Absolute value functions

What you’ll learn

Evaluating limits of piecewise-defined and absolute value functions with one-sided limits

Piecewise-defined functions

A piecewise function uses different formulas on different intervals of $x$ . To evaluate a limit at a breakpoint (a value of $x$ where the formula changes), find the one-sided limits.

The overall limit exists only if both one-sided limits are equal.

Example

Let $f (x)$ be the piecewise function defined by

$f (x) = ⎩ ⎨ ⎧ x^{2} 2 x 4 if x \leq - 1 if - 1 < x \leq 2 if x > 2$

Evaluate:

a) $x \to - 1 lim f (x)$

b) $x \to 2 lim f (x)$

c) $x \to 3 lim f (x)$

Answers

(spoiler)

a) Does not exist
b) $4$
c) $4$

Solutions

a) $x \to - 1 lim f (x)$

(spoiler)

$x$ approaches $- 1$ , which is a breakpoint. Comparing the one-sided limits,

Left-hand limit:

For $x \leq - 1$ , the function is $f (x) = x^{2}$ , so

$x \to - 1^{-} lim x^{2} = 1$

Right-hand limit:

For $- 1 < x \leq 2$ , the function is $f (x) = 2 x$ , so

$x \to - 1^{+} lim 2 x = - 2$

Since

$x \to - 1^{-} lim f (x) \neq = x \to - 1^{+} lim f (x)$

the overall limit $x \to - 1 lim f (x)$ does not exist.

b) $x \to 2 lim f (x)$

(spoiler)

Here, $x$ approaches $2$ , which is also a breakpoint.

Left-hand limit:

For $- 1 < x \leq 2$ , $f (x) = 2 x$ , so

$x \to 2^{-} lim 2 x = 4$

Right-hand limit:

For $x > 2$ , $f (x) = 4$ , so

$x \to 2^{+} lim 4 = 4$

Both one-sided limits match, so

$x \to 2 lim f (x) = 4$

c) $x \to 3 lim f (x)$

(spoiler)

Since $x$ approaches $3$ which is not a breakpoint, one-sided limits are not necessary.

As $x$ approaches $3$ , $f (x)$ is defined by the third branch, $f (x) = 4$ for $x \geq 2$ . Therefore,

$x \to 3 lim 4 = 4$

When evaluating the limit of a rational expression with absolute values, direct substitution may result in the indeterminate form $\frac{0}{0}$ . In these cases, rewriting the absolute value function as a piecewise-defined one helps evaluate the limit by considering each side separately.

For example, the absolute value function $y = ∣ x ∣$ is defined piecewise as

$∣ x ∣ = ⎩ ⎨ ⎧ x - x if x \geq 0 if x < 0$

Rewrite

$f (x) = ∣3 - x ∣$

as a piecewise function.

1. Find the breakpoint(s)

Set the expression inside the absolute value equal to $0$ :

$3 - x x = 0 = 3$

This divides the domain into two intervals: $x < 3$ and $x \geq 3$ .

2. Test the intervals

To determine how the expression behaves on each interval, test a value from each side:

For $x < 3$ : Test $x = 2$ .
- Since $3 - 2 > 0$ , use the positive version.
For $x > 3$ : Test $x = 4$ .
- Since $3 - 4 < 0$ , use the negative version.

At $x = 3$ , both pieces evaluate to $0$ . By convention, the equality is included with the greater than sign unless the problem states otherwise.

Therefore in piecewise form,

$f (x) = ⎩ ⎨ ⎧ 3 - x - (3 - x) if x < 3 if x \geq 3$

Examples

Use the piecewise-defined function above to evaluate

$x \to 3 lim \frac{∣3 - x ∣}{x - 3}$

Solution

(spoiler)

In piecewise form,

$\frac{∣3 - x ∣}{x - 3} = ⎩ ⎨ ⎧ \frac{3 - x}{x - 3} \frac{- ( 3 - x )}{x - 3} if x < 3 if x \geq 3$

Next, consider the one-sided limits:

Left-hand limit:

$= = x \to 3^{-} lim \frac{3 - x}{x - 3} x \to 3^{-} lim \frac{( 3 - x )}{- ( 3 - x )} - 1$

Right-hand limit:

$= = x \to 3^{+} lim \frac{- ( 3 - x )}{x - 3} x \to 3^{-} lim \frac{- ( 3 - x )}{- ( 3 - x )} 1$

Since the one-sided limits don’t match, $x \to 3 lim f (x)$ does not exist.

Find

$x \to 1 lim \frac{x ^{2} - 3 x + 2}{∣ x - 1∣}$

Solution

(spoiler)

In piecewise form,

$f (x) = ⎩ ⎨ ⎧ \frac{x ^{2} - 3 x + 2}{x - 1} \frac{x ^{2} - 3 x + 2}{- ( x - 1 )} if x \geq 1 if x < 1$

Next, evaluate the one-sided limits.

Left-hand limit:

$= = = x \to 1^{-} lim \frac{x ^{2} - 3 x + 2}{x - 1} x \to 1^{-} lim \frac{( x - 2 ) ( x - 1 )}{( x - 1 )} x \to 1^{-} lim (x - 2) - 1$

Right-hand limit:

$= = = x \to 1^{+} lim \frac{x ^{2} - 3 x + 2}{- ( x - 1 )} x \to 1^{+} lim \frac{( x - 2 ) ( x - 1 )}{- ( x - 1 )} x \to 1^{+} lim \frac{( x - 2 )}{- 1} 1$

Since the one-sided limits don’t match, $x \to 1 lim f (x)$ does not exist.

What you’ll learn

Evaluating limits of piecewise-defined and absolute value functions with one-sided limits

Piecewise-defined functions

A piecewise function uses different formulas on different intervals of $x$ . To evaluate a limit at a breakpoint (a value of $x$ where the formula changes), find the one-sided limits.

The overall limit exists only if both one-sided limits are equal.

Example

Let $f (x)$ be the piecewise function defined by

$f (x) = ⎩ ⎨ ⎧ x^{2} 2 x 4 if x \leq - 1 if - 1 < x \leq 2 if x > 2$

Evaluate:

a) $x \to - 1 lim f (x)$

b) $x \to 2 lim f (x)$

c) $x \to 3 lim f (x)$

Answers

(spoiler)

a) Does not exist
b) $4$
c) $4$

Solutions

a) $x \to - 1 lim f (x)$

(spoiler)

$x$ approaches $- 1$ , which is a breakpoint. Comparing the one-sided limits,

Left-hand limit:

For $x \leq - 1$ , the function is $f (x) = x^{2}$ , so

$x \to - 1^{-} lim x^{2} = 1$

Right-hand limit:

For $- 1 < x \leq 2$ , the function is $f (x) = 2 x$ , so

$x \to - 1^{+} lim 2 x = - 2$

Since

$x \to - 1^{-} lim f (x) \neq = x \to - 1^{+} lim f (x)$

the overall limit $x \to - 1 lim f (x)$ does not exist.

b) $x \to 2 lim f (x)$

(spoiler)

Here, $x$ approaches $2$ , which is also a breakpoint.

Left-hand limit:

For $- 1 < x \leq 2$ , $f (x) = 2 x$ , so

$x \to 2^{-} lim 2 x = 4$

Right-hand limit:

For $x > 2$ , $f (x) = 4$ , so

$x \to 2^{+} lim 4 = 4$

Both one-sided limits match, so

$x \to 2 lim f (x) = 4$

c) $x \to 3 lim f (x)$

(spoiler)

Since $x$ approaches $3$ which is not a breakpoint, one-sided limits are not necessary.

As $x$ approaches $3$ , $f (x)$ is defined by the third branch, $f (x) = 4$ for $x \geq 2$ . Therefore,

$x \to 3 lim 4 = 4$