Limit at a Point¶

Definition¶

Let $X\subseteq\mathbb{R}\text{,}$ let $f\colon X\to\mathbb{R}$ be a function, and let $x_0$ be a limit point of $X\text{.}$ We say that $f$ converges to $L \in \mathbb{R}$ as $x$ approaches $x_0\text{,}$ and write

if for every $\varepsilon > 0$ there exists a number $\delta>0$ such that, for every $x\in X\text{,}$

Some remarks:

• The point $x_0$ must be a limit point of the domain; $x_0$ itself need not belong to $X\text{.}$
• The condition $0<|x-x_0|$ excludes $x=x_0\text{,}$ so the value $f(x_0)$ is irrelevant to the limit.
• If the limit exists, it is unique.
• The $\varepsilon$-$\delta$ definition contains no motion: it is a precise implication between two inequalities. An opponent chooses the output tolerance $\varepsilon>0\text{;}$ we must answer with a valid input tolerance $\delta>0\text{.}$

Interpretation¶

The limit describes the value that $f(x)$ approaches as $x$ gets arbitrarily close to $x_0\text{,}$ regardless of whether $f(x_0)$ is defined. The definition turns "approaches" into a test: every output tolerance $\varepsilon$ around $L$ must be met by some input tolerance $\delta$ around $x_0$.

In the figure the point $a$ plays the role of $x_0$, and $f(x)=x+1$ is drawn with an open circle at $(1,2)$, so the value at $a=1$ is left out. The horizontal band is $|y-L|<\varepsilon$ around $L=2$$2$, the vertical band is $|x-a|<\delta$$\delta$. Since $|f(x)-2|=|x-1|$$1|$, the choice $\delta=\varepsilon$ works for every $\varepsilon>0$$0$, so $\lim_{x\to1}f(x)=2$$2$.

The order of the quantifiers matters: $\varepsilon$ is given first, and $\delta$ may depend on it. Once a positive $\delta$ works for a given $\varepsilon$, every smaller positive radius works too. Valid radii need not be unique; a tighter output tolerance does not force every choice of radius to become strictly smaller. The limit exists only if an answer $\delta$ exists for every $\varepsilon>0$$0$, however small. The requirement that $x_0$ is a limit point of the domain guarantees that there are points $x\in X$ with $0<|x-x_0|<\delta$ to test. At an isolated point the implication would hold vacuously for every $L$, and the limit would not be unique.

The same pattern defines the limit of a sequence, with an index $N$ in place of $\delta$, and the limit at infinity, with a bound $M$. Using only one side of $x_0$ gives the one-sided limits, and requiring $L=f(x_0)$ gives continuity. The idea returns later in the derivative, which is defined as a limit of this kind.

A common slip is to compute $f(x_0)$ and call it the limit. The two agree only when $f$ is continuous at $x_0$: $f(x)=(x^2-1)/(x-1)$ is not defined at $1$, yet $\lim_{x\to1}f(x)=2$$2$.

Key Results¶

• Limits, if they exist, are unique.
• A function may have a limit at a point where it is not defined.

Examples¶

• $\displaystyle\lim_{x \to 0} x^2 = 0$
• $\displaystyle\lim_{x \to 0} \frac{\sin x}{x} = 1$

Prerequisites¶

• Function
• Punctured Neighbourhood

Related Concepts¶

• L'Hôpital's Rule
• Signum Function
• Singularity

Sources¶

• Math I: Calculus · Lecture 5

Aliases: limit of a function at a point Keywords: function limit Symbols: delta, element of, epsilon, real numbers, subset

Last updated: 27 October 2026 11:52