One-sided limit

If ${\displaystyle I}$ represents some interval that is contained in the domain of ${\displaystyle f}$ and if ${\displaystyle a}$ is point in ${\displaystyle I}$ then the right-sided limit as ${\displaystyle x}$ approaches ${\displaystyle a}$ can be rigorously defined as the value ${\displaystyle R}$ that satisfies:[3][4]

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

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

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

Example 1: The limits from the left and from the right of ${\displaystyle g(x):=-{\frac {1}{x}}}$ as ${\displaystyle x}$ approaches ${\displaystyle a:=0}$ are

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

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

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

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

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

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

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

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

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

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

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