Sometimes a function behaves completely differently depending on which side you approach from. One-sided limits let you describe each side independently, and tell you exactly when a two-sided limit can and cannot exist.
Left-hand & right-hand notationExistence theoremPiecewise functionsAbsolute valueFloor function10 worked examples
Share this page
§ 01Why We Need One-Sided Limits
A standard limit asks what a function approaches as x closes in from both sides simultaneously. But what if the function looks completely different on the left of a point than it does on the right?
Consider the sign function: sgn(x) = −1 for negative x and +1 for positive x. As x approaches 0 from the left, sgn(x) is hovering at −1. As x approaches 0 from the right, sgn(x) is hovering at +1. These two behaviours are irreconcilable, there is no single value that the function is "approaching" at x = 0, so the ordinary two-sided limit does not exist.
One-sided limits were invented precisely to handle this situation. Rather than asking about both sides at once, a one-sided limit asks: what does f(x) approach if we only allow x to come from one specific direction? This gives us a finer instrument for analysing function behaviour, and, as we will see, the relationship between the two one-sided limits is exactly what determines whether the ordinary two-sided limit exists.
One-sided limits are not just a technicality
They are essential for understanding piecewise-defined functions, removable and jump discontinuities, the behaviour of functions at endpoints of their natural domains, and the definition of the derivative itself — which is defined as a two-sided limit that secretly depends on two one-sided limits agreeing.
If you are comfortable with the basic idea of a limit (that limx→a f(x) = L means f(x) gets arbitrarily close to L as x approaches a) then one-sided limits are a small notational step: you simply restrict the direction of approach.
§ 02The Notation — Reading the Superscript ⁻ and ⁺
One tiny superscript, a minus sign or a plus sign after the value a, carries all the directional information. Getting comfortable reading these symbols out loud is the first step.
Left-Hand Limit (LHL)
limx→a⁻ f(x) = L
Read: "the limit of f(x) as x approaches a from the left." The superscript minus means x comes from values less than a, it slides up to a along the negative direction.
Right-Hand Limit (RHL)
limx→a⁺ f(x) = L
Read: "the limit of f(x) as x approaches a from the right." The superscript plus means x comes from values greater than a, it slides down to a from above.
A common source of confusion: the ⁻ and ⁺ are not signs attached to the value a. They are directional arrows, indicating which side of a the x-values live on. So x → 3⁻ means x takes values like 2.9, 2.99, 2.999… getting closer to 3 from below. And x → 3⁺ means x takes values like 3.1, 3.01, 3.001… approaching 3 from above.
Memory Aid
x → a⁻ : x < a, x increasing toward a (from the left)
x → a⁺ : x > a, x decreasing toward a (from the right)
What the Limit Value Actually Means
When we write limx→a⁻ f(x) = L, we are saying: for every positive tolerance ε, there is a strip of x-values immediately to the left of a (say (a − δ, a)) such that f(x) stays within ε of L for all x in that strip. The function does not need to equal L at x = a, and x is never allowed to actually equal a (the limit describes approach, not arrival).
This means a one-sided limit can exist even if f(a) is undefined, defined differently, or a completely different value. The limit is purely about the journey, not the destination.
§ 03The Two-Sided Limit Existence Theorem
This is the most important result on this page. It is the bridge between one-sided and two-sided limits, and it is used constantly throughout calculus.
Theorem — Two-Sided Limit Existence
limx→a f(x) exists ⟺ limx→a⁻ f(x) and limx→a⁺ f(x) both exist and are equal.
When both one-sided limits exist and are equal to the same value L, then the two-sided limit also equals L. If the one-sided limits are unequal, or if either one fails to exist, the two-sided limit does not exist.
This theorem has two directions, both equally important:
To prove a limit exists: compute both one-sided limits and check they are equal.
To prove a limit does not exist: show the one-sided limits differ, or show one of them fails to exist.
Left: LHL = RHL (limit exists, even with hole). Centre: LHL ≠ RHL (jump, no limit). Right: vertical asymptote (one or both sides → ∞).
§ 04Ten Worked Examples
From simple polynomials to piecewise functions, absolute values, and the floor function, each example applies the checklist: find LHL, find RHL, compare.
Example 01Find the one-sided limits of f(x) = x² + 1 at x = 2
A polynomial, continuous everywhere. Both one-sided limits equal the function value.
Left-hand limit
limx→2⁻ (x² + 1) = 4 + 1 = 5
Right-hand limit
limx→2⁺ (x² + 1) = 4 + 1 = 5
LHL = 5RHL = 5Two-sided limit = 5 ✓
Example 02Piecewise: f(x) = { x + 1 if x < 2, x² − 1 if x ≥ 2 } at x = 2
Use the appropriate branch for each side. Left uses the x < 2 branch; right uses x ≥ 2.
Example 03Piecewise: f(x) = { 2x if x < 1, x + 3 if x ≥ 1 } at x = 1
A classic jump discontinuity setup. Watch the one-sided limits diverge.
Left-hand limit — use (2x)
limx→1⁻ 2x = 2
Right-hand limit — use (x + 3)
limx→1⁺ (x + 3) = 4
LHL = 2 ≠ RHL = 4. The function jumps by 2 units at x = 1.
LHL = 2RHL = 4Two-sided limit DNE (jump discontinuity)
Example 04Absolute value: find the one-sided limits of f(x) = |x − 3| / (x − 3) at x = 3
The absolute value creates a split: |x − 3| = x − 3 when x > 3, and |x − 3| = −(x − 3) when x < 3.
Left side (x < 3): |x−3| = −(x−3)
f(x) = −(x−3)/(x−3) = −1
Left-hand limit
limx→3⁻ f(x) = −1
Right side (x > 3): |x−3| = x−3
f(x) = (x−3)/(x−3) = 1
Right-hand limit
limx→3⁺ f(x) = 1
LHL = −1RHL = 1Two-sided limit DNE (jump of 2 units)
Example 05Find the one-sided limits of f(x) = 1/(x − 2) at x = 2
A vertical asymptote at x = 2. The function blows up, but in which direction on each side?
Left side (x < 2): (x − 2) is small and negative
limx→2⁻ 1/(x−2) = −∞
Right side (x > 2): (x − 2) is small and positive
limx→2⁺ 1/(x−2) = +∞
LHL = −∞RHL = +∞Two-sided limit DNE (infinite, opposite directions)
Example 06Find the one-sided limits of f(x) = 1/(x − 2)² at x = 2
The square in the denominator means both sides blow up in the same direction.
Left side: (x−2)² is small and positive (square is always ≥ 0)
limx→2⁻ 1/(x−2)² = +∞
Right side: same reasoning
limx→2⁺ 1/(x−2)² = +∞
Both one-sided limits are +∞, they are equal in the extended sense. We write limx→2 f(x) = +∞, acknowledging that the two-sided "limit" is ∞ (even though the ordinary finite limit still does not exist).
LHL = +∞RHL = +∞Both sides agree: lim = +∞
Example 07Three-piece function: f(x) = { x² if x < 0, 1 if x = 0, 2x + 1 if x > 0 } at x = 0
Left-hand limit — use x²
limx→0⁻ x² = 0
Right-hand limit — use 2x + 1
limx→0⁺ (2x + 1) = 1
Function value f(0) = 1 (explicitly defined)
f(0) = 1
LHL ≠ RHL, so the two-sided limit does not exist. Note that f(0) = 1 = RHL: the function agrees with the right side, but the jump from the left prevents a full limit.
LHL = 0RHL = 1Two-sided limit DNE
Example 08Floor function: f(x) = ⌊x⌋ at x = 3
The floor function (greatest integer function) ⌊x⌋ rounds down to the nearest integer. It has a jump discontinuity at every integer.
Left side: x just below 3, e.g. 2.99, 2.999 → ⌊x⌋ = 2
limx→3⁻ ⌊x⌋ = 2
Right side: x just above 3, e.g. 3.001 → ⌊x⌋ = 3
limx→3⁺ ⌊x⌋ = 3
Function value f(3) = ⌊3⌋ = 3
f(3) = 3
The floor function has a jump of 1 at every integer. The LHL is always one less than the RHL.
LHL = 2RHL = 3Two-sided limit DNE at every integer
Example 09Find k so that limx→4 f(x) exists, where f(x) = { kx − 1 if x < 4, x² − k if x ≥ 4 }
For the two-sided limit to exist, we need LHL = RHL.
Left-hand limit (x < 4 branch)
limx→4⁻ (kx − 1) = 4k − 1
Right-hand limit (x ≥ 4 branch)
limx→4⁺ (x² − k) = 16 − k
Set LHL = RHL
4k − 1 = 16 − k ⟹ 5k = 17 ⟹ k = 17/5
k = 17/5 = 3.4
Example 10Does limx→0 |x|/x exist?
This is the sign function sgn(x) rewritten. Use the definition of |x|: it equals x when x > 0 and −x when x < 0.
Left side (x < 0): |x| = −x, so |x|/x = −x/x = −1
limx→0⁻ |x|/x = −1
Right side (x > 0): |x| = x, so |x|/x = x/x = 1
limx→0⁺ |x|/x = 1
The one-sided limits −1 and +1 are unequal, so the two-sided limit does not exist. This result is used frequently to show that |x| is not differentiable at x = 0.
LHL = −1RHL = 1Two-sided limit DNE
§ 05Common Mistakes
Mistake 1 — Using the wrong branch of a piecewise function
For the left-hand limit at x = a, always use the branch that applies when x < a (approaching from below). For the right-hand limit, use the branch for x > a. Students frequently use the "x = a" branch for one of the one-sided limits, which gives the function value, not the limit.
Mistake 2 — Concluding the limit exists because the function value is defined
f(a) being defined says nothing about whether limx→a f(x) exists. The limit is about the approach. A function can have f(a) defined yet still have a jump discontinuity at a because LHL ≠ RHL.
Mistake 3 — Forgetting the sign split for absolute values
|x − a| is not simply "x − a." It equals x − a when x > a, and −(x − a) when x < a. Missing this split means you compute the same thing on both sides and falsely conclude the limit exists.
Mistake 4 — Assuming lim = +∞ and lim = −∞ means the limit exists
Two one-sided limits that are both infinite do NOT make the two-sided limit exist (in the finite sense). However, if both equal +∞ (or both equal −∞), we conventionally write lim = ±∞, acknowledging the infinite divergence is consistent from both sides.
§ 06Practice Quiz — 10 Questions
Sharpen your one-sided limit skills with this mixed quiz. Calculate LHL and RHL, determine whether the two-sided limit exists, and find values of k.
Score: 0 / 10
Question 1 of 10 — Multiple Choice
What does the notation limx→5⁻ f(x) mean?
B. The superscript ⁻ means x approaches 5 from values less than 5 — from the left. Think of x = 4.9, 4.99, 4.999… The ⁺ superscript is for the right (x > 5).
Question 2 of 10 — Multiple Choice
For f(x) = { 3x if x < 2, 7 if x ≥ 2 }, what is limx→2⁻ f(x)?
C — 6. For the left-hand limit (x < 2), use the branch 3x. As x → 2⁻: 3(2) = 6. The value 7 is what f(2) equals, but that is the function value, not the left-hand limit.
Question 3 of 10 — Multiple Choice
For the same f(x) above, does limx→2 f(x) exist?
C. LHL = 6 (from Q2), RHL = limx→2⁺ 7 = 7. Since 6 ≠ 7, the two-sided limit does not exist — this is a jump discontinuity of size 1.
Question 4 of 10 — Fill in the Blank
For f(x) = |x + 2| / (x + 2), find limx→−2⁺ f(x).
Answer:
For x > −2: (x + 2) > 0, so |x + 2| = x + 2. Simplify the fraction.
For x > −2: |x+2| = x+2, so |x+2|/(x+2) = 1. The RHL = 1.
Question 5 of 10 — Fill in the Blank
For the same f(x) = |x + 2| / (x + 2), find limx→−2⁻ f(x).
Answer:
For x < −2: (x + 2) < 0, so |x + 2| = −(x + 2). Simplify carefully.
For x < −2: |x+2| = −(x+2), so the fraction = −(x+2)/(x+2) = −1. The LHL = −1.
Question 6 of 10 — Multiple Choice
The floor function ⌊x⌋ at any non-integer value c has:
A. At a non-integer c (e.g. c = 2.7), the floor function is constant at ⌊c⌋ = 2 in a whole neighbourhood around c. Both one-sided limits equal 2. Jumps only happen at integer values of x.
Question 7 of 10 — Multiple Choice
If limx→a⁻ f(x) = 4 and limx→a⁺ f(x) = 4, which conclusion is valid?
C. By the existence theorem: LHL = RHL = 4 implies the two-sided limit equals 4. We cannot conclude f(a) = 4 (the function value might differ or be undefined), and we certainly cannot conclude continuity or differentiability without more information.
Question 8 of 10 — Fill in the Blank
Find k so that limx→3 f(x) exists, where f(x) = { 2x + k if x < 3, x² − 1 if x ≥ 3 }.
Answer: k =
Set LHL = RHL: limx→3⁻(2x+k) = limx→3⁺(x²−1). Solve for k.
LHL = 2(3)+k = 6+k. RHL = 9−1 = 8. Set equal: 6+k = 8 → k = 2.
Question 9 of 10 — Multiple Choice
limx→0⁺ √x = ?
A — 0. As x → 0 through positive values, √x → 0. The right-hand limit exists and equals 0. Note: the left-hand limit limx→0⁻ √x does not exist (√x is not defined for negative x), but the right-hand limit is perfectly well-defined.
Question 10 of 10 — Multiple Choice
For f(x) = { x² + 1 if x ≤ 0, 2x + 1 if x > 0 }, does limx→0 f(x) exist?
B — Yes, limit = 1. LHL: limx→0⁻(x²+1) = 0+1 = 1. RHL: limx→0⁺(2x+1) = 0+1 = 1. Both equal 1, so by the existence theorem limx→0 f(x) = 1. (In fact f is also continuous at 0 since f(0) = 1 = limit.)
Next StepsContinue the Limits Series
One-sided limits are the lens through which every discontinuity and every behaviour at a boundary is analysed. The topics that follow depend on them directly:
Limits at Infinity: a different kind of one-sidedness: only x → +∞ or x → −∞ is meaningful, so every infinite limit is inherently one-sided.
Continuity, the three conditions for continuity use two-sided limits, which (as you now know) depend on matching one-sided limits.
Indeterminate Forms, 0/0 can produce one-sided limits that differ, which is another way to show the two-sided limit fails.