Locate a hole and its limiting value
A square root with a gap at x = 9
The chalk curve shows \(f(x)=\dfrac{x-9}{\sqrt{x}-3}\). The orange dot rides on the curve at the input \(x\) that you choose, and the faint vertical line marks \(x=9\).
At \(x=9\) the bottom is \(\sqrt{9}-3=0\), so \(f\) has no value there. Move \(x\) toward 9 from either side, and the height of the dot approaches 6.
For \(x\ge 0\) the top equals \((\sqrt{x}-3)(\sqrt{x}+3)\), so the fraction equals \(\sqrt{x}+3\) at every input except 9, and \(\sqrt{9}+3=6\).
The open ring marks the missing point, which is called a hole. Its input is 9 and its height is the limit 6, so the hole is at (9, 6).
The hole and the dot
The hole is now an open ring at any input \(c\), and the orange line marks the height \(L\) that the curve approaches.
A filled dot on the line \(x=c\) shows the value \(f(c)\), if \(f\) has one. Move the dot up and down, and the curve and the ring stay in place.
The hole is the point \((c,\ L)\). The value \(f(c)\) is a different fact, and the limit does not use it.
Notation
Find the input \(c\) where the formula has no value. For a fraction, \(c\) makes the bottom 0.
\[\text{bottom}=0\ \text{at}\ x=c\]Cancel the factor that is 0 at \(c\), because a limit uses only inputs \(x\neq c\). The result is a simpler formula \(g\), the orange dashed curve.
\[f(x)=g(x)\quad (x\neq c)\]Substitute \(c\) into \(g\). Because \(g\) equals \(f\) near \(c\), the result is the limit \(L\), the height of the orange line.
\[\lim_{x\to c} f(x)=\cy{g(c)}=\cy{L}\]The hole is the point with input \(c\) and height \(L\).
\[(c,\ \cy{L})\]The value \(f(c)\) is separate. It may be missing, or it may be a dot at another height.
\[f(c)\ \text{is not used}\]
When it fails
The left side of the curve still approaches \(\cy{L}\), but the right side now approaches a height 2 above it.
No single height fits both sides, so the limit does not exist and there is no hole to locate. A hole needs both sides to approach the same height.
Example 1: an expanded square
Find the hole of \(h(t)=\dfrac{(5+t)^2-25}{t}\).
The bottom is 0 at \(t=0\), so \(h\) has no value at 0. The top is also 0 there.
\[\frac{(5+0)^2-25}{0}=\frac{0}{0}\]Expand the top. It has the factor t.
\[\begin{gathered}(5+t)^2-25\\ =10t+t^2=t(10+t)\end{gathered}\]Cancel \(t\), because \(t\) is not 0 in a limit.
\[\frac{t(10+t)}{t}=10+t\quad (t\neq 0)\]At t = 0 the simpler formula equals 10, so the hole is the point with input 0 and height 10.
\[(0,\ \cy{10})\]
Example 2: a dot above the hole
Let \(j(x)=\dfrac{x^3-1}{x-1}\) for \(x\neq 1\), and \(j(1)=5\). Find the hole.
Factor the top and cancel \(x-1\), because \(x\) is not 1 in a limit.
\[\begin{gathered}\frac{(x-1)(x^2+x+1)}{x-1}\\ =x^2+x+1\quad (x\neq 1)\end{gathered}\]A tempting wrong move: The problem says \(j(1)=5\), so the hole is at \((1,5)\).
\[(1,\ \cr{5})\]The dot at height 5 is not on the curve. Near x = 1 the curve on both sides is close to height 3, where the ring is.
The hole uses the limit, not the value. Substitute 1 into the simpler formula to find the limit 3, so the hole is \((1,3)\). The dot at \((1,5)\) shows the value \(j(1)\), a separate point.
\[1^2+1+1=\cy{3}\]
Practice
Find the hole of \(k(x)=\dfrac{x^2+3x-10}{x+5}\).
The ring is at the input where the bottom is 0. Cancel the common factor. Then substitute that input into the simpler formula.
Solution
The bottom is 0 at \(x=-5\), so \(k\) has no value there.
Factor the top and cancel \(x+5\).
\[\begin{gathered}\frac{(x+5)(x-2)}{x+5}\\ =x-2\quad (x\neq -5)\end{gathered}\]At \(x=-5\) the simpler formula equals \(-7\), so the hole is the point with input \(-5\) and height \(-7\).
\[(-5,\ \cy{-7})\]
Questions
The graph shows \(f(x)=\dfrac{x^2-1}{x+1}\) for \(x\neq -1\), with \(f(-1)=1\). It has an open ring and a separate filled dot at \(x=-1\).
Which point is the hole in the graph?
- \((-1,\,-2)\)
- \((-1,\,1)\)
- \((-2,\,-1)\)
- \((-1,\,0)\)
Answer
\((-1,\,-2)\)
What is the value \(f(-1)\)?
- 1
- \(-2\)
- \(f\) has no value at \(-1\)
- 0
Answer
1
Find the hole of \(m(x)=\dfrac{x^3+27}{x+3}\).
- \((-3,\,27)\)
- \((-3,\,0)\)
- \((3,\,27)\)
- \((-3,\,9)\)
Answer
\((-3,\,27)\)
Summary
You find the input where the formula has no value, cancel the factor that is 0 there, and substitute the input into the simpler formula.
This works because the simpler formula has the same values near the input, and the value at the input does not change the limit.
Use it to give the coordinates of a hole in the graph of a fraction or a square root.
This is one of four skills in topic 1.9 of Which & Why: AP Calculus.