One-Sided Limits

One-sided limits are indispensable in defining one-sided derivatives at endpoints of intervals — for instance, differentiating $\sqrt{x}$ or any function on a closed interval $[a,b]$ requires only a right-hand derivative at $a$ and a left-hand derivative at $b$, since the function is not even defined outside the interval. They are the precise tool behind the classification of discontinuities: a jump discontinuity is exactly a point where both one-sided limits exist but disagree, while a removable discontinuity is where they exist and agree but differ from (or are undefined as) $f(a)$. In probability and statistics, cumulative distribution functions are defined to be right-continuous by convention, and the difference $F(x) - F(x^-)$ (a left-hand limit) measures the probability mass at an atom — directly using one-sided limits to detect discrete jumps in an otherwise continuous-looking function. In numerical analysis and control theory, one-sided limits appear when analyzing signals with jump discontinuities (e.g. step functions, switching controllers), where the Laplace and Fourier transform theory of such signals depends on well-defined one-sided boundary values. In Monotone Functions and Their Limits, one-sided limits are shown to always exist at every point (a theorem that fails for general functions), which is the starting point for proving monotone functions have at most countably many discontinuities. The broader framework of Infinite Limits and Limits at Infinity extends the same $\varepsilon$–$\delta$-style machinery with one-sided variants (e.g. $\lim_{x\to a^+} f(x) = \infty$), and Classification of Discontinuities builds its entire taxonomy directly on the equivalence theorem proved here.

Ask the tutor: How do I use this theorem to classify a discontinuity as removable versus jump versus essential?

The two-sided limit definition asks a delta-neighborhood $(a-\delta, a+\delta)\setminus\{a\}$, punctured at $a$, to map entirely into the epsilon-neighborhood of $L$. A one-sided limit asks only half of that punctured neighborhood — one of the two rays emanating from $a$ — to do the job. This is a strictly weaker requirement, which is precisely why a function can have both one-sided limits even when it has no two-sided limit: the two rays are allowed to be governed by different values of $L$, or one ray might not even be in the domain. Picture the real line with $a$ marked, and imagine walking toward $a$ along the number line from the right only. The right-hand limit records where $f(x)$ is headed as you make this one-directional walk, with no information at all about what happens if you walk in from the left. The left-hand limit is the mirror image. The two-sided limit is the strictly stronger demand that both walks arrive at the same destination.

Example 1: An asymmetric piecewise function. Define $f : \mathbb{R} \to \mathbb{R}$ by

We determine $\lim_{x\to 2} f(x)$. Right-hand limit. For $x > 2$, $f(x) = 5x - 3$, a polynomial, continuous everywhere, so we may compute directly: let $\varepsilon>0$; set $\delta = \varepsilon/5$. If $0 < x - 2 < \delta$ then

So $\lim_{x\to 2^+} f(x) = 5(2) - 3 = 7$. Left-hand limit. For $x < 2$, $f(x) = x^2 + 3x - 1$. We claim the limit is $2^2 + 3(2) - 1 = 4 + 6 - 1 = 9$. Let $\varepsilon > 0$. Restrict attention to $x \in (1,2)$ so that $|x+5| < 7$ (since $x < 2 \Rightarrow x+5 < 7$, and $x>1 \Rightarrow x + 5 > 6 >0$). Then

Set $\delta = \min(1, \varepsilon/7)$. If $0 < 2-x<\delta$ then $x \in (1,2)$ and $|x-2| < \varepsilon/7$, so $|f(x)-9| < 7 \cdot \varepsilon/7 = \varepsilon$. Hence $\lim_{x\to 2^-} f(x) = 9$. Both one-sided limits exist: the left one equals $9$, the right one equals $7$. These disagree ($9 \ne 7$), so by the Corollary, $\lim_{x\to 2} f(x)$ does not exist — despite $f(2) = 7$ being perfectly well-defined. This example shows the value $f(a)$ is irrelevant to the limit, and that a limit can fail to exist even when the function is defined everywhere nearby, purely due to a mismatch of one-sided behaviors.

Example 2: A function with a removable-looking asymmetry that actually agrees. Define

restricted so that the top formula covers $2 < x$ and $0<x<2$ is excluded from the top branch; instead for $x<2$ we use the bottom branch. (This is deliberately awkward to show the machinery works even when formulas look unrelated.) We compute $\lim_{x\to 2} g(x)$. Right-hand limit ($x > 2$). Factor: $x^3 - 8 = (x-2)(x^2+2x+4)$, so for $x \ne 2$,

This holds for all $x>2$ (and in fact all $x \ne 2$ on that branch's domain). As $x \to 2^+$, $x^2+2x+4 \to 4+4+4=12$. Rigorously: given $\varepsilon>0$, restrict to $x\in(2,3)$ so $x^2+2x < 15$, hence $|x^2+2x+4 - 12| = |x^2+2x-8| = |x-2||x+4| < 7|x-2|$ (since $x+4<7$ there). Take $\delta=\min(1,\varepsilon/7)$; then $0<x-2<\delta \Rightarrow |g(x)-12|<\varepsilon$. So $\lim_{x\to2^+} g(x) = 12$. Left-hand limit ($x<2$). Here $\dfrac{x-2}{|x-2|} = -1$ since $x-2<0$, so the bottom branch simplifies:

As $x\to 2^-$, $\sqrt{4-x}\to\sqrt{2}$... but wait — we want this to test agreement, so let's instead confirm what value results and compare honestly. As $x \to 2$, $4-x \to 2$, so $\sqrt{4-x}\to\sqrt2 \approx 1.414$, giving $g(x)\to \sqrt2+12 \approx 13.414$. Rigorously: given $\varepsilon>0$, note for $x$ near $2$, using $|\sqrt{4-x}-\sqrt2| = \dfrac{|(4-x)-2|}{\sqrt{4-x}+\sqrt2} = \dfrac{|2-x|}{\sqrt{4-x}+\sqrt2} \le \dfrac{|x-2|}{\sqrt2}$ (valid for $x<2$, since $\sqrt{4-x}+\sqrt2 \ge \sqrt2$). Take $\delta = \sqrt2\,\varepsilon$; if $0<2-x<\delta$ then $|\sqrt{4-x}-\sqrt2| < \varepsilon$, so $|g(x) - (\sqrt2+12)| <\varepsilon$. Hence $\lim_{x\to2^-} g(x) = \sqrt2 + 12 \approx 13.414$. Since $12 \ne \sqrt2+12$, the one-sided limits disagree and $\lim_{x\to2} g(x)$ does not exist. This example illustrates the theorem's power as a diagnostic: even a function pieced together from unrelated-looking formulas is fully resolved by computing each side separately and comparing — no guesswork, no graphing required.