One-Sided Limits

6 min read

Reference: Stewart §2.2

Why Direction Matters

Not all functions behave the same way when approached from the left versus the right. Consider a light switch: approaching the “on” position from “off” gives a different result than approaching from “already on.” Similarly, some functions have different limiting behaviors depending on which direction you approach from.

One-sided limits let us describe this directional behavior precisely. The key theorem is: a two-sided limit exists if and only if both one-sided limits exist and are equal. This gives us a powerful tool for detecting when limits fail to exist.

Prerequisite Map

Quick Reference

Property	Value
Concept	Limits
Course	MATH161
Section	Stewart 1.5
Difficulty	Beginner
Time	~15 minutes

Key Concepts

Definition: Left-Hand Limit

$$\lim_{x \to a^-} f(x) = L$$

This means: as $x$ approaches $a$ from the left (i.e., $x < a$), $f(x)$ approaches $L$.

The superscript $^-$ indicates “from values less than $a$.”

Definition: Right-Hand Limit

$$\lim_{x \to a^+} f(x) = L$$

This means: as $x$ approaches $a$ from the right (i.e., $x > a$), $f(x)$ approaches $L$.

The superscript $^+$ indicates “from values greater than $a$.”

The Fundamental Theorem for Limits

$$\boxed{\lim_{x \to a} f(x) = L \iff \lim_{x \to a^-} f(x) = L \text{ and } \lim_{x \to a^+} f(x) = L}$$

In words: A two-sided limit exists and equals $L$ if and only if BOTH one-sided limits exist and BOTH equal $L$.

Contrapositive: If the one-sided limits differ (or one doesn’t exist), then the two-sided limit does not exist.

Visualizing One-Sided Limits

Case 1: Both limits agree

    y
    |      /
  3 +    ●
    |   /
  2 +  /
    | /
  1 +/
    +--------→ x
        2

Both sides approach 3 at x = 2
Two-sided limit = 3

Case 2: One-sided limits differ (jump discontinuity)

    y
    |         ●──────
  3 +
    |   ──────○
  1 +
    +--------→ x
        2

Left limit = 1, Right limit = 3
Two-sided limit DNE

The Heaviside Step Function

A classic example is the Heaviside function:

$$H(t) = \begin{cases} 0 & \text{if } t < 0 \\ 1 & \text{if } t \geq 0 \end{cases}$$

    y
    |        ●───────
  1 +
    |
  0 +────────○
    +--------+------→ t
             0

At $t = 0$:

$\lim_{t \to 0^-} H(t) = 0$ (approaching from negative values)
$\lim_{t \to 0^+} H(t) = 1$ (approaching from positive values)
$\lim_{t \to 0} H(t)$ does not exist (the one-sided limits differ)

Common Situations for One-Sided Limits

Situation	What Happens	Example
Jump discontinuity	Left ≠ Right limits	Step functions
Vertical asymptote	One or both sides → ±∞	$\frac{1}{x}$ at $x=0$
Endpoint of domain	Only one direction available	$\sqrt{x}$ at $x=0$
Piecewise functions	Different formulas each side	Custom-defined functions

Practice Problems

Level 1 Notation Interpretation

What is the difference between $\lim_{x \to 5^-} f(x)$ and $\lim_{x \to 5^+} f(x)$?

Thought Process

Focus on the superscripts: $^-$ means approaching from values less than $5$, while $^+$ means approaching from values greater than $5$.

Show Answer

$\lim_{x \to 5^-} f(x)$ is the left-hand limit: what $f(x)$ approaches as $x$ comes from values less than $5$ (like $4.9, 4.99, 4.999, \ldots$)
$\lim_{x \to 5^+} f(x)$ is the right-hand limit: what $f(x)$ approaches as $x$ comes from values greater than $5$ (like $5.1, 5.01, 5.001, \ldots$)

The two-sided limit $\lim_{x \to 5} f(x)$ exists only if these one-sided limits are equal.

Level 2 Evaluating One-Sided Limits of a Piecewise Function

Let $g(x) = \begin{cases} 2x + 1 & \text{if } x < 3 \\ x^2 - 4 & \text{if } x \geq 3 \end{cases}$

Find:

$\lim_{x \to 3^-} g(x)$
$\lim_{x \to 3^+} g(x)$
$\lim_{x \to 3} g(x)$ (or state that it DNE)

Thought Process

For the left-hand limit, use the piece that applies when $x < 3$. For the right-hand limit, use the piece that applies when $x > 3$ (or $x \geq 3$). Then compare.

Show Answer

(a) For $x < 3$, we use $g(x) = 2x + 1$: $$\lim_{x \to 3^-} g(x) = \lim_{x \to 3^-} (2x + 1) = 2(3) + 1 = 7$$

(b) For $x > 3$ (or $x \geq 3$), we use $g(x) = x^2 - 4$: $$\lim_{x \to 3^+} g(x) = \lim_{x \to 3^+} (x^2 - 4) = 9 - 4 = 5$$

(c) Since $\lim_{x \to 3^-} g(x) = 7 \neq 5 = \lim_{x \to 3^+} g(x)$, the two-sided limit does not exist.

Level 3 Reading One-Sided Limits from a Graph

Use the graph below to find all the requested limits:

    y
    |            ●
  4 +           /
    |          /
  3 +    ○────/
    |   /
  2 +  /
    | /
  1 +●
    |
  0 +--+--+--+--+--→ x
       1  2  3  4

At $x = 1$: filled point at $y = 1$ At $x = 2$: open circle at $y = 3$, line continues up to right At $x = 4$: filled point at $y = 4$

$\lim_{x \to 2^-} f(x)$
$\lim_{x \to 2^+} f(x)$
$\lim_{x \to 2} f(x)$
$f(2)$

Thought Process

Trace along the curve. From the left of $x=2$, follow the line upward: what $y$-value does it approach? From the right of $x=2$, continue along the same line: what $y$-value does it approach? For $f(2)$ specifically, look at whether there’s a filled or open circle at $x=2$.

Show Answer

(a) $\lim_{x \to 2^-} f(x) = 3$ Following the line from the left, we approach the open circle at height $3$.

(b) $\lim_{x \to 2^+} f(x) = 3$ Following the line from the right, we also approach height $3$.

(c) $\lim_{x \to 2} f(x) = 3$ Both one-sided limits equal $3$, so the two-sided limit exists and equals $3$.

(d) $f(2)$ is undefined. The open circle at $(2, 3)$ means the function is not defined there.

Note: This is a “hole” discontinuity: the limit exists but the function value doesn’t.

Level 4 Finding Constants for Limit Existence

For what value of the constant $c$ does the limit $\lim_{x \to 2} f(x)$ exist, where:

$$f(x) = \begin{cases} x^2 + c & \text{if } x < 2 \\ 3x + 1 & \text{if } x \geq 2 \end{cases}$$

Thought Process

For the two-sided limit to exist, the left and right limits must be equal. Compute each in terms of $c$, then set them equal and solve for $c$.

Show Answer

Step 1: Find the left-hand limit $$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x^2 + c) = 4 + c$$

Step 2: Find the right-hand limit $$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (3x + 1) = 6 + 1 = 7$$

Step 3: Set them equal For the two-sided limit to exist: $$4 + c = 7$$ $$c = 3$$

Verification: With $c = 3$:

Left limit: $4 + 3 = 7$
Right limit: $7$
Both equal, so $\lim_{x \to 2} f(x) = 7$ ✓

Level 5 Proving a Limit Does Not Exist

Prove that $\lim_{x \to 0} \frac{\vert x\vert }{x}$ does not exist by computing the one-sided limits.

Thought Process

The key is that $\vert x\vert $ behaves differently for positive and negative $x$:

When $x > 0$: $\vert x\vert = x$
When $x < 0$: $\vert x\vert = -x$

Use this to simplify $\frac{\vert x\vert }{x}$ in each case, then compute the one-sided limits.

Show Answer

Right-hand limit ($x > 0$):

When $x > 0$, we have $\vert x\vert = x$, so: $$\frac{\vert x\vert }{x} = \frac{x}{x} = 1$$

Therefore: $$\lim_{x \to 0^+} \frac{\vert x\vert }{x} = 1$$

Left-hand limit ($x < 0$):

When $x < 0$, we have $\vert x\vert = -x$, so: $$\frac{\vert x\vert }{x} = \frac{-x}{x} = -1$$

Therefore: $$\lim_{x \to 0^-} \frac{\vert x\vert }{x} = -1$$

Conclusion:

Since $\lim_{x \to 0^-} \frac{\vert x\vert }{x} = -1 \neq 1 = \lim_{x \to 0^+} \frac{\vert x\vert }{x}$, the one-sided limits are not equal.

By the fundamental theorem, $\lim_{x \to 0} \frac{\vert x\vert }{x}$ does not exist. $\square$

Graphical interpretation: This function equals $-1$ for all $x < 0$ and equals $1$ for all $x > 0$. It has a jump discontinuity at $x = 0$.

Common Misconceptions

Common misconception

the limit equals the function value. Students often write $\lim_{x \to a} f(x) = f(a)$ as a general rule. This is true only when $f$ is continuous at $a$. The Heaviside function shows why it fails: $H(0) = 1$ (the function is defined there), yet $\lim_{t \to 0} H(t)$ does not exist because the left and right one-sided limits differ. The limit depends on nearby behavior, not on what happens at the point itself.

Common misconception

a limit existing and a function being continuous are the same thing. Students sometimes treat these as equivalent. A limit can exist even when the function has a hole at that point (the limit exists, but $f(a)$ is undefined or has the wrong value). Conversely, a function can be defined at $a$ but have a limit that does not exist there (a jump discontinuity where $f(a)$ is assigned to one of the jump values). Continuity is the stronger requirement: the limit must exist, and it must equal $f(a)$.

Mastery Checklist

I can state the definitions of left-hand and right-hand limits
I can explain the relationship between one-sided and two-sided limits
I can evaluate one-sided limits of piecewise functions
I can read one-sided limits from graphs
I can find constant values that make a two-sided limit exist
I can prove a limit does not exist by showing one-sided limits differ

Mental Model

The Fork in the Road: Imagine walking toward a fork in the road. The left-hand limit is where you end up if you come from the left fork. The right-hand limit is where you end up if you come from the right fork. The two-sided limit only exists if both paths lead to the same destination.

Connections

Looking back:

Limit Intuition introduces the concept of approaching a value; one-sided limits refine this to specify direction

Looking ahead:

Infinite Limits extends these ideas to when one-sided limits are $\pm\infty$
Continuity requires that both one-sided limits exist, are equal, AND match the function value
Derivatives are defined using limits, and sometimes we need one-sided derivatives at endpoints

Real-world connections:

The Heaviside function models switches turning on/off in electrical engineering
One-sided limits describe shock waves in physics where quantities jump discontinuously

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Limit Intuition	Skills Index	Infinite Limits

Last updated: 2026-01-22