One-Sided Limits: Left-Hand and Right-Hand Limits

What Are One-Sided Limits?

One-sided limits describe the value a function approaches as $x$ moves toward a point $a$ from just one side. The left-hand limit (LHL) uses only values $x < a$, and the right-hand limit (RHL) uses only values $x > a$. Neither one looks at $f(a)$ itself.

A Quora reader asks the definition question directly: "What is the definition of a one-sided limit?" In symbols, the two are written

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

• $x \to a^{-}$ means $x$ approaches $a$ through values smaller than $a$ (from the left on the number line).

• $x \to a^{+}$ means $x$ approaches $a$ through values larger than $a$ (from the right).

• $L$ is the left-hand limit and $R$ is the right-hand limit. Both must be real numbers for the one-sided limit to exist.

The precise version uses the same $\varepsilon$-$\delta$ idea as the ordinary limit of a function, with one change. For the right-hand limit, $\lim_{x \to a^{+}} f(x) = R$ means: for every $\varepsilon > 0$ there is a $\delta > 0$ such that

$$0 < x - a < \delta \quad \Longrightarrow \quad \lvert f(x) - R \rvert < \varepsilon$$

For the left-hand limit, the condition becomes $0 < a - x < \delta$. The only difference from the two-sided definition is that $\lvert x - a \rvert$ is replaced by a one-directional gap, so $x$ is never allowed onto the other side.

What Is The Difference Between Left-Hand And Right-Hand Limits And A Two-Sided Limit?

A two-sided limit asks one question about both sides at once, while left-hand and right-hand limits split that question in two. The Quora version of this, "What is the difference between one sided and two sided limits?", has a short answer: a two-sided limit is a promise that both one-sided limits exist and agree.

One notation trap deserves its own line. The minus in $a^{-}$ is a direction, not a sign. So $x \to 2^{-}$ means "toward $2$ from below", with values like $1.9$, $1.99$, $1.999$, and it has nothing to do with $-2$. The same holds for $x \to 0^{-}$, which uses small negative numbers simply because those are the numbers to the left of $0$.

When Does A Limit Exist? The Left-Hand And Right-Hand Limit Test

The two-sided limit exists exactly when both one-sided limits exist and are equal. This is the existence criterion every course uses (OpenStax states it as Theorem 2.2 in Calculus Volume 1, Section 2.2):

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

The test runs in both directions. If the limit exists, both sides share its value. If both sides share a value, the limit exists and equals it.

A Quora question asks why the second direction forces "does not exist": "Why is it that when both sides of the limit is not equal making the limit 'do not exist'?" The reason sits in the definition. A limit must be a single number $L$ that $f(x)$ stays close to for every $x$ near $a$. If values just left of $a$ crowd around $1$ and values just right crowd around $3$, no single number can be close to both crowds at once, so no $L$ works.

There are three outcomes worth naming:

1. Both sides exist and agree. The limit exists. Example: $f(x) = x^2$ at $a = 1$, where both sides give $1$.

2. Both sides exist but disagree. The limit does not exist, and the graph has a jump discontinuity there.

3. At least one side fails. The limit does not exist. A side can fail by oscillating, by growing without bound, or because the function is not defined on that side at all.

The value $f(a)$ never enters this test. A function can have equal one-sided limits at $a$ and still have $f(a)$ undefined, or defined as something else. That gap between "the limit exists" and "the function behaves well at $a$" is exactly what continuity of a function checks.

How Do You Find One-Sided Limits Algebraically?

To find one-sided limits algebraically, keep only the formula or sign pattern that holds on the chosen side, then evaluate its limit as usual. That answers the Quora question "How to find one-sided limits algebraically" in one line, but the details depend on what kind of expression you have.

• Piece selection. If $f$ is defined by different formulas on each side, use only the formula for $x < a$ (left) or $x > a$ (right), then substitute or simplify.

• Sign resolution. If $\lvert \cdot \rvert$, $\sqrt{\cdot}$ or an even power hides a sign, replace it with the version that is true on that side. For $x < 0$, $\lvert x \rvert = -x$, and for $x > 0$, $\lvert x \rvert = x$.

• Integer part. For the floor function, decide which integer $\lfloor x \rfloor$ equals on a small interval beside $a$. That value is constant there, so it is the one-sided limit.

• Direction of a small denominator. When a denominator tends to $0$, its sign on each side decides whether the expression heads to $+\infty$ or $-\infty$. For example, $x - 2 > 0$ when $x > 2$, so $\lim_{x \to 2^{+}} \frac{1}{x - 2} = +\infty$.

A table of values is a useful check, not a proof. For $f(x) = \frac{\lvert x - 2 \rvert}{x - 2}$:

The left column settles on $-1$ and the right column on $1$. The algebra agrees: for $x < 2$ the numerator is $-(x - 2)$, so the quotient is $-1$, and for $x > 2$ it is $+1$.

How Do One-Sided Limits Work For Piecewise, Absolute Value And Greatest Integer Functions?

These three families are where one-sided limits earn their keep, because each one is built from different rules on the two sides of a point. The method is the same in every case: rewrite the function so each side has one plain formula, then take the limit of that formula.

Piecewise Functions

A function defined in pieces switches formula at a break point. The left-hand limit at the break uses the piece that covers values just below it, and the right-hand limit uses the piece just above it, regardless of which piece "owns" the point itself. If you need a refresher on reading the notation, see piecewise functions.

Absolute Value Functions

The absolute value is piecewise in disguise: $\lvert x - a \rvert = -(x - a)$ for $x < a$ and $x - a$ for $x \geq a$. The classic case is the sign quotient

$$\lim_{x \to 0^{-}} \frac{\lvert x \rvert}{x} = -1 \qquad \lim_{x \to 0^{+}} \frac{\lvert x \rvert}{x} = 1$$

so $\lim_{x \to 0} \frac{\lvert x \rvert}{x}$ does not exist. Graphically, the V-shape of absolute value functions has slope $-1$ on the left of its corner and $+1$ on the right, and this quotient records those two slopes.

Greatest Integer Function

The greatest integer function $\lfloor x \rfloor$ gives the largest integer less than or equal to $x$. It is constant between consecutive integers and jumps up by $1$ at each integer. For any integer $n$:

$$\lim_{x \to n^{-}} \lfloor x \rfloor = n - 1 \qquad \lim_{x \to n^{+}} \lfloor x \rfloor = n$$

So the two-sided limit fails at every integer. At a non-integer such as $2.5$, both sides sit on the same step and the limit is simply $\lfloor 2.5 \rfloor = 2$.

Examples Of One-Sided Limits

Example 1

Let $f(x) = 2x + 1$ for $x < 3$ and $f(x) = 10 - x$ for $x \geq 3$. Find both one-sided limits at $x = 3$ and decide whether $\lim_{x \to 3} f(x)$ exists.

Left piece:

$$\lim_{x \to 3^{-}} (2x + 1) = 2(3) + 1 = 7$$

Right piece:

$$\lim_{x \to 3^{+}} (10 - x) = 10 - 3 = 7$$

Both one-sided limits equal 7.

Final answer: $\lim_{x \to 3} f(x) = 7$.

Example 2

Find $\lim_{x \to 3} \lfloor x \rfloor$.

The tempting move is direct substitution, because $\lfloor 3 \rfloor$ is easy to compute:

$$\lfloor 3 \rfloor = 3 \quad \text{so the limit is } 3\text{?}$$

Test it with a value just to the left. At $x = 2.99$, $\lfloor 2.99 \rfloor = 2$, not 3, and every $x$ in $(2, 3)$ gives the same 2. The substitution only described the right side and the point itself.

Take each side separately instead. On $(2, 3)$ the function equals 2, and on $(3, 4)$ it equals 3:

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

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

Final answer: the limit does not exist, because the left-hand limit 2 differs from the right-hand limit 3.

Example 3

Find $\lim_{x \to 0^{+}} \frac{x^2 + 3\lvert x \rvert}{x}$ and $\lim_{x \to 0^{-}} \frac{x^2 + 3\lvert x \rvert}{x}$.

For $x > 0$, $\lvert x \rvert = x$:

$$\frac{x^2 + 3x}{x} = x + 3$$

$$\lim_{x \to 0^{+}} (x + 3) = 3$$

For $x < 0$, $\lvert x \rvert = -x$:

$$\frac{x^2 - 3x}{x} = x - 3$$

$$\lim_{x \to 0^{-}} (x - 3) = -3$$

Final answer: the right-hand limit is 3 and the left-hand limit is $-3$, so the two-sided limit does not exist.

Example 4

Find $\lim_{x \to 1^{+}} \left(\sqrt{x - 1} + x^2\right)$.

The square root needs $x - 1 \geq 0$, so the function only lives on $x \geq 1$. The right-hand limit is the natural question:

$$\lim_{x \to 1^{+}} \sqrt{x - 1} = 0$$

$$\lim_{x \to 1^{+}} x^2 = 1$$

$$\lim_{x \to 1^{+}} \left(\sqrt{x - 1} + x^2\right) = 0 + 1 = 1$$

Final answer: 1. The left-hand limit is not defined, because the function has no values to the left of 1. Textbooks differ on how to report the two-sided limit at such an endpoint: many say it does not exist, while others treat the right-hand limit as the limit when $a$ is an endpoint of the domain and range of a function. Follow your course's convention.

Example 5

Let $f(x) = \frac{x^2 - 9}{x - 3}$ for $x < 3$ and $f(x) = kx$ for $x > 3$. Find $k$ so that $\lim_{x \to 3} f(x)$ exists.

The left piece gives $\frac{0}{0}$ at $x = 3$, so factor first:

$$\frac{x^2 - 9}{x - 3} = \frac{(x - 3)(x + 3)}{x - 3} = x + 3$$

$$\lim_{x \to 3^{-}} (x + 3) = 6$$

Right piece:

$$\lim_{x \to 3^{+}} kx = 3k$$

Set the one-sided limits equal:

$$3k = 6$$

$$k = 2$$

Final answer: $k = 2$, and then $\lim_{x \to 3} f(x) = 6$.

Example 6

Find $\lim_{x \to 0^{+}} \frac{1}{1 + e^{1/x}}$ and $\lim_{x \to 0^{-}} \frac{1}{1 + e^{1/x}}$.

From the right, $x$ is small and positive, so $\frac{1}{x} \to +\infty$ and $e^{1/x} \to +\infty$:

$$\lim_{x \to 0^{+}} \frac{1}{1 + e^{1/x}} = 0$$

From the left, $x$ is small and negative, so $\frac{1}{x} \to -\infty$ and $e^{1/x} \to 0$:

$$\lim_{x \to 0^{-}} \frac{1}{1 + e^{1/x}} = \frac{1}{1 + 0} = 1$$

A quick numerical check agrees: at $x = 0.1$ the value is about $0.0000454$, and at $x = -0.1$ it is about $0.999955$.

Final answer: the right-hand limit is 0 and the left-hand limit is 1, so the two-sided limit does not exist. This function has no visible break in its formula, which is why it catches people who only look for pieces.

Why Do One-Sided Limits Matter? "A Jump Has Two Honest Values"

One-sided limits matter because many real quantities switch abruptly, and a single "nearby value" cannot describe a switch. At a jump, the left and right limits are the two facts that describe it. Every major use of the idea comes back to that.

• Signals that switch. A circuit switch, a digital bit or a sharp edge in an image changes value at an instant. Engineers describe the state "just before" and "just after" the switch with exactly these two limits, written $f(t^{-})$ and $f(t^{+})$.

• Fourier series at a jump. In the late 1820s, Peter Gustav Lejeune Dirichlet proved conditions under which a Fourier series represents a function. At a jump, the series converges to the average $\frac{f(a^{-}) + f(a^{+})}{2}$ of the two one-sided limits, not to $f(a)$. For a square wave jumping from 0 to 1, that midpoint is $\frac{1}{2}$.

• Slopes at corners. Whether a graph has a tangent line at a corner is decided by comparing the left-hand and right-hand limits of the difference quotient, the route into differentiability of a function. At the corner of $\lvert x \rvert$, those one-sided slopes are $-1$ and $1$.

• Endpoints of an interval. Continuity on a closed interval $[a, b]$ only asks for the right-hand limit at $a$ and the left-hand limit at $b$, because the function has no values beyond its ends.

The Dirichlet result is worth a second look. The square wave is a perfectly ordinary signal, yet the formula that builds it from sine waves lands on a value the signal never actually takes. Only the one-sided limits explain why.

Where Do Students Trip Up On One-Sided Limits?

Most lost marks on one-sided limits come from treating the point $a$ as if it decided everything, or from carrying the right-side formula over to the left. These three errors cover nearly all of it.

Mistake 1: Reading The Minus In The Notation As A Negative Number

Where it slips in:

Any limit written $x \to a^{-}$, especially $x \to 3^{-}$ or $x \to 0^{-}$.

Don't do this:

Do not substitute $-3$ for $3^{-}$, or treat $0^{-}$ as "negative zero" in a calculation.

The correct way:

Read $a^{-}$ as "toward $a$ from below". For $3^{-}$, test numbers such as $2.9$ and $2.99$.

The first instinct on seeing $x \to 3^{-}$ for the first time is to look for a $-3$ somewhere in the problem. Saying the notation aloud as "from the left" for the first few problems removes that habit quickly.

Mistake 2: Substituting The Point Instead Of Approaching It

Where it slips in:

Floor functions and piecewise functions where one formula owns $x = a$, as in Example 2.

Don't do this:

Do not compute $f(a)$ and report it as the left-hand or right-hand limit.

The correct way:

Pick a small interval on the chosen side, find the formula that holds on that whole interval, and take its limit.

The rusher makes this one most often: the point is right there, substitution is fast, and the left side never gets checked. Writing both one-sided limits on separate lines before computing either is the habit that stops it.

Mistake 3: Dropping The Sign When Removing An Absolute Value

Where it slips in:

Left-hand limits of expressions such as $\frac{\lvert x \rvert}{x}$ or $\frac{x^2 + 3\lvert x \rvert}{x}$.

Don't do this:

Do not replace $\lvert x \rvert$ with $x$ on the left of 0. That gives the right-side answer twice.

The correct way:

On $x < 0$, write $\lvert x \rvert = -x$, then simplify. Example 3 shows the two sides giving $3$ and $-3$.

The memorizer tends to carry "absolute value makes it positive" into this step and writes $\lvert x \rvert = x$ everywhere. I'll admit the notation does not help: $-x$ looks negative even though, for $x < 0$, it is a positive number. Checking with one concrete value, such as $x = -0.1$, settles it.

Mistake 2 has a real-world twin in imaging. When an MRI reconstruction treats a sharp tissue edge as if one value described both sides, the Gibbs phenomenon produces ringing lines next to the edge, an overshoot of about 9% of the jump that does not shrink as more terms are added. In spinal scans, those lines can mimic a real condition called syringomyelia.

Practice Problems On One-Sided Limits

1. Let $f(x) = x^2 - 1$ for $x < 4$ and $f(x) = 3x + 3$ for $x \geq 4$. Find both one-sided limits at $x = 4$. Does $\lim_{x \to 4} f(x)$ exist?

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

3. Find $\lim_{x \to 2.5^{-}} \lfloor 2x \rfloor$ and $\lim_{x \to 2.5^{+}} \lfloor 2x \rfloor$.

4. Find $\lim_{x \to 9^{-}} \left(\sqrt{9 - x} + 2\right)$. Can you take the right-hand limit?

5. Let $f(x) = cx + 4$ for $x < 1$ and $f(x) = 2x^2 - c$ for $x > 1$. Find $c$ so that $\lim_{x \to 1} f(x)$ exists, and give the limit.

Answer to Problem 1: Both sides equal 15 ($16 - 1$ and $12 + 3$), so the limit exists and equals 15.

Answer to Problem 2: The left-hand limit is $-1$ and the right-hand limit is $1$.

Answer to Problem 3: Just left of 2.5, $2x$ is just below 5, so $\lfloor 2x \rfloor = 4$. Just right of 2.5, it is 5. The one-sided limits are 4 and 5.

Answer to Problem 4: The left-hand limit is 2. The right-hand limit cannot be taken, because $\sqrt{9 - x}$ is not real for $x > 9$.

Answer to Problem 5: $c + 4 = 2 - c$ gives $c = -1$, and the limit is 3.

Bottom Line

• One-sided limits follow a function toward $a$ from one side: $x \to a^{-}$ uses values below $a$, and $x \to a^{+}$ uses values above it.

• The two-sided limit exists exactly when the left-hand and right-hand limits both exist and are equal.

• The value $f(a)$ never decides a limit, so substituting the point is not a method.

• Piecewise, absolute value and greatest integer functions need each side rewritten with the formula that holds there.

• At an integer $n$, the floor function has left-hand limit $n - 1$ and right-hand limit $n$.

Your Next Move

Work through practice problems 1 to 5; if Problem 3 trips you up, return to the greatest integer function subsection and redo Example 2.

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Read More

• Limit laws

• Standard limits

• Types of discontinuity

• Floor and ceiling function

• Limits at infinity