Removing a Discontinuity — AP® Calculus AB Notes

The last three topics taught you to spot and classify breaks. Now you repair them. Exam questions phrase this two ways: "what value should $f(c)$ be?" and "for what value of $k$ is $f$ continuous?" Both come down to one idea: the value at the point has to equal the limit there.

Filling a hole

A removable discontinuity has a limit, $L = \lim_{x \to c} f(x)$ , but the value $f(c)$ is missing or different. To remove it, define (or redefine) the function at $c$ to be $L$ . Then all three conditions hold: $f(c)$ is defined, the limit exists, and they are equal.

Worked example 1. $b(x) = \frac{\sqrt{x + 9} - 3}{x}$ is undefined at 0. What value of $b(0)$ makes $b$ continuous at 0?

Find the limit. Substitution gives $\frac{0}{0}$ , so multiply by the conjugate:

$\frac{\sqrt{x + 9} - 3}{x} \cdot \frac{\sqrt{x + 9} + 3}{\sqrt{x + 9} + 3} = \frac{x}{x(\sqrt{x + 9} + 3)} = \frac{1}{\sqrt{x + 9} + 3}$

So $\lim_{x \to 0} b(x) = \frac{1}{6}$ . Define $b(0) = \frac{1}{6}$ .

Often you can do better than patching one point: write a single formula that has no hole. $\frac{x^2 - 9}{x + 3}$ equals $x - 3$ for $x \ne -3$ , and $g(x) = x - 3$ is a polynomial, continuous everywhere, with $g(-3) = -6$ exactly where the hole was.

Choosing a constant in a piecewise function

When a piecewise function contains a constant $k$ , continuity at the break point turns into an equation for $k$ . Find each side's limit using its own rule, and remember which rule gives the value at the break point.

Two graphs of f(x) = kx + 1 for x < 2 and f(x) = x² − 1 for x ≥ 2. Left, with k = 2, the line reaches an open circle at (2, 5) while the parabola starts at a closed dot at (2, 3): a jump. Right, with k = 1, the line reaches (2, 3) exactly where the parabola starts, so the pieces meet and f is continuous at 2. The condition is 2k + 1 = 3, so k = 1. — **Figure 1.** With the wrong $k$ the pieces miss; with the right $k$ they meet. LevlPrep original diagram.

Worked example 2. Let $f(x) = kx + 1$ for $x < 2$ and $f(x) = x^2 - 1$ for $x \ge 2$ . Find $k$ so that $f$ is continuous at 2.

Left: $\lim_{x \to 2^-} f(x)$ $= 2k + 1$ .

Right and value: $\lim_{x \to 2^+} f(x)$ $= 2^2 - 1$ $= 3$ , and $f(2) = 3$ from the same rule.

Set them equal: $2k + 1 = 3$ , so $k = 1$ .

Check: with $k = 1$ the left side approaches $2 + 1 = 3$ , which matches.

Each piece here is a polynomial, continuous on its own interval, so the break point is the only place to check. With $k = 1$ , $f$ is continuous on $(-\infty, \infty)$ .

Two constants, two break points

A function with three pieces has two break points. Each one gives an equation, and together they fix two constants.

Worked example 3. $f(x) = x + 2$ for $x < 0$ , $ax + b$ for $0 \le x \le 2$ , and $3x^2 - 2$ for $x > 2$ . Find $a$ and $b$ so that $f$ is continuous everywhere.

At $x = 0$ : the left side approaches $0 + 2 = 2$ ; the middle rule gives $f(0) = b$ . So $b = 2$ .

At $x = 2$ : the middle rule gives $f(2) = 2a + b = 2a + 2$ ; the right side approaches $3(4) - 2 = 10$ . So $2a + 2 = 10$ , and $a = 4$ .

When no value works

A constant can only move its own piece. Suppose $f(x) = x^2 + k$ for $x < 3$ , $f(3) = 5$ , and $f(x) = 2x + 1$ for $x > 3$ . The right side approaches 7, but the value is fixed at 5. No choice of $k$ changes either of those, so no $k$ makes $f$ continuous at 3. Always compare all three: left limit, right limit and value.

Jumps and infinite discontinuities cannot be removed either. At a jump the one-sided limits disagree, so there is no single limit for the value to match. Near a vertical asymptote the function is unbounded, so again there is no limit. Only removable discontinuities live up to their name.

Writing it up

A complete answer names the condition and shows the numbers: "For $f$ to be continuous at 2, $\lim_{x \to 2^-} f(x)$ $= \lim_{x \to 2^+} f(x)$ $= f(2)$ . The left rule gives $2k + 1$ and the right rule gives 3, so $2k + 1 = 3$ and $k = 1$ ." Then check by substituting back.