Decide whether you can fill a break

A gap in a line

The graph of \(f(x) = \dfrac{x^2 + x - 20}{x - 4}\) is a straight line with one point missing at \(x = 4\). The formula gives 0/0 there, so \(f(4)\) is not defined, and the ring marks the break.

Move the two orange points toward \(x = 4\), one from each side. At \(x = 3.9\) the height is 8.9, and at \(x = 4.1\) it is 9.1. As the points come closer to 4, both heights come closer to 9.

Because both sides approach the same height, the one point \((4, 9)\) fills the break. This lesson shows when one value can fill a break.

Any break

The curve has a break at the input \(c\). The orange dot sits at the height that the left piece approaches, and the red segment shows the gap up to the right piece.

Set the gap to 0. Both pieces approach the same height, so the dot joins them.

Raise the gap again. The dot meets only the left piece, because the right piece approaches a different height.

Press the button. The right piece rises without bound, so it approaches no height and no dot can meet it.

One value can fill the break only if both sides approach the same real number.

Notation

1. The height that the graph approaches from the left of \(c\) is the left limit, and the height from the right is the right limit. The orange lines mark these two heights.

   \[\cy{\lim_{x\to c^-} f(x)} \qquad \cy{\lim_{x\to c^+} f(x)}\]

2. If the two limits are equal, the graph approaches one height \(L\) from both sides. Then the two-sided limit exists and equals \(L\).

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

3. Give the function the value \(L\) at \(c\). The value then equals the limit, so the function is continuous at \(c\).

   \[f(c) = \cy{L}\]

4. A break can be filled only if the two limits are equal and their common value \(L\) is a real number, not \(\infty\).

   A break that one value can fill is a removable discontinuity.

When it fails

If the left and right limits are different, the two-sided limit does not exist. One value cannot equal two different heights.

If one side rises or falls without bound, its limit is not a real number. No value can meet a side that approaches no height.

Example 1: a gap that can be filled

Can the break of \(f(x) = \dfrac{x^2 - 9x + 14}{x - 2}\) at \(x = 2\) be filled? If it can, find the value.

1. At \(x = 2\) the top is \(4 - 18 + 14 = 0\), and the bottom is 0. The formula gives 0/0, so \(f(2)\) is not defined.

2. Factor the top. A limit uses only values of \(f\) near 2, never at 2, so \(x - 2\) is not 0 and can cancel.

   \[\begin{aligned} f(x) &= \frac{(x-2)(x-7)}{x-2} \\ &= x - 7 \quad (x \neq 2) \end{aligned}\]

3. The two orange points approach the height \(2 - 7 = -5\) from both sides. The limits are equal, and \(-5\) is a real number.

   \[\lim_{x\to 2} f(x) = \cy{-5}\]

4. The break can be filled with the value \(f(2) = -5\).

Example 2: two different heights

Let \(g(x) = x + 3\) for \(x < 1\), and \(g(x) = 3x - 1\) for \(x > 1\). Can one value fill the break at \(x = 1\)?

1. From the left, \(g(x) = x + 3\) approaches \(1 + 3 = 4\).

   \[\lim_{x\to 1^-} g(x) = \cy{4}\]

2. From the right, \(g(x) = 3x - 1\) approaches \(3 - 1 = 2\).

   \[\lim_{x\to 1^+} g(x) = \cy{2}\]

3. A tempting wrong move: The heights are 4 and 2, so choose the value halfway between them.

   \[g(1) = 3\]

   A dot at height 3 touches neither piece. A red gap of 1 remains between the dot and the left ring, and another between the dot and the right ring.

4. The limits 4 and 2 are different, so the two-sided limit does not exist. No value fills this break.

Practice

Decide whether the break of \(h(x) = \dfrac{x^2 - 10x + 16}{x - 8}\) at \(x = 8\) can be filled. If it can, find the value.

Move the two points toward \(x = 8\) and read the height that each side approaches. Then compare the two heights.

Questions

The graph shows \(f\). It has breaks at \(x = 1\), \(x = 3\) and \(x = 5\).

1. Can the break at \(x = 1\) be filled by one value?

   • Yes, with the value 2
   • Yes, with the value 1
   • No, because \(f(1)\) is not defined
   Answer

   Yes, with the value 2

2. Can the break at \(x = 3\) be filled by one value?

   • No, because the left side approaches 4 and the right side approaches \(\tfrac12\)
   • Yes, because \(f(3) = 4\) is already defined
   • Yes, with the value 2.25, halfway between 4 and \(\tfrac12\)
   Answer

   No, because the left side approaches 4 and the right side approaches \(\tfrac12\)

3. Find the value that fills the break of \(k(x) = \dfrac{x^2 - 2x - 15}{x + 3}\) at \(x = -3\).

   • \(-8\)
   • 0
   • \(-3\)
   • The break cannot be filled
   Answer

   \(-8\)

Summary

You find the limit from each side of the break and check that both are the same real number.

One value fills the break only then, because a single point can meet the graph only at a height that both sides approach.

The test applies to any break, whether the formula gives 0/0 or the graph shows a gap.

This is one of four skills in topic 1.13 of Which & Why: AP Calculus.