One-sided limit

In calculus, a one-sided limit refers to either one of the two limits of a function ${\displaystyle f(x)}${\displaystyle f(x)} of a real variable ${\displaystyle x}${\displaystyle x} as ${\displaystyle x}${\displaystyle x} approaches a specified point either from the left or from the right.^{[1] [2]}

The limit as ${\displaystyle x}${\displaystyle x} decreases in value approaching ${\displaystyle a}${\displaystyle a} (${\displaystyle x}${\displaystyle x} approaches ${\displaystyle a}${\displaystyle a} "from the right"^[3] or "from above") can be denoted:^{[1] [2]}

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

The limit as ${\displaystyle x}${\displaystyle x} increases in value approaching ${\displaystyle a}${\displaystyle a} (${\displaystyle x}${\displaystyle x} approaches ${\displaystyle a}${\displaystyle a} "from the left"^{[4] [5]} or "from below") can be denoted:^{[1] [2]}

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

If the limit of ${\displaystyle f(x)}${\displaystyle f(x)} as ${\displaystyle x}${\displaystyle x} approaches ${\displaystyle a}${\displaystyle a} exists then the limits from the left and from the right both exist and are equal. In some cases in which the limit $${\displaystyle \lim _{x\to a}f(x)}$${\displaystyle \lim _{x\to a}f(x)} does not exist, the two one-sided limits nonetheless exist. Consequently, the limit as ${\displaystyle x}${\displaystyle x} approaches ${\displaystyle a}${\displaystyle a} is sometimes called a "two-sided limit".^{[citation needed ]}

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.

If ${\displaystyle I}${\displaystyle I} represents some interval that is contained in the domain of ${\displaystyle f}${\displaystyle f} and if ${\displaystyle a}${\displaystyle a} is a point in ${\displaystyle I}${\displaystyle I} then the right-sided limit as ${\displaystyle x}${\displaystyle x} approaches ${\displaystyle a}${\displaystyle a} can be rigorously defined as the value ${\displaystyle R}${\displaystyle R} that satisfies:^{[6] [verification needed ]} $${\displaystyle {\text{for all }}\varepsilon >0\;{\text{ there exists some }}\delta >0\;{\text{ such that for all }}x\in I,{\text{ if }}\;0<x-a<\delta {\text{ then }}|f(x)-R|<\varepsilon ,}$${\displaystyle {\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 ${\displaystyle x}${\displaystyle x} approaches ${\displaystyle a}${\displaystyle a} can be rigorously defined as the value ${\displaystyle L}${\displaystyle L} that satisfies: $${\displaystyle {\text{for all }}\varepsilon >0\;{\text{ there exists some }}\delta >0\;{\text{ such that for all }}x\in I,{\text{ if }}\;0<a-x<\delta {\text{ then }}|f(x)-L|<\varepsilon .}$${\displaystyle {\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 ${\displaystyle I}${\displaystyle I} represent an interval, where ${\displaystyle I\subseteq \mathrm {domain} (f)}${\displaystyle I\subseteq \mathrm {domain} (f)}, and ${\displaystyle a\in I}${\displaystyle a\in I}.

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

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

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:

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

$${\displaystyle |x-a|=|(-1)(-x+a)|=|(-1)(a-x)|=|(-1)||a-x|=|a-x|.}$${\displaystyle |x-a|=|(-1)(-x+a)|=|(-1)(a-x)|=|(-1)||a-x|=|a-x|.}

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

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

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

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

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

The one-sided limit to a point ${\displaystyle p}${\displaystyle 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 ${\displaystyle p.}${\displaystyle p.}^{[1] [verification needed ]} Alternatively, one may consider the domain with a half-open interval topology.^{[citation needed ]}

A noteworthy theorem treating one-sided limits of certain power series at the boundaries of their intervals of convergence is Abel's theorem.^{[citation needed ]}

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