Wemazi

Topic 1.10

Recognize a removable hole

Builds on locating a discontinuity. 7 minutes.

Can one value repair the break?

The curve shows \(f(x)=\dfrac{x^2+3x-4}{x-1}\). The denominator is 0 at \(x=1\), so \(f\) has no value there and the graph has an open ring. The numerator factors as \((x+4)(x-1)\), so for every other \(x\) the fraction equals \(x+4\) and the curve heads to height 5 from both sides.

Move the orange dot at x = 1 up and down with the slider.

At every height except 5 the dot is off the curve, so the pen must still lift.

At height 5 the dot fills the ring and the break is gone, so one value, the height of the ring, repairs the break.

A finite limit makes a break removable

The graph now shows a break at an input \(c\) where both sides head to the same height \(L\), a finite number.

Such a break is a removable discontinuity, or a hole, because one value repairs it: set the value at \(c\) equal to \(L\).

Move the dot with the slider, or press the button to fill the ring. The repaired function equals \(f\) at every other input and is continuous at \(c\).

The limit decides whether a repair exists. The old value at \(c\), missing or wrong, does not matter.

Notation

  1. Find the two-sided limit at \(c\), the height that both sides head to.

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

  2. If \(L\) is finite and \(f(c)\) is missing or is not \(L\), the break is removable. The dot shows a value that is not \(L\).

    \[f(c)\ \text{missing, or}\ f(c)\neq L\]

  3. Define a new function \(F\). It equals \(f\) at every input except \(c\), and at \(c\) it equals the limit.

    \[F(x)=\begin{cases} f(x) & x\neq c\\ \cy{L} & x=c\end{cases}\]

  4. The new function \(F\) is continuous at \(c\), because \(F(c)\) exists, the limit at \(c\) exists, and the limit equals \(F(c)\).

When it fails

Press the button to lift the right side of the curve. On the right of \(c\) the curve now heads to a higher height than on the left.

The two sides head to different heights, so no single value fills the gap, and the break is not removable.

Example 1: a missing value

Show that \(k(x)=\dfrac{2x^2-18}{x-3}\) has a removable break at \(x=3\). Find the value that repairs it.

  1. The denominator is 0 at \(x=3\), so \(k\) has no value there, and the graph has a ring. The numerator is also 0 at 3, so \(x-3\) is a factor of it.

    \[2(3)^2-18=0\]

  2. Factor the numerator and cancel \(x-3\). This is allowed because \(x\) is not 3 in a limit.

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

  3. The two orange points approach height 12 from both sides. The limit at 3 is a finite number, so the break is removable.

    \[\lim_{x\to 3} k(x)=2(3+3)=\cy{12}\]

  4. Define \(K(3)=12\), and let \(K\) equal \(k\) at every other input. The dot fills the ring, so \(K\) is continuous at 3.

Example 2: two breaks, one removable

Let \(n(x)=\dfrac{x^2-x-6}{x^2-2x-3}\). Decide which breaks of \(n\) are removable.

  1. Factor the denominator. It is 0 at \(x=3\) and at \(x=-1\), so \(n\) has no value at either input.

    \[x^2-2x-3=(x-3)(x+1)\]

  2. A tempting wrong move: The function \(n\) has no value at either input, so both breaks are removable.

    \[x=3,\ x=-1:\ \text{removable}\]

    A removable break needs a finite limit. On the right of \(x=-1\) the graph climbs beside the dashed line, and at \(x=-0.99\) the value is already 101. No height can fill that gap.

  3. Factor the numerator and cancel \(x-3\). For \(x\neq 3\) the result is a simpler fraction.

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

  4. At \(x=3\) the simpler fraction equals \(\frac{5}{4}\), a finite number, so the break at 3 is removable. At \(x=-1\) its numerator is 1 and its denominator is 0, so the values grow without bound in size, and that break is not removable.

    \[\frac{3+2}{3+1}=\cy{\frac{5}{4}}\]

Practice

Let \(q(x)=\dfrac{3x^2+6x}{x+2}\) for \(x\neq -2\). Which value of \(q(-2)\) makes \(q\) continuous at \(-2\)?

Find the limit at \(-2\). If it is a finite number, the repair value is that limit.

Solution

  1. The denominator is 0 at \(-2\). Factor the numerator and cancel \(x+2\).

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

  2. The limit at \(-2\) is a finite number, so the break is removable. The orange line marks its height.

    \[\lim_{x\to -2} q(x)=3(-2)=\cy{-6}\]

  3. Setting \(q(-2)=-6\) fills the hole, so \(q\) is continuous at \(-2\).

Questions

The graph shows a function \(F\) with a break at each of the inputs \(-2\), 1 and 4.

  1. At which input can the break be removed by changing the value of F at that one input?

    • \(-2\)
    • 1
    • 4
    • All three inputs
    Answer

    \(-2\)

  2. Which value of \(F(-2)\) removes the break at \(-2\)?

    • 3
    • \(-2\)
    • 6
    • No value works
    Answer

    3

  3. Let \(g(x)=\dfrac{x^2-9x+14}{x-7}\) for \(x\neq 7\). Which value of \(g(7)\) makes \(g\) continuous at 7?

    • 5
    • 0
    • 7
    • No value works, because the denominator is 0
    Answer

    5

Summary

You find the two-sided limit at the break and check that it is a finite number.

This works because a finite limit is the height of the missing point, so one value equal to the limit fills the hole.

Use it to tell a hole from a jump or an infinite break, and to repair the hole.

This is one of five skills in topic 1.10 of Which & Why: AP Calculus.