One Sided Limits

From the left and from the right

Take f(x) = −1 for x < 0 and f(x) = 1 for $x\ge 0$x ≥ 0. At x = −0.1, −0.01 and −0.001 the output is −1 every time. At x = 0.1, 0.01 and 0.001 it is 1 every time.

A one-sided limit uses inputs from one side only. $x\to 0^{-}$x → 0⁻ means x approaches 0 from below, through negative numbers, and $x\to 0^{+}$x → 0⁺ means from above. Here the limit of f(x) as $x\to 0^{-}$x → 0⁻ is −1, the left-hand limit, and the limit as $x\to 0^{+}$x → 0⁺ is 1, the right-hand limit.

f(x) = −1 for x < 0 and 1 for $x\ge 0$x ≥ 0. Coming in from the left the graph runs along −1 to a hollow circle; from the right it runs along 1 to the filled point at (0, 1), which is f(0).

Two answers mean no limit

The limit of f(x) as $x\to 0$x → 0 would have to be one number that the outputs approach from both sides. The outputs on the left stay at −1 and the outputs on the right stay at 1, and no single number is near both. So the limit as $x\to 0$x → 0 does not exist.

The average, 0, is not the limit: f(x) is never near 0 for any input near 0. Nor is f(0) = 1: the value at the point never decides a limit, and here the left side disagrees with it.

The two one-sided limits on one line: −1 from the left and +1 from the right, 2 apart. No single number is approached from both sides.

The condition

The limit of f(x) as $x\to c$x → c exists exactly when both one-sided limits exist and are equal, and then it is their common value. If they differ, or either fails to exist, there is no limit.

They can agree. Take f(x) = x + 1 for x < 2 and f(x) = 5 − x for $x\ge 2$x ≥ 2. From the left, f(1.9) = 2.9 and f(1.99) = 2.99, heading to 3. From the right, f(2.1) = 2.9 and f(2.01) = 2.99, also heading to 3. Both sides give 3, so the limit as $x\to 2$x → 2 is 3, even though the formula changes there.

They can disagree at a join. Take f(x) = x + 1 for x < 2 and f(x) = x + 2 for $x\ge 2$x ≥ 2. From the left the outputs 2.9, 2.99, 2.999 approach 3; from the right 4.1, 4.01, 4.001 approach 4. The limit as $x\to 2$x → 2 does not exist, even though f(2) = 4. The gap 4 − 3 = 1 is the height of the jump.

x + 1 below 2 and x + 2 from 2 on. The left piece runs up to the hollow circle at height 3, and the right piece starts at the filled point at height 4. The two sides disagree, so there is no limit at 2.

Two probes walk toward x = 2, one along each piece. The left piece lands on 1 and the right piece on $4+\Delta$4 + Δ. Walking the probes closer cannot change where they land; only the handle beside the right piece, which raises or lowers it by $\Delta$Δ, can. Set $\Delta$Δ so the two landings agree.

One-sided limits that do not exist

A one-sided limit can fail too. $f\left(x\right)=\frac{1}{x}$f(x) = 1/x gives 10, 100 and 1000 at x = 0.1, 0.01 and 0.001: as $x\to 0^{+}$x → 0⁺ the outputs grow without bound, so there is no right-hand limit. At x = −0.1, −0.01 and −0.001 they are −10, −100 and −1000, so there is no left-hand limit either.

$y=\frac{1}{x}$y = 1/x. As $x\to 0^{+}$x → 0⁺ the curve climbs past every height, and as $x\to 0^{-}$x → 0⁻ it falls past every depth, so neither one-sided limit exists at 0.

At the end of a domain

$\sqrt{x}$√x has no inputs below 0, so at x = 0 only the right-hand limit can be taken. $\sqrt{0.01}=0.1$√0.01 = 0.1, $\sqrt{0.0001}=0.01$√0.0001 = 0.01 and $\sqrt{0.000001}=0.001$√0.000001 = 0.001, so the limit as $x\to 0^{+}$x → 0⁺ is 0.

The usual mistakes

Averaging the two sides. Left 2 and right 5 do not give a limit of 3.5; the outputs are never near 3.5. When the sides disagree, there is no limit.

Using one side only. A right-hand limit of 5 is not the limit if the left-hand limit is 2.

Using the value at the point. f(2) = 4 for the jump above, but the left side heads to 3, so there is no limit.

Doubting an agreement. Left 3 and right 3 settle it: the limit exists and is 3, whatever happens at the point itself.

A car park and a slippery road

In the first application below, the charge is a constant on each side of the three-hour mark, so each one-sided limit is that constant. In the second, the grip $\mu$μ can only be positive, so the stopping distance is followed as $\mu \to 0^{+}$μ → 0⁺, from above.