One-Sided Limits

Learn to find left-hand and right-hand limits from graphs and piecewise formulas, and decide when a two-sided limit exists.

Advanced30 min18 problems
Practice Problems

Definition

A one-sided limit looks at a function from only one side of a point.
  • The left-hand limit lim⁡x→a−f(x)=L\lim_{x \to a^-} f(x) = L means: as xx gets close to aa through values less than aa (like 1.9,1.99,1.9991.9, 1.99, 1.999 for a=2a = 2), f(x)f(x) gets close to LL.
  • The right-hand limit lim⁡x→a+f(x)=M\lim_{x \to a^+} f(x) = M means: as xx gets close to aa through values greater than aa (like 2.1,2.01,2.0012.1, 2.01, 2.001), f(x)f(x) gets close to MM.
You know this notation from the lesson Limit Notation: the small sign shows the side and is not part of the number, so 3−3^- is not −3-3. In this lesson you compute these limits and use them to decide whether lim⁡x→af(x)\lim_{x \to a} f(x) exists.
The key rule connects one-sided and two-sided limits:
lim⁡x→af(x)=L  ⟺  lim⁡x→a−f(x)=L and lim⁡x→a+f(x)=L\lim_{x \to a} f(x) = L \iff \lim_{x \to a^-} f(x) = L \text{ and } \lim_{x \to a^+} f(x) = L
If the two one-sided limits are different, the two-sided limit does not exist (DNE). The value f(a)f(a) itself never matters for any of these limits: the function may be undefined at aa, or have a completely different value there.

Try it now

What does x→3−x \to 3^- mean?

Worked Examples

The graph of ff shows the line y=x+1y = x + 1 for x<1x < 1 (open circle at (1,2)(1, 2)) and the line y=x+3y = x + 3 for x≥1x \ge 1 (closed dot at (1,4)(1, 4)). Find lim⁡x→1−f(x)\lim_{x \to 1^-} f(x), lim⁡x→1+f(x)\lim_{x \to 1^+} f(x) and lim⁡x→1f(x)\lim_{x \to 1} f(x).
1

Follow the graph from the left

Trace the left part toward x=1x = 1: the heights approach the open circle at y=2y = 2 → lim⁡x→1−f(x)=2\lim_{x \to 1^-} f(x) = 2
2

Follow the graph from the right

Trace the right part toward x=1x = 1: the heights approach y=4y = 4 → lim⁡x→1+f(x)=4\lim_{x \to 1^+} f(x) = 4
3

Compare the two sides

2≠42 \neq 4, so the graph jumps at x=1x = 1 → The one-sided limits differ
4

Decide about the two-sided limit

Different one-sided limits mean no single value is approached. The closed dot f(1)=4f(1) = 4 does not change this → lim⁡x→1f(x)\lim_{x \to 1} f(x) does not exist

Common Mistakes

Using the closed dot f(a)f(a) as the limit

Why it's wrong: A limit describes the heights the graph approaches near aa, not the height at aa. In Example 1, f(1)=4f(1) = 4, but from the left the graph approaches 22.

Correct: Trace each side of the graph toward aa and ignore the dot at x=ax = a. Use f(a)f(a) only when the question asks for the value.

Reading x→3−x \to 3^- as "xx approaches −3-3"

Why it's wrong: The small minus sign is written after the number, as a superscript. It describes the direction, not the sign of the number.

Correct: x→3−x \to 3^- means x=2.9,2.99,2.999,…x = 2.9, 2.99, 2.999, \dots (values below 33). x→3+x \to 3^+ means x=3.1,3.01,3.001,…x = 3.1, 3.01, 3.001, \dots

Using the piece that contains x=ax = a for both sides

Why it's wrong: In f(x)=x+1f(x) = x + 1 for x<1x < 1 and f(x)=x+3f(x) = x + 3 for x≥1x \ge 1, the piece x+3x + 3 only describes the right side. The left side is always given by the piece for x<1x < 1.

Correct: For x→a−x \to a^- use the formula for values just below aa; for x→a+x \to a^+ use the formula for values just above aa. The ≤\le or ≥\ge sign only decides f(a)f(a).

Saying the limit exists because both one-sided limits exist

Why it's wrong: Both one-sided limits in Example 1 exist (22 and 44), but they are different numbers, so there is no single value LL.

Correct: The two-sided limit exists only if both one-sided limits exist and are equal.

Interactive Visual

Function graph

-3-2-112345-2-1123456780xy

Tap or hover over the graph to read off values.

f
  • At x = 1: from the left f(x) → 2, from the right f(x) → 4. The values differ, so the limit does not exist (jump). f(1) = 4.

Table of values

x0.90.9911.011.1
y1.91.9944.014.1

Function graph

-4-3-2-11234-3-2-11230xy

Tap or hover over the graph to read off values.

y = |x|/x
  • At x = 0: from the left y(x) → -1, from the right y(x) → 1. The values differ, so the limit does not exist (jump). y(0) is undefined.

Table of values

x-0.1-0.010.010.1
y-1-111

Function graph

-2-10123456-505xy

Tap or hover over the graph to read off values.

y = (x^2 - 4)/|x - 2|
  • At x = 2: from the left y(x) → -4, from the right y(x) → 4. The values differ, so the limit does not exist (jump). y(2) is undefined.

Table of values

x1.91.992.012.1
y-3.9-3.994.014.1

Interactive Sandbox

Interactive Grapher

-10-50510-10-50510xy

Tap or hover over the graph to read off values.

Try these examples:

y = 2x + 1

m=2, b=1

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Practice Problems

18 problems
Problem 1 of 18
Easy
What does x→3−x \to 3^- mean?

Why It Matters

One-sided limits are the tool for functions that change their rule at a point:
  • Piecewise functions: tax brackets, phone plans and shipping prices switch formulas at certain values, and you must look at each side separately
  • Jumps: a parking fee that jumps from 3 to 5 dollars after one hour has different left-hand and right-hand limits
  • Absolute values: expressions like ∣x∣x\frac{|x|}{x} behave differently for negative and positive xx
  • Continuity: in the lessons Continuity and Types of Discontinuity you will check one-sided limits to classify every break in a graph
  • Calculus: derivatives of functions like ∣x∣|x| are studied with left and right limits
Whenever a graph has a break, the first question is always: what happens on the left, and what happens on the right?

Real World Applications

Parking Fees

Many parking garages charge by started hour, so the price jumps at each full hour. One-sided limits describe the price just before and just after the jump.

Example:

A garage charges P(t)=3P(t) = 3 dollars for 0<t≤10 < t \le 1 hour and P(t)=5P(t) = 5 dollars for 1<t≤21 < t \le 2 hours. Just before one hour you pay 3 dollars, just after it 5 dollars.

1Try It Yourself

The garage uses P(t)=3P(t) = 3 for 0<t≤10 < t \le 1 and P(t)=5P(t) = 5 for 1<t≤21 < t \le 2 (tt in hours, price in dollars).

By how many dollars does the price jump at t=1t = 1?

Step 1: Write the mathematical expression

Right-hand limit minus left-hand limit at t=1t = 1:

Building a Smooth Ramp

Engineers who join a ramp to a platform need the heights from both sides to meet. Otherwise there is a step that people can trip over.

Example:

A ramp has height h(x)=0.2xh(x) = 0.2x meters for 0≤x<50 \le x < 5, and the platform starts at x=5x = 5 at a constant height cc. The ramp and the platform meet without a step exactly when lim⁡x→5−h(x)=c\lim_{x \to 5^-} h(x) = c.

2Try It Yourself

The ramp has height h(x)=0.2xh(x) = 0.2x for 0≤x<50 \le x < 5 and the platform has height cc for x≥5x \ge 5 (all lengths in meters).

Which platform height cc gives no step at x=5x = 5?

Step 1: Write the mathematical expression

The platform height must equal the left-hand limit at x=5x = 5:

Mobile Data Plans

A data plan often has a fixed price up to a limit and an extra charge per gigabyte after it. One-sided limits show whether the price jumps at the limit.

Example:

A plan costs C(g)=10C(g) = 10 dollars for 0≤g≤50 \le g \le 5 GB and C(g)=10+2(g−5)C(g) = 10 + 2(g - 5) dollars for g>5g > 5. Just above 5 GB the price starts to grow, but it starts from 10.

3Try It Yourself

The plan costs C(g)=10C(g) = 10 for 0≤g≤50 \le g \le 5 and C(g)=10+2(g−5)C(g) = 10 + 2(g - 5) for g>5g > 5 (gg in GB, price in dollars).

What is lim⁡g→5C(g)\lim_{g \to 5} C(g)?

Step 1: Write the mathematical expression

Find both one-sided limits at g=5g = 5 and compare them:

Key Takeaways

  • 1lim⁡x→a−f(x)\lim_{x \to a^-} f(x) is the left-hand limit: xx approaches aa through values less than aa
  • 2lim⁡x→a+f(x)\lim_{x \to a^+} f(x) is the right-hand limit: xx approaches aa through values greater than aa
  • 3lim⁡x→af(x)=L\lim_{x \to a} f(x) = L exactly when both one-sided limits exist and both equal LL
  • 4If the one-sided limits are different (a jump), the two-sided limit does not exist
  • 5For a piecewise function, use the piece for values just below aa on the left and the piece for values just above aa on the right
  • 6The value f(a)f(a) (open circle or closed dot) never changes a limit

Frequently Asked Questions

Yes, and this is the typical case at a jump. For ∣x∣x\frac{|x|}{x} at 00 both one-sided limits exist (−1-1 and 11), but the two-sided limit does not, because they are different.
Yes, and this is the typical case at a jump. For ∣x∣x\frac{|x|}{x} at 00 both one-sided limits exist (−1-1 and 11), but the two-sided limit does not, because they are different.
Neither. Circles and dots only show the value f(a)f(a). The limits come from the curve on each side as it approaches x=ax = a.
Then only one one-sided limit makes sense. For x\sqrt{x} at 00 there are no values to the left, so we use lim⁡x→0+x=0\lim_{x \to 0^+} \sqrt{x} = 0. At the end of a domain, the one-sided limit is the natural question.
Yes. For example 1x\frac{1}{x} grows without bound as x→0+x \to 0^+. These cases are covered in the lesson Infinite Limits; in this lesson all one-sided limits are numbers.

Glossary

One-sided limit
A limit that uses values of xx from only one side of aa
Left-hand limit
The value f(x)f(x) approaches as x→a−x \to a^-, through values less than aa
Right-hand limit
The value f(x)f(x) approaches as x→a+x \to a^+, through values greater than aa
Two-sided limit
The ordinary limit lim⁡x→af(x)\lim_{x \to a} f(x); it exists only if the left-hand and right-hand limits are equal
Piecewise function
A function given by different formulas on different intervals of xx
Jump discontinuity
A break where both one-sided limits exist but are different, so the graph jumps

Formula Card

Left-hand limit

lim⁡x→a−f(x)\lim_{x \to a^-} f(x)

$x$ approaches $a$ from the left (values less than $a$)

Right-hand limit

lim⁡x→a+f(x)\lim_{x \to a^+} f(x)

$x$ approaches $a$ from the right (values greater than $a$)

When the limit exists

lim⁡x→af(x)=L  ⟺  lim⁡x→a−f(x)=lim⁡x→a+f(x)=L\lim_{x \to a} f(x) = L \iff \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L

Both one-sided limits must exist and be equal

Absolute value by sides

∣x−a∣=x−a|x - a| = x - a for x>ax > a, ∣x−a∣=−(x−a)|x - a| = -(x - a) for x<ax < a

Use this to simplify expressions with absolute values on each side

Today's challengeOne problem, three tries

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