Removing Discontinuities | Math Coach Amy

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set \(f(a) = \lim_{x \to a} f(x)\)

only when that limit exists and is finite

What to Remember

Fill the hole

If \(\lim_{x \to a} f(x) = L\) but \(f\) is missing at \(a\), define \(f(a) = L\). That removes a removable discontinuity.

Piecewise: match the sides

Set the left piece at \(a\) equal to the right piece at \(a\). Solve for the unknown constant.

Jumps cannot be removed

If the one-sided limits disagree, no single y-value makes the function continuous there.

Trig holes

For \(\dfrac{\sin kx}{mx}\), the limit is \(\dfrac{k}{m}\). That is the y-value that fills the hole at 0.

Level 1

Easy

Name the y-value that fills a hole.

Example

Let \(f(x) = \dfrac{x^2 - 9}{x - 3}\) for \(x \neq 3\). What value should be assigned to \(f(3)\) to make \(f\) continuous at 3?

\(\dfrac{(x - 3)(x + 3)}{x - 3} = x + 3\) for \(x \neq 3\).

\(\lim_{x \to 3} f(x) = 6\)

Set \(f(3) = 6\).

Answer: \(6\)

Let \(f(x) = \dfrac{x^2 - 16}{x + 4}\) for \(x \neq -4\). What value of \(f(-4)\) makes \(f\) continuous at \(-4\)?

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\(x - 4\) after canceling.

\(-4 - 4 = -8\)

\(-8\)

Answer: \(-8\)

Let \(f(x) = \dfrac{x^2 - 5x + 6}{x - 2}\) for \(x \neq 2\). Find \(f(2)\) that makes \(f\) continuous at 2.

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\(\dfrac{(x - 2)(x - 3)}{x - 2} = x - 3\)

\(2 - 3 = -1\)

\(-1\)

Answer: \(-1\)

Can \(f(x) = \dfrac{|x|}{x}\) be made continuous at 0 by assigning a value to \(f(0)\)?

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Left-hand limit is \(-1\). Right-hand limit is \(1\).

The two-sided limit does not exist.

No.

Answer: no

Can \(f(x) = \dfrac{1}{x - 5}\) be made continuous at 5 by assigning a value to \(f(5)\)?

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Vertical asymptote. The limit is infinite.

No.

Answer: no

Let \(f(x) = \dfrac{x}{x}\) for \(x \neq 0\). What value of \(f(0)\) makes \(f\) continuous at 0?

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\(\dfrac{x}{x} = 1\) for \(x \neq 0\).

\(1\)

Answer: \(1\)

Level 2

Medium

Find the constant so two pieces meet.

Example

Find \(a\) so that \(f(x) = \begin{cases} x^2 + 2 & x < 1 \\ ax + 3 & x \geq 1 \end{cases}\) is continuous at \(x = 1\).

Set the pieces equal at \(x = 1\).

\(1^2 + 2 = a(1) + 3\)

\(3 = a + 3\)

\(a = 0\)

Answer: \(0\)

Find \(a\) so that \(f(x) = \begin{cases} 3x - 1 & x \leq 4 \\ ax - 5 & x > 4 \end{cases}\) is continuous at 4.

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\(3(4) - 1 = a(4) - 5\)

\(12 - 1 = 4a - 5\)

\(11 = 4a - 5\)

\(16 = 4a\)

\(a = 4\)

Answer: \(4\)

Find \(a\) so that \(f(x) = \begin{cases} 4 - x^2 & x < -2 \\ ax^2 + 1 & x \geq -2 \end{cases}\) is continuous at \(-2\).

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\(4 - 4 = a(4) + 1\)

\(0 = 4a + 1\)

\(4a = -1\)

\(a = -\dfrac{1}{4}\)

Answer: \(-\dfrac{1}{4}\)

Find \(a\) so that \(f(x) = \begin{cases} x^2 + x + a & x < 2 \\ x^3 - 2 & x \geq 2 \end{cases}\) is continuous at 2.

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\(4 + 2 + a = 8 - 2\)

\(6 + a = 6\)

\(a = 0\)

Answer: \(0\)

Find \(a\) so that \(f(x) = \begin{cases} 2x + 5 & x \leq -1 \\ ax + 2 & x > -1 \end{cases}\) is continuous at \(-1\).

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\(2(-1) + 5 = a(-1) + 2\)

\(-2 + 5 = -a + 2\)

\(3 = -a + 2\)

\(1 = -a\)

\(a = -1\)

Answer: \(-1\)

Find \(c\) so that \(f(x) = \begin{cases} \dfrac{\sin 6x}{8x} & x \neq 0 \\ c & x = 0 \end{cases}\) is continuous at 0.

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\(\dfrac{\sin 6x}{8x} = \dfrac{6}{8} \cdot \dfrac{\sin 6x}{6x}\)

\(\lim_{x \to 0} f(x) = \dfrac{3}{4}\)

\(c = \dfrac{3}{4}\)

Answer: \(\dfrac{3}{4}\)

Level 3

Hard

A quadratic in the unknown, or a trig hole.

Example

Find every value of \(a\), if any, so that \(f(x) = \begin{cases} a^2 + x^2 & x < 4 \\ a(x + 4) & x \geq 4 \end{cases}\) is continuous at \(x = 4\).

\(\lim_{x \to 4^-} f(x) = a^2 + 16\)

\(\lim_{x \to 4^+} f(x) = a(8) = 8a\)

\(f(4) = 8a\)

\(a^2 + 16 = 8a\)

\(a^2 - 8a + 16 = 0\)

\((a - 4)^2 = 0\)

\(a = 4\) only.

Answer: \(4\)

Find \(c\) so that \(f(x) = \begin{cases} \dfrac{\sin 5x}{10x} & x \neq 0 \\ c & x = 0 \end{cases}\) is continuous at 0.

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\(\dfrac{5}{10} \cdot \dfrac{\sin 5x}{5x}\)

\(c = \dfrac{1}{2}\)

Answer: \(\dfrac{1}{2}\)

Find \(a\) so that \(f(x) = \begin{cases} x^2 - 4 & x < 2 \\ ax & x \geq 2 \end{cases}\) is continuous at 2.

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\(4 - 4 = 2a\)

\(0 = 2a\)

\(a = 0\)

Answer: \(0\)

Find every \(a\) so that \(f(x) = \begin{cases} a^2 + x^2 & x < 2 \\ a(x + 2) & x \geq 2 \end{cases}\) is continuous at 2.

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\(a^2 + 4 = 4a\)

\(a^2 - 4a + 4 = 0\)

\((a - 2)^2 = 0\)

\(a = 2\)

Answer: \(2\)

Let \(f(x) = \dfrac{x^2 - 2x - 8}{x + 2}\) for \(x \neq -2\). Find \(f(-2)\) that removes the discontinuity.

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\(x^2 - 2x - 8 = (x - 4)(x + 2)\)

\(f(x) = x - 4\) for \(x \neq -2\).

\(-2 - 4 = -6\)

\(-6\)

Answer: \(-6\)

Is it possible to extend \(f(x) = \dfrac{x^2 - 1}{x^2 - 2x + 1}\) to be continuous at \(x = 1\)?

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\(x^2 - 2x + 1 = (x - 1)^2\)

\(\dfrac{(x - 1)(x + 1)}{(x - 1)^2} = \dfrac{x + 1}{x - 1}\) for \(x \neq 1\).

The simplified form still has a vertical asymptote at 1.

No.

Answer: no

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