One-sided limit

At $x=0,$ the function $f(x)=x^2+\mathrm{sign}(x),$ where $\mathrm{sign}(x)$ denotes the sign function, has a left limit of $-1,$ a right limit of $+1,$ and a function value of $0.$

In calculus, a one-sided limit refers to either one of the two limits of a function $f(x)$ of a real variable $x$ as $x$ approaches a specified point either from the left or from the right.

The limit, as $x$ decreases in value approaching $a$ ($x$ approaches $a$ "from the right" or "from above"), is denoted:

$$\underset{x\to a^{+}}{lim}f(x)\text{ or }\underset{x\downarrow a}{lim}f(x)\text{ or }\underset{x\searrow a}{lim}f(x)\text{ or }f(a+).$$

The limit, as $x$ increases in value approaching $a$ ($x$ approaches $a$ "from the left" or "from below"), is denoted:

$$\underset{x\to a^{-}}{lim}f(x)\text{ or }\underset{x\uparrow a}{lim}f(x)\text{ or }\underset{x\nearrow a}{lim}f(x)\text{ or }f(a-).$$

If the limits from the left and right both exist and are equal, then the limit of $f(x)$ as $x$ approaches $a$ exists. Conversely, if the limit of $f(x)$ as $x$ approaches $a$ exists, then the limits from left and right both exist and are equal. Consequently, the limit as $x$ approaches $a$ is sometimes called a "two-sided limit".Template:Citation needed It is denoted: $$\underset{x\to a}{lim}f(x).$$

In some cases in which the two-sided limit does not exist, the two individual one-sided limits nonetheless exist and they are then necessarily unequal.

It is possible for only one of the two one-sided limits to exist. It is also possible for neither of the two one-sided limits to exist.

Formal definition

Definition

If $I$ represents some interval that is contained in the domain of a function $f$ and if $a$ is a point in $I$, then the right-sided limit as $x$ approaches $a$ can be rigorously defined as the value $R$ that satisfies: Template:Blockquote and the left-sided limit as $x$ approaches $a$ can be rigorously defined as the value $L$ that satisfies: Template:Blockquote

These definitions can be represented more symbolically as follows: Let $I$ represent an interval, where $I\subseteq \mathrm{d}\mathrm{o}\mathrm{m}\mathrm{a}\mathrm{i}\mathrm{n}(f)$ and $a\in I$, then $$\begin{array}{cc}\underset{x\to a^{+}}{lim}f(x)=R& ⟺\mathrm{\forall }\epsilon \in {ℝ}_{+},\mathrm{\exists }\delta \in {ℝ}_{+},\mathrm{\forall }x\in I,0<x-a<\delta ⟶|f(x)-R|<\epsilon ,\\ \underset{x\to a^{-}}{lim}f(x)=L& ⟺\mathrm{\forall }\epsilon \in {ℝ}_{+},\mathrm{\exists }\delta \in {ℝ}_{+},\mathrm{\forall }x\in I,0<a-x<\delta ⟶|f(x)-L|<\epsilon .\end{array}$$

Intuition

In comparison to the formal definition for the limit of a function at a point, the one-sided limit (as the name would suggest) only deals with input values to one side of the approached input value.

For reference, the formal definition for the limit of a function at a point is as follows:

$$\underset{x\to a}{lim}f(x)=L⟺\mathrm{\forall }\epsilon \in {ℝ}_{+},\mathrm{\exists }\delta \in {ℝ}_{+},\mathrm{\forall }x\in I,0<|x-a|<\delta ⟹|f(x)-L|<\epsilon .$$

To define a one-sided limit, we must modify this inequality. Note that the absolute distance between $x$ and $a$ is

$$|x-a|=|(-1)(-x+a)|=|(-1)(a-x)|=|(-1)||a-x|=|a-x|.$$

For the limit from the right, we want $x$ to be to the right of $a$, which means that $a<x$, so $x-a$ is positive. From above, $x-a$ is the distance between $x$ and $a$. We want to bound this distance by our value of $\delta$, giving the inequality $x-a<\delta$. Putting together the inequalities $0<x-a$ and $x-a<\delta$ and using the transitivity property of inequalities, we have the compound inequality $0<x-a<\delta$.

Similarly, for the limit from the left, we want $x$ to be to the left of $a$, which means that $x<a$. In this case, it is $a-x$ that is positive and represents the distance between $x$ and $a$. Again, we want to bound this distance by our value of $\delta$, leading to the compound inequality $0<a-x<\delta$.

Now, when our value of $x$ is in its desired interval, we expect that the value of $f(x)$ is also within its desired interval. The distance between $f(x)$ and $L$, the limiting value of the left sided limit, is $|f(x)-L|$. Similarly, the distance between $f(x)$ and $R$, the limiting value of the right sided limit, is $|f(x)-R|$. In both cases, we want to bound this distance by $\epsilon$, so we get the following: $|f(x)-L|<\epsilon$ for the left sided limit, and $|f(x)-R|<\epsilon$ for the right sided limit.

Examples

Example 1. The limits from the left and from the right of $g(x):=-\frac{1}{x}$ as $x$ approaches $a:=0$ are, respectively $$\underset{x\to 0^{-}}{lim}-\frac{1}{x}=+\mathrm{\infty }\text{ and }\underset{x\to 0^{+}}{lim}-1/x=-\mathrm{\infty }.$$ The reason why $\underset{x\to 0^{-}}{lim}-\frac{1}{x}=+\mathrm{\infty }$ is because $x$ is always negative (since $x\to 0^{-}$ means that $x\to 0$ with all values of $x$ satisfying $x<0$), which implies that $-1/x$ is always positive so that $\underset{x\to 0^{-}}{lim}-\frac{1}{x}$ diverges to $+\mathrm{\infty }$ (and not to $-\mathrm{\infty }$) as $x$ approaches $0$ from the left. Similarly, $\underset{x\to 0^{+}}{lim}-\frac{1}{x}=-\mathrm{\infty }$ since all values of $x$ satisfy $x>0$ (said differently, $x$ is always positive) as $x$ approaches $0$ from the right, which implies that $-1/x$ is always negative so that $\underset{x\to 0^{+}}{lim}-\frac{1}{x}$ diverges to $-\mathrm{\infty }.$

Plot of the function $f(x)=\frac{1}{1+2^{-1/x}}$.

Example 2. One example of a function with different one-sided limits is $f(x)=\frac{1}{1+2^{-1/x}}$, where the limit from the left is $\underset{x\to 0^{-}}{lim}f(x)=0$ and the limit from the right is $\underset{x\to 0^{+}}{lim}f(x)=1.$ To calculate these limits, first show that $$\underset{x\to 0^{-}}{lim}{2}^{-1/x}=\mathrm{\infty }\text{ and }\underset{x\to 0^{+}}{lim}{2}^{-1/x}=0,$$ which is true because $\underset{x\to 0^{-}}{lim}-1/x=+\mathrm{\infty }$ and $\underset{x\to 0^{+}}{lim}-1/x=-\mathrm{\infty }$ so that consequently, $$\underset{x\to 0^{+}}{lim}\frac{1}{1+2^{-1/x}}=\frac{1}{1+\underset{x\to 0^{+}}{lim}{2}^{-1/x}}=\frac{1}{1+0}=1$$ whereas $\underset{x\to 0^{-}}{lim}\frac{1}{1+2^{-1/x}}=0$ because the denominator diverges to infinity; that is, because $\underset{x\to 0^{-}}{lim}1+2^{-1/x}=\mathrm{\infty }$. Since $\underset{x\to 0^{-}}{lim}f(x)\ne \underset{x\to 0^{+}}{lim}f(x)$, the limit $\underset{x\to 0}{lim}f(x)$ does not exist.

Relation to topological definition of limit

The one-sided limit to a point $p$ corresponds to the general definition of limit, with the domain of the function restricted to one side, by either allowing that the function domain is a subset of the topological space, or by considering a one-sided subspace, including $p.$Template:Verify source Alternatively, one may consider the domain with a half-open interval topology.Template:Citation needed

Abel's theorem

A noteworthy theorem treating one-sided limits of certain power series at the boundaries of their intervals of convergence is Abel's theorem.Template:Citation needed

See also

• Projectively extended real line
• Semi-differentiability
• Limit superior and limit inferior