-4}\) near \(x=4\).

1. The bottom is 0 at \(x=4\), so \(f\) has no value there. The top is also 0, so it has the factor \(x-4\).

   \[\frac{4^2-4-12}{4-4}=\frac{0}{0}\]

2. Factor the top.

   \[\frac{(x-4)(x+3)}{x-4}\]

3. Cancel \(x-4\), because \(x\) is not 4 in a limit.

   \[x+3\quad (x\neq 4)\]

4. The simpler formula equals 7 at \(x=4\), so the curve approaches 7 from both sides. The open ring is at \((4,7)\).

   \[\lim_{x\to 4} f(x)=4+3=\cy{7}\]

Example 2: two fractions, two graphs

Which of \(P(x)=\dfrac{x^2-36}{x-6}\) and \(Q(x)=\dfrac{x+6}{x-6}\) approaches a height as \(x\) approaches 6?

1. Neither \(P\) nor \(Q\) has a value at 6, because the bottom is 0 in both.

2. A tempting wrong move: Both fractions have the same bottom, so both graphs behave alike at 6.

   \[\text{same bottom}\Rightarrow\text{same graph}\]

   \(P\) is a line with one ring at height 12. The orange curve \(Q\) rises without bound on the right of 6 and falls without bound on the left.

3. Substitute 6 into the top of each fraction. The top of \(P\) is 0, so \(P\) has the factor \(x-6\). The top of \(Q\) is 12, so \(Q\) has no such factor.

   \[6^2-36=0,\qquad 6+6=12\]

4. \(P\) cancels to \(x+6\), so its graph approaches 12. Near 6 the top of \(Q\) is close to 12 and its bottom is close to 0, so its values grow without bound.

   \[\begin{gathered}P(x)=x+6\ \ (x\neq 6)\\ \lim_{x\to 6}P(x)=\cy{12}\end{gathered}\]

Practice

Describe the graph of \(f(x)=\dfrac{x^2-9x+20}{x-5}\) near \(x=5\). Give the height that it approaches.

Substitute 5 into the top. If the top is 0, factor out x − 5 and read the height from the simpler formula.

Questions

The graph shows \(h(x)=\dfrac{x^2+2x-3}{x^2-4x+3}\). It has an open ring at \(x=1\) and a dashed vertical line at \(x=3\).

1. As \(x\) approaches 1, which height does the curve approach?

   • \(-2\)
   • 0, because the top is 0 at 1
   • The limit does not exist, because h has no value at 1
   • 1
   Answer

   \(-2\)

2. As \(x\) approaches 3 from the right, what do the values of \(h\) do?

   • They grow without bound, so there is no limit at 3
   • They approach 12, the value of the top at 3
   • They approach \(-2\), as at \(x=1\)
   • They approach 0, because the bottom is 0
   Answer

   They grow without bound, so there is no limit at 3

3. Find \(\displaystyle\lim_{x\to -2}\frac{x^2+x-2}{x+2}\).

   • \(-3\)
   • 0
   • 3
   • The limit does not exist
   Answer

   \(-3\)

Summary

If the top and bottom of a fraction are both 0 at an input, you cancel the shared factor and evaluate the simpler formula there.

This works because a limit uses only the nearby values, and there the fraction and the simpler formula are equal.

Use it when a fraction has no value at an input, to decide whether the graph approaches a height.

This is one of four skills in topic 1.9 of Which & Why: AP Calculus.