Lesson 1.10 · Unit 1 · CED topic 1.13
Removing discontinuities
If a graph has a hole, we can imagine filling it in. If it has a jump or an asymptote, no single dot can repair it. This lesson turns that picture into two skills the exam asks for again and again: redefining a function value to make a function continuous, and solving for an unknown parameter so that a piecewise function is continuous.
Learning objective.
- LIM-2.C. Determine values of x or solve for parameters that make discontinuous functions continuous, if possible.
Essential knowledge.
- LIM-2.C.1. If the limit of a function exists at a discontinuity in its graph, then it is possible to remove the discontinuity by defining or redefining the value of the function at that point, so it equals the value of the limit of the function as \(x\) approaches that point.
- LIM-2.C.2. In order for a piecewise-defined function to be continuous at a boundary to the partition of its domain, the value of the expression defining the function on one side of the boundary must equal the value of the expression defining the other side of the boundary, as well as the value of the function at the boundary.
Removing a removable discontinuity
If the limit of a function exists at a discontinuity, then it is possible to remove the discontinuity by defining or redefining the value of the function at that point so that it equals the limit. In symbols: if \(\displaystyle\lim_{x\to c}f(x)=L\) exists but \(f(c)\ne L\) (or \(f(c)\) is undefined), the function \[F(x)=\begin{cases}f(x),&x\ne c\\ L,&x=c\end{cases}\] is continuous at \(c\). Only one value changes. The graph of \(F\) is the graph of \(f\) with the hole filled in (or with a misplaced point moved to the hole).
Redefine \(f(x)=\dfrac{x^2-x-6}{x-3}\) at \(x=3\) so that it becomes continuous there.
Factor: \(x^2-x-6=(x-3)(x+2)\), so \(f(x)=x+2\) for \(x\ne3\) and \(\displaystyle\lim_{x\to3}f(x)=5\). Define \(f(3)=5\). With that value, \(f(x)=x+2\) for all \(x\), a polynomial, which is continuous everywhere.
Jump and infinite discontinuities cannot be removed this way. At a jump the two one-sided limits are different numbers, so no single value of \(f(c)\) can equal both; at a vertical asymptote there is no real number to equal the limit.
Continuity of piecewise functions
A piecewise function can be forced to be continuous at a boundary by choosing parameters. The recipe: at each boundary \(x=b\), set the value of the left expression equal to the value of the right expression (and to the function value, if it is given there separately). Each boundary produces one equation.
Find \(k\) so that \(g(x)=\begin{cases}\dfrac{x^2-9}{x-3},&x\ne3\\ k,&x=3\end{cases}\) is continuous at \(x=3\).
\(\displaystyle\lim_{x\to3}\dfrac{(x-3)(x+3)}{x-3}=6\). For continuity, \(g(3)=\displaystyle\lim_{x\to3}g(x)\), so \(k=6\).
For \(h(x)=\begin{cases}\dfrac{\sin2x}{x},&x\lt0\\ 3x+k,&x\ge0\end{cases}\), find \(k\) so that \(h\) is continuous at \(0\).
Left limit: \(\displaystyle\lim_{x\to0^-}\dfrac{\sin2x}{x}=2\cdot\lim\dfrac{\sin2x}{2x}=2\). Right side: \(h(0)=3(0)+k=k\), and the right limit is also \(k\). Continuity requires \(k=2\).
Find \(a\) and \(b\) so that \(f(x)=\begin{cases}2x,&x\lt1\\ ax+b,&1\le x\le3\\ x^2-1,&x\gt3\end{cases}\) is continuous everywhere.
Each piece is a polynomial, so the only trouble is at \(x=1\) and \(x=3\).
At \(x=1\). The left limit is \(2(1)=2\) and the middle expression gives \(a+b\). So \(a+b=2\).
At \(x=3\). The middle expression gives \(3a+b\); the right limit is \(3^2-1=8\). So \(3a+b=8\).
Subtract the first equation from the second: \(2a=6\), so \(a=3\), and then \(b=2-3=-1\). Check: at \(x=1\), \(3(1)-1=2\) matches; at \(x=3\), \(3(3)-1=8\) matches.
Notice that each of these is the three-condition test of Lesson 1.11 with an unknown in it. When you write a solution, show the one-sided limits (or the value from each side), set them equal, and solve.
- A removable discontinuity can be removed by (re)defining \(f(c)=\lim_{x\to c}f(x)\). Jump and infinite discontinuities cannot.
- To make a piecewise function continuous at a boundary, set the two sides (and the function value) equal and solve for the parameter; use one equation per boundary.
- Evaluate the limit that involves the unknown-free piece first; trigonometric and rational pieces may need Lesson 1.6 techniques.
- Check your parameter values by substituting them back at each boundary.
- What value should be assigned to \(f(4)\) so that \(f(x)=\dfrac{x^2-16}{x-4}\) is continuous at \(x=4\)?
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\(\dfrac{(x-4)(x+4)}{x-4}=x+4\) for \(x\ne4\), so the limit is 8. Define \(f(4)=8\).
- Find \(k\) so that \(f(x)=\begin{cases}x^2-k,&x\lt2\\ 3x+1,&x\ge2\end{cases}\) is continuous at \(x=2\).
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Left limit: \(4-k\). Right side: \(3(2)+1=7\). Set \(4-k=7\), so \(k=-3\).
- For each, say whether the discontinuity can be removed: (a) a hole at \(x=2\); (b) a jump at \(x=0\); (c) a vertical asymptote at \(x=1\).
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(a) Yes, define the value at \(x=2\) to be the limit. (b) No, the one-sided limits differ. (c) No, the limit is not a real number.
- Find \(a\) and \(b\) so that \(f(x)=\begin{cases}x+2,&x\lt-1\\ ax+b,&-1\le x\le2\\ 3x+1,&x\gt2\end{cases}\) is continuous everywhere.
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At \(x=-1\): \(-a+b=1\). At \(x=2\): \(2a+b=7\). Subtract: \(3a=6\), so \(a=2\) and \(b=3\).
- How should \(f(0)\) be defined so that \(f(x)=\dfrac{\sin3x}{x}\) is continuous at \(x=0\)?
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\(\displaystyle\lim_{x\to0}\dfrac{\sin3x}{x}=3\), so define \(f(0)=3\).
- A student claims the discontinuity of \(f(x)=\dfrac{1}{x^2}\) at \(x=0\) can be removed by defining \(f(0)=0\). Explain the error.
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\(\dfrac1{x^2}\) is unbounded near 0, so there is no real limit for \(f(0)\) to equal. Defining \(f(0)=0\) does not make the function continuous: the discontinuity is infinite, not removable.