One-sided limit

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

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

$$\lim _{x\to a^{+}}f(x)\quad {\text{ or }}\quad \lim _{x\,\downarrow \,a}\,f(x)\quad {\text{ or }}\quad \lim _{x\searrow a}\,f(x)\quad {\text{ or }}\quad f(x+)$$

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

$$\lim _{x\to a^{-}}f(x)\quad {\text{ or }}\quad \lim _{x\,\uparrow \,a}\,f(x)\quad {\text{ or }}\quad \lim _{x\nearrow a}\,f(x)\quad {\text{ or }}\quad f(x-)$$

If the limit of $f(x)$ as $x$ approaches $a$ exists then the limits from the left and from the right both exist and are equal. In some cases in which the limit

$$\lim _{x\to a}f(x)$$

does not exist, the two one-sided limits nonetheless exist. Consequently, the limit as $x$ approaches $a$ is sometimes called a "two-sided limit".

It is possible for exactly one of the two one-sided limits to exist (while the other does not 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 $f$ 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:

$${\text{for all }}\varepsilon >0\;{\text{ there exists some }}\delta >0\;{\text{ such that for all }}x\in I,{\text{ if }}\;0<x-a<\delta {\text{ then }}|f(x)-R|<\varepsilon ,$$

and the left-sided limit as $x$ approaches $a$ can be rigorously defined as the value $L$ that satisfies:

$${\text{for all }}\varepsilon >0\;{\text{ there exists some }}\delta >0\;{\text{ such that for all }}x\in I,{\text{ if }}\;0<a-x<\delta {\text{ then }}|f(x)-L|<\varepsilon .$$

We can represent the same thing more symbolically, as follows.

Let $I$ represent an interval, where $I\subseteq \mathrm {domain} (f)$, and $a\in I$.

$$\lim _{x\to a^{+}}f(x)=R~~~\iff ~~~(\forall \varepsilon \in \mathbb {R} _{+},\exists \delta \in \mathbb {R} _{+},\forall x\in I,(0<x-a<\delta \longrightarrow |f(x)-R|<\varepsilon ))$$

$$\lim _{x\to a^{-}}f(x)=L~~~\iff ~~~(\forall \varepsilon \in \mathbb {R} _{+},\exists \delta \in \mathbb {R} _{+},\forall x\in I,(0<a-x<\delta \longrightarrow |f(x)-L|<\varepsilon ))$$

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:

$$\lim _{x\to a}f(x)=L~~~\iff ~~~\forall \varepsilon \in \mathbb {R} _{+},\exists \delta \in \mathbb {R} _{+},\forall x\in I,0<|x-a|<\delta \implies |f(x)-L|<\varepsilon .$$

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 $\varepsilon$, so we get the following: $|f(x)-L|<\varepsilon$ for the left sided limit, and $|f(x)-R|<\varepsilon$ 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

$$\lim _{x\to 0^{-}}{-1/x}=+\infty \qquad {\text{ and }}\qquad \lim _{x\to 0^{+}}{-1/x}=-\infty$$

The reason why $\lim _{x\to 0^{-}}{-1/x}=+\infty$ 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 $\lim _{x\to 0^{-}}{-1/x}$ diverges to $+\infty$ (and not to $-\infty$) as $x$ approaches $0$ from the left. Similarly, $\lim _{x\to 0^{+}}{-1/x}=-\infty$ 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 $\lim _{x\to 0^{+}}{-1/x}$ diverges to $-\infty .$

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

Example 2: One example of a function with different one-sided limits is $f(x)={\frac {1}{1+2^{-1/x}}},$ (cf. picture) where the limit from the left is $\lim _{x\to 0^{-}}f(x)=0$ and the limit from the right is $\lim _{x\to 0^{+}}f(x)=1.$ To calculate these limits, first show that

$$\lim _{x\to 0^{-}}2^{-1/x}=\infty \qquad {\text{ and }}\qquad \lim _{x\to 0^{+}}2^{-1/x}=0$$

(which is true because $\lim _{x\to 0^{-}}{-1/x}=+\infty {\text{ and }}\lim _{x\to 0^{+}}{-1/x}=-\infty$) so that consequently,

$$\lim _{x\to 0^{+}}{\frac {1}{1+2^{-1/x}}}={\frac {1}{1+\displaystyle \lim _{x\to 0^{+}}2^{-1/x}}}={\frac {1}{1+0}}=1$$

whereas $\lim _{x\to 0^{-}}{\frac {1}{1+2^{-1/x}}}=0$ because the denominator diverges to infinity; that is, because $\lim _{x\to 0^{-}}1+2^{-1/x}=\infty .$ Since $\lim _{x\to 0^{-}}f(x)\neq \lim _{x\to 0^{+}}f(x),$ the limit $\lim _{x\to 0}f(x)$ does not exist.

Relation to topological definition of limit

See also: Filters in topology

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 $p.$ Alternatively, one may consider the domain with a half-open interval topology.

Abel's theorem

Main article: 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.

See also

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