Removing Discontinuities (AP Calculus AB 1.13)

The function $f(x) = \dfrac{x^2 - 2x - 8}{x - 4}$ has no value at $x = 4$ , so it is discontinuous there. Everywhere else it equals $x + 2$ , because the numerator factors as $(x - 4)(x + 2)$ . Its graph is the line $y = x + 2$ with a single point missing at $(4, 6)$ , and declaring $f(4) = 6$ fills that point and makes $f$ continuous. Repairs like this, by choosing one function value or a constant inside a formula, work for some discontinuities and are impossible for others, and the limit at the point decides which.

A limit that exists can be matched

Topic 1.11 defined continuity at $x = a$ with three conditions: $f(a)$ is defined, $\lim\limits_{x \to a} f(x)$ exists, and the two are equal. When you repair a function at $a$ , you are allowed to choose $f(a)$ . You are not able to change the limit by doing so, because the limit as $x \to a$ depends only on the values of $f$ at inputs near $a$ , never on the value at $a$ itself. So the repair works exactly when there is a finite limit to match.

That splits the three kinds of discontinuity from topic 1.10 cleanly:

Discontinuity at $a$	Limit at $a$	Can choosing $f(a)$ fix it?
Removable (a hole, or a point moved off the curve)	exists, equal to some number $L$	Yes: set $f(a) = L$
Jump	left and right limits are different numbers	No: $f(a)$ can equal at most one of them
Infinite	at least one side is unbounded	No: no number matches

This is where the name "removable" comes from: it is the one kind of discontinuity that a single new function value removes.

There are two versions of the repair. If $f(a)$ is undefined (a hole), you define $f(a) = L$ . If $f(a)$ is defined but is the wrong number (a point displaced from the curve), you redefine it. For instance, let $p(x) = \dfrac{\sin(5x)}{x}$ for $x \neq 0$ and $p(0) = 1$ . Writing $\dfrac{\sin(5x)}{x} = 5\cdot\dfrac{\sin(5x)}{5x}$ and using $\lim\limits_{u \to 0} \dfrac{\sin u}{u} = 1$ (topic 1.8) gives $\lim\limits_{x \to 0} p(x) = 5$ . The value $p(0) = 1$ is defined but wrong, and changing it to $p(0) = 5$ makes $p$ continuous at 0.

Finding the number to fill in

For a quotient at an input $a$ where the denominator is 0, substitute $a$ into the numerator first.

Numerator not 0. The quotient behaves like a nonzero number divided by something shrinking to 0, so it grows without bound: an infinite discontinuity, which cannot be removed (topic 1.14 covers this behavior).
Numerator also 0. The form $\frac{0}{0}$ says nothing yet. Simplify with the algebra of topic 1.6 (factor and cancel, multiply by a conjugate, use a trig limit) and find the limit. If it exists, that number is the value to fill in.

For example, $q(x) = \dfrac{\sqrt{x+7} - 3}{x - 2}$ gives $\frac{0}{0}$ at $x = 2$ . Multiplying the top and bottom by the conjugate $\sqrt{x+7} + 3$ turns the numerator into $(x + 7) - 9 = x - 2$ , so for $x \neq 2$ ,

q(x) = \frac{x - 2}{(x - 2)\left(\sqrt{x+7} + 3\right)} = \frac{1}{\sqrt{x+7} + 3} \;\longrightarrow\; \frac{1}{3 + 3} = \frac{1}{6}.

Defining $q(2) = \frac{1}{6}$ removes the discontinuity. The simplified formula agrees with $q$ only for $x \neq 2$ , so cancelling finds the limit, and stating $q(2) = \frac{1}{6}$ is the separate step that does the repair.

Matching the pieces at a boundary

A piecewise-defined function uses one formula on one side of a boundary point $c$ and another formula on the other side:

f(x) = \begin{cases} g(x), & x < c \\ h(x), & x \geq c \end{cases}

Suppose $g$ and $h$ are each continuous at $c$ , as polynomials, exponentials, sines and cosines are. Near $c$ on the left, $f$ uses $g$ , so $\lim\limits_{x \to c^-} f(x) = g(c)$ ; near $c$ on the right, $f$ uses $h$ , so $\lim\limits_{x \to c^+} f(x) = h(c)$ . The value at $c$ itself comes from whichever piece includes $c$ , here $f(c) = h(c)$ . You evaluate $g$ at $c$ even though $f$ never uses $g$ at $c$ : the one-sided limit only looks at inputs near $c$ , and $g$ is continuous there, so its limit is its value.

Continuity at $c$ needs the left limit, the right limit and $f(c)$ to be one number. With two pieces that becomes one equation, $g(c) = h(c)$ . If the function also lists a separate value at $c$ , say $f(c) = v$ , all three must agree: $g(c) = h(c) = v$ .

Each boundary gives one equation, so one unknown constant can usually be pinned down by one boundary, and two constants need two boundaries (as in Example 3). For example,

f(x) = \begin{cases} x^2 + k, & x < 3 \\ 4x - k, & x \geq 3 \end{cases}

is continuous at 3 when $9 + k = 12 - k$ , that is, $k = \frac{3}{2}$ . Both pieces then give $10.5$ at $x = 3$ .

The matching condition is about the boundary only. Each piece also has to be continuous on its own interval, which fails if, say, a rational piece has a zero denominator inside the interval where it is used. Check that before you report that a function is continuous everywhere.

When no value works

"Make $f$ continuous, if possible" allows for the answer "it isn't possible." Two things can cause that.

The discontinuity is not removable. If the limit at $a$ is infinite or the one-sided limits differ, no function value repairs it, as in the table above.

The equation for the constant has no solution. In

f(x) = \begin{cases} 3x + k, & x < 0 \\ e^x + k, & x \geq 0 \end{cases}

matching at 0 requires $k = 1 + k$ , which no number satisfies. The constant $k$ shifts both pieces up or down by the same amount, so the gap of 1 between them at $x = 0$ never closes; $f$ has a jump at 0 for every $k$ . An equation for $k$ that is quadratic can fail the same way when its discriminant is negative. When you find no solution, say so and say why: "no value of $k$ makes the one-sided limits equal."

Worked examples

Worked exampleExample 1: which discontinuities can be removed

Let $h(x) = \dfrac{x^3 - 8}{x^2 - 4}$ . Find every value of $x$ where $h$ is discontinuous. For each one, either define $h$ there to make it continuous or explain why that is impossible.

Show solution

A rational function is continuous everywhere it is defined, so the only candidates are the zeros of the denominator, $x^2 - 4 = 0$ , which are $x = 2$ and $x = -2$ .

Factor both parts to see what is shared. The difference of cubes gives $x^3 - 8 = (x - 2)(x^2 + 2x + 4)$ , and $x^2 - 4 = (x - 2)(x + 2)$ . For $x \neq \pm 2$ ,

h(x) = \frac{x^2 + 2x + 4}{x + 2}.

At $x = 2$ : the original is $\frac{0}{0}$ , and the simplified form gives

\lim_{x \to 2} h(x) = \frac{4 + 4 + 4}{2 + 2} = 3.

The limit exists, so the discontinuity is removable. Define $h(2) = 3$ .

At $x = -2$ : the numerator of the simplified form is $4 - 4 + 4 = 4$ , not 0, while the denominator $x + 2$ goes to 0. The quotient is unbounded near $-2$ , so this is an infinite discontinuity, and no value of $h(-2)$ can make $h$ continuous there.

The factor $x - 2$ appeared in both numerator and denominator, which is what made $x = 2$ removable. The factor $x + 2$ appeared only in the denominator, which is what made $x = -2$ an asymptote.

Worked exampleExample 2: a constant that appears squared

Find all values of $k$ for which the function

f(x) = \begin{cases} k^2x - 3, & x < 2 \\ k(x + 3), & x \geq 2 \end{cases}

is continuous at $x = 2$ .

Show solution

Both pieces are polynomials in $x$ , so each is continuous, and the one-sided limits are the pieces evaluated at 2:

\lim_{x \to 2^-} f(x) = 2k^2 - 3, \qquad \lim_{x \to 2^+} f(x) = 5k = f(2).

Continuity at 2 requires $2k^2 - 3 = 5k$ , or $2k^2 - 5k - 3 = 0$ . This factors as $(2k + 1)(k - 3) = 0$ , so $k = 3$ or $k = -\frac{1}{2}$ .

Both roots solve the continuity equation itself, so both are answers; substituting them back only confirms the arithmetic. With $k = 3$ , the left piece gives $18 - 3 = 15$ and the right gives $15$ . With $k = -\frac{1}{2}$ , the left gives $\frac{1}{2} - 3 = -\frac{5}{2}$ and the right gives $-\frac{5}{2}$ . Both values of $k$ work, and the answer must list both.

Worked exampleExample 3: two constants, two boundaries (exam style, no calculator)

Let $f$ be the function defined by

f(x) = \begin{cases} \dfrac{\sin(4x)}{x}, & x < 0 \\ ax + b, & 0 \leq x \leq 3 \\ \dfrac{x^2 + 7x - 30}{x - 3}, & x > 3 \end{cases}

where $a$ and $b$ are constants.

(a) Find $\lim\limits_{x \to 0^-} f(x)$ .

(b) Find the values of $a$ and $b$ for which $f$ is continuous at both $x = 0$ and $x = 3$ . Justify your answer.

(d) Suppose instead that $b$ has the value from part (b) but $a = 1$ . What kind of discontinuity does $f$ have at $x = 3$ , and can it be removed by changing only the value of $f(3)$ ?

Show solution

(a) For $x < 0$ , write $\dfrac{\sin(4x)}{x} = 4\cdot\dfrac{\sin(4x)}{4x}$ . As $x \to 0^-$ , $4x \to 0$ , so

\lim_{x \to 0^-} f(x) = 4\cdot 1 = 4.

(b) At $x = 0$ : the left limit is 4 from part (a). The middle piece is linear, so the right limit is $a\cdot 0 + b = b$ , and $f(0) = b$ . Continuity at 0 requires $b = 4$ .

At $x = 3$ : the middle piece gives $\lim\limits_{x \to 3^-} f(x) = f(3) = 3a + b$ . For $x > 3$ the numerator factors as $(x + 10)(x - 3)$ , so $f(x) = x + 10$ there and

\lim_{x \to 3^+} f(x) = 13.

Continuity at 3 requires $3a + b = 13$ . With $b = 4$ , $3a = 9$ and $a = 3$ .

Justification, as you would write it: with $a = 3$ and $b = 4$ , $\lim\limits_{x \to 0^-} f(x) = 4 = \lim\limits_{x \to 0^+} f(x) = f(0)$ and $\lim\limits_{x \to 3^-} f(x) = 13 = \lim\limits_{x \to 3^+} f(x) = f(3)$ , so $f$ is continuous at $x = 0$ and $x = 3$ .

(c) Yes. On $x < 0$ , $f$ is a quotient of continuous functions whose denominator $x$ is never 0 there. On $0 < x < 3$ it is a linear function. On $x > 3$ it is a rational function whose denominator $x - 3$ is never 0 there. So $f$ is continuous on each open piece, and part (b) makes it continuous at the two boundaries.

(d) With $a = 1$ and $b = 4$ , the left limit at 3 is $3 + 4 = 7$ and the right limit is still 13. Both one-sided limits are finite and unequal, so $f$ has a jump discontinuity at $x = 3$ . Changing $f(3)$ cannot remove it: $\lim\limits_{x \to 3} f(x)$ does not exist, and continuity requires that limit to exist and equal $f(3)$ .

Where students lose points

On the exam

Questions that ask for a constant making a piecewise function continuous expect you to show the equation you solved, built from the one-sided limits and the function value at the boundary, and to state your constants. A justification that the result is continuous names the three quantities and shows they are equal, as in Example 3(b). Topic 2.4 adds a second condition when a question asks for a function that is both continuous and differentiable at a boundary; this topic's equation is always the first of the two.

Check yourself

Question 1
Let $f(x) = \dfrac{x^2 - 5x + 6}{x^2 - 4}$ for $x \neq 2$ and $x \neq -2$ . What value of $f(2)$ makes $f$ continuous at $x = 2$ ?
1. A
  $-\dfrac{1}{4}$
2. B
  $\dfrac{1}{4}$
3. C
  $0$
4. D
  No value works, because $f$ has a vertical asymptote at $x = 2$ .
Show answer
A
Both the numerator and the denominator are 0 at $x = 2$ , so factor: $f(x) = \dfrac{(x-2)(x-3)}{(x-2)(x+2)} = \dfrac{x-3}{x+2}$ for $x \neq 2$ , and the limit at 2 is $\dfrac{-1}{4}$ . Choice D reacts to the zero denominator without checking the numerator; the asymptote is at $x = -2$ , where only the denominator is 0.
Question 2
The function $f(x) = \begin{cases} x^2 + a, & x < 1 \\ b, & x = 1 \\ 3x - a, & x > 1 \end{cases}$ is continuous at $x = 1$ . What is the value of $b$ ?
1. A
  $1$
2. B
  $2$
3. C
  $3$
4. D
  $4$
Show answer
B
The left limit is $1 + a$ and the right limit is $3 - a$ . Setting them equal gives $a = 1$ , so both limits are 2, and continuity also needs $f(1) = b$ to equal that common limit: $b = 2$ . Choice A is the value of $a$ , which is where students stop when they match only the two formulas.
Question 3
Is there a real number $k$ that makes $f(x) = \begin{cases} kx + 3, & x < 1 \\ k^2 + x^2 + 3, & x \geq 1 \end{cases}$ continuous at $x = 1$ ? If so, find it.
Show answer
No real value of $k$ works.
Matching the pieces at 1 requires $k + 3 = k^2 + 4$ , that is, $k^2 - k + 1 = 0$ . Its discriminant is $(-1)^2 - 4 = -3 < 0$ , so the equation has no real solution, and for every $k$ the one-sided limits at 1 differ (a jump).
Question 4
Let $g(x) = \dfrac{\sqrt{x + k} - 2}{x - 1}$ for $x \neq 1$ , where $k$ is a constant. Find the value of $k$ and the value of $g(1)$ that make $g$ continuous at $x = 1$ .
Show answer
$k = 3$ and $g(1) = \dfrac{1}{4}$ .
The denominator goes to 0, so the limit can exist only if the numerator also goes to 0: $\sqrt{1 + k} = 2$ , so $k = 3$ . Otherwise the discontinuity is infinite and cannot be removed. With $k = 3$ , multiplying by the conjugate gives $g(x) = \dfrac{1}{\sqrt{x+3} + 2}$ for $x \neq 1$ , which approaches $\dfrac{1}{4}$ , so define $g(1) = \dfrac{1}{4}$ .

Go deeper

OpenStax, Calculus Volume 1, 2.4 Continuity