One-Sided Limits: Left & Right Limits

One-sided limits describe what a function approaches from just one side of a number. The left-hand limit $\lim_{x \to a^-} f(x)$ uses x-values less than a, and the right-hand limit $\lim_{x \to a^+} f(x)$ uses x-values greater than a. The two-sided limit exists only when both one-sided limits exist and are equal.

Before this lesson: Introduction to Limits, Piecewise Functions and Absolute Value Functions

Why one side at a time

Some graphs break apart at a point. A parking fee jumps from $7 to $10 the moment you pass 2 hours. Near the jump, the values just before and just after are different, so “the” value the function approaches depends on the direction you come from.

One-sided limits handle this. The small minus sign in $a^-$ means “from the left,” and the plus sign in $a^+$ means “from the right.” Neither sign has anything to do with a being negative or positive.

The rule that connects them: $\lim_{x \to a} f(x) = L$ exactly when $\lim_{x \to a^-} f(x) = L$ and $\lim_{x \to a^+} f(x) = L$ . If the two sides differ, or either side grows without bound, the two-sided limit does not exist.

Graph of f(x) = x² for x less than 1 and f(x) = x + 2 for x greater than or equal to 1. The left piece ends at an open circle at (1, 1) and the right piece starts at a filled dot at (1, 3). Approaching from the left the heights go to 1, and from the right they go to 3, so the two-sided limit at x = 1 does not exist. — A jump at x = 1: the left side heads to 1 and the right side heads to 3, so there is no two-sided limit there.

How to find one-sided limits

Situation	What to do
Piecewise function	For the left-hand limit, use the piece for x less than a; for the right-hand limit, use the piece for x greater than a. Substitute x = a into that piece.
Absolute value over the same expression	Rewrite without bars on each side: \|u\| = −u when u is negative and \|u\| = u when u is positive.
Division by something near 0	Check the sign of the numerator and the denominator on each side to decide between growing without bound positively or negatively.
A graph	Trace toward x = a from each side and read the height. Ignore the dot exactly at x = a.

One-sided limits examples

Example 1: A jump in a piecewise function

Let $f(x) = x^2$ for $x \lt 1$ and $f(x) = x + 2$ for $x \ge 1$ . Find both one-sided limits at x = 1 and decide whether $\lim_{x \to 1} f(x)$ exists.

\lim_{x \to 1^-} f(x) = 1^2 = 1

From the left, x is less than 1, so use the piece $x^2$ . Substitute 1.

\lim_{x \to 1^+} f(x) = 1 + 2 = 3

From the right, x is greater than 1, so use the piece x + 2. Substitute 1.

1 \ne 3

Compare the two sides. They are different.

Answer: the left-hand limit is 1 and the right-hand limit is 3, so $\lim_{x \to 1} f(x)$ does not exist. (Also, f(1) = 3, from the second piece.)

Example 2: Two pieces that meet

Let $g(x) = 2x + 1$ for $x \lt 3$ and $g(x) = x^2 - 2$ for $x \ge 3$ . Find $\lim_{x \to 3} g(x)$ .

\lim_{x \to 3^-} g(x) = 2(3) + 1

From the left, use 2x + 1.

\lim_{x \to 3^-} g(x) = 7

Simplify.

\lim_{x \to 3^+} g(x) = 3^2 - 2

From the right, use $x^2 - 2$ .

\lim_{x \to 3^+} g(x) = 7

Simplify: 9 − 2 = 7.

Answer: both sides equal 7, so $\lim_{x \to 3} g(x) = 7$ . The pieces meet, and the graph has no break at x = 3.

Example 3: Absolute value over x

Find the one-sided limits of $h(x) = \frac{|x|}{x}$ at x = 0.

\frac{|x|}{x} = \frac{-x}{x} = -1

For x less than 0, |x| = −x (see absolute value). Simplify.

\lim_{x \to 0^-} h(x) = -1

The function equals −1 for every negative x.

\frac{|x|}{x} = \frac{x}{x} = 1

For x greater than 0, |x| = x.

\lim_{x \to 0^+} h(x) = 1

The function equals 1 for every positive x.

Answer: the left-hand limit is −1 and the right-hand limit is 1, so $\lim_{x \to 0} \frac{|x|}{x}$ does not exist. The graph is two flat rays with a jump at 0.

Example 4: Growing without bound on each side

Find $\lim_{x \to 2^-} \frac{1}{x - 2}$ and $\lim_{x \to 2^+} \frac{1}{x - 2}$ .

x	1.9	1.99	1.999	2	2.001	2.01	2.1
$\frac{1}{x - 2}$	−10	−100	−1000	undefined	1000	100	10

\frac{1}{1.99 - 2} = \frac{1}{-0.01} = -100

Find one value from the left. The denominator is a small negative number.

\frac{1}{2.01 - 2} = \frac{1}{0.01} = 100

Find one value from the right. The denominator is a small positive number.

Answer: from the left the values decrease without bound, $\lim_{x \to 2^-} \frac{1}{x - 2} = -\infty$ . From the right they increase without bound, $\lim_{x \to 2^+} \frac{1}{x - 2} = \infty$ . Neither is a number, and the two-sided limit does not exist. The line x = 2 is a vertical asymptote.

Example 5: Word problem with a parking fee

A garage charges $4 for up to 1 hour, $7 for more than 1 hour up to 2 hours, and $10 for more than 2 hours up to 3 hours. Let C(t) be the cost for t hours. Find the one-sided limits at t = 2 and C(2).

\lim_{t \to 2^-} C(t) = 7

Just under 2 hours, such as 1.9 or 1.99 hours, the fee is $7.

\lim_{t \to 2^+} C(t) = 10

Just over 2 hours, such as 2.01 hours, the fee is $10.

C(2) = 7

Exactly 2 hours is in the “more than 1 up to 2” group.

Answer: the left-hand limit is $7, the right-hand limit is $10, and C(2) = $7. The limit as t approaches 2 does not exist, which matches the sudden jump in price.

Common mistakes

Mistake	Example of the error	Fix
Reading the minus sign as a negative number	Thinking $x \to 1^-$ means x approaches −1	It means x approaches 1 from the left, through values like 0.99.
Using the wrong piece	Using x + 2 for the left-hand limit in Example 1	Left means x less than a, so use the piece for x less than 1.
Using the function value as the limit	Saying $\lim_{x \to 1^-} f(x) = 3$ because f(1) = 3	One-sided limits look near a, not at a. The left side heads to 1.
Calling a limit equal when only one side is checked	Saying $\lim_{x \to 1} f(x) = 1$ from the left side alone	Check both sides; here they differ, so the limit does not exist.
Getting the sign of an infinite limit wrong	$\lim_{x \to 2^-} \frac{1}{x - 2} = \infty$	On the left, x − 2 is negative, so the values go to $-\infty$ .
Ignoring where a function is defined	Looking for $\lim_{x \to 0^-} \sqrt{x}$	$\sqrt{x}$ is not defined for x less than 0, so only the right-hand limit makes sense.

Related methods

Endpoints of a domain

At the edge of a domain, only one side is available. For $\sqrt{x}$ at 0, the right-hand limit is $\lim_{x \to 0^+} \sqrt{x} = 0$ , and there is no left-hand limit to check.

Looking ahead to continuity

A function is continuous at a when its two one-sided limits agree and equal f(a). In Example 2, both sides are 7 and g(3) = 7, so g is continuous at 3. In Example 1 the jump breaks continuity. Calculus 1 studies this in detail.

Practice problems

Let f(x) = 2x for $x \lt 3$ and f(x) = x + 4 for $x \ge 3$ . Find both one-sided limits at 3. Does the limit exist?
Let f(x) = $x^2 + 1$ for $x \lt 0$ and f(x) = 1 − x for $x \ge 0$ . Find $\lim_{x \to 0} f(x)$ .
Find the one-sided limits of $\frac{|x - 4|}{x - 4}$ at x = 4.
Find $\lim_{x \to 5^-} \frac{1}{x - 5}$ and $\lim_{x \to 5^+} \frac{1}{x - 5}$ .
Find $\lim_{x \to 0^+} \sqrt{x}$ . Why is there no left-hand limit?
Find the value of k so that f(x) = kx + 1 for $x \lt 2$ and f(x) = $x^2$ for $x \ge 2$ has a limit at x = 2.
The floor function $\lfloor x \rfloor$ gives the greatest whole number less than or equal to x. Find $\lim_{x \to 3^-} \lfloor x \rfloor$ and $\lim_{x \to 3^+} \lfloor x \rfloor$ .
Let h(x) = 3 − x for $x \le 1$ and h(x) = 2x for $x \gt 1$ . Find both one-sided limits at 1, the limit and h(1).
Find $\lim_{x \to -1^-} \frac{x + 3}{x + 1}$ and $\lim_{x \to -1^+} \frac{x + 3}{x + 1}$ .
Shipping costs $5 for packages up to 2 pounds and $8 for packages over 2 pounds up to 5 pounds. Find the one-sided limits of the cost at 2 pounds.

Show answers

\lim_{x \to 3^-} f(x) = 2(3) = 6

From the left, use 2x.

\lim_{x \to 3^+} f(x) = 3 + 4 = 7

From the right, use x + 4.

Answer: 6 and 7. They differ, so the limit does not exist.

\lim_{x \to 0^-} f(x) = 0^2 + 1 = 1

From the left, use $x^2 + 1$ .

\lim_{x \to 0^+} f(x) = 1 - 0 = 1

From the right, use 1 − x.

Answer: both sides are 1, so the limit is 1.

\frac{|x - 4|}{x - 4} = \frac{-(x - 4)}{x - 4} = -1

For x less than 4, x − 4 is negative, so $|x - 4| = -(x - 4)$ .

\frac{|x - 4|}{x - 4} = \frac{x - 4}{x - 4} = 1

For x greater than 4, x − 4 is positive.

Answer: the left-hand limit is −1 and the right-hand limit is 1, so the limit does not exist.

\frac{1}{4.99 - 5} = -100

From the left the denominator is a small negative number.

\frac{1}{5.01 - 5} = 100

From the right it is a small positive number.

Answer: $-\infty$ from the left and $\infty$ from the right.

\lim_{x \to 0^+} \sqrt{x} = \sqrt{0} = 0

Square roots of small positive numbers are close to 0.

Answer: 0. There is no left-hand limit because $\sqrt{x}$ is not a real number for x less than 0 (see square roots).

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

From the left, substitute 2 into kx + 1.

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

From the right, substitute 2 into $x^2$ .

2k + 1 = 4

For a limit to exist, the sides must be equal.

2k = 3

Subtract 1 from both sides.

k = 1.5

Divide both sides by 2.

Answer: $k = \frac{3}{2}$

\lim_{x \to 3^-} \lfloor x \rfloor = 2

Just below 3, such as 2.99, the floor is 2.

\lim_{x \to 3^+} \lfloor x \rfloor = 3

Just above 3, such as 3.01, the floor is 3.

Answer: 2 from the left and 3 from the right, so the limit at 3 does not exist.

\lim_{x \to 1^-} h(x) = 3 - 1 = 2

From the left, use 3 − x.

\lim_{x \to 1^+} h(x) = 2(1) = 2

From the right, use 2x.

h(1) = 3 - 1 = 2

x = 1 belongs to the first piece.

Answer: both one-sided limits are 2, the limit is 2, and h(1) = 2.

\frac{-1.01 + 3}{-1.01 + 1} = \frac{1.99}{-0.01} = -199

From the left, the top is near 2 and the bottom is a small negative number.

\frac{-0.99 + 3}{-0.99 + 1} = \frac{2.01}{0.01} = 201

From the right, the bottom is a small positive number.

Answer: $-\infty$ from the left and $\infty$ from the right.

10.

\lim_{w \to 2^-} C(w) = 5

Just under 2 pounds, the cost is $5.

\lim_{w \to 2^+} C(w) = 8

Just over 2 pounds, the cost is $8.

Answer: $5 from the left and $8 from the right, so the limit at 2 pounds does not exist.

One-sided limits FAQ

What is a one-sided limit?

It is the value a function approaches as x gets close to a number from only one side, either the left (values less than a) or the right (values greater than a).

What does x → a⁻ mean?

It means x approaches a from the left, through values smaller than a. The superscript minus does not make a negative.

When does a two-sided limit exist?

When the left-hand and right-hand limits both exist and are equal. Their shared value is the limit.

Can a one-sided limit be infinite?

Yes. Near a vertical asymptote, a one-sided limit can be $\infty$ or $-\infty$ , which describes how the values grow. It is still not a real-number limit.

Summary

$\lim_{x \to a^-} f(x)$ uses x-values less than a; $\lim_{x \to a^+} f(x)$ uses x-values greater than a.
The two-sided limit exists only when both one-sided limits exist and are equal.
For piecewise functions, use the piece that applies on each side.
Jumps give different one-sided limits; vertical asymptotes give $\infty$ or $-\infty$ .
At the edge of a domain, only one side can be checked.

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