Limit of a Function

You already know how sequences converge in a metric space — a sequence $(x_n)$ converges to $L$ if its terms eventually land arbitrarily close to $L$. Functions are different: $f(x)$ is defined for all real $x$ near a point, not just a discrete list of them. The limit of a function captures what value $f(x)$ approaches as $x$ moves continuously toward a target, without requiring $f$ to be defined at that target or to agree with the limiting value there.

The ε–δ definition

Let $f$ be a real-valued function defined on some set $D \subseteq \mathbb{R}$, and let $a$ be a limit point of $D$ — meaning every neighborhood of $a$ contains a point of $D$ other than $a$ itself. A real number $L$ is the limit of $f$ at $a$ if for every $\varepsilon > 0$ there exists $\delta > 0$ such that

$$0 < |x - a| < \delta \text{ and } x \in D \implies |f(x) - L| < \varepsilon.$$

When this holds, you write

$$\lim_{x \to a} f(x) = L \qquad \text{or} \qquad f(x) \to L \text{ as } x \to a.$$

The condition $0 < |x - a| < \delta$ says that $x$ is within distance $\delta$ of $a$, but not equal to $a$. The behavior of $f$ at $a$ itself is irrelevant to the limit — $f$ might not even be defined there.

What ε–δ is actually saying

Given a tolerance $\varepsilon > 0$, you must produce a radius $\delta > 0$ such that whenever $x$ is within distance $\delta$ of $a$ (and $x \neq a$), the output $f(x)$ is within distance $\varepsilon$ of $L$. The smaller you make the tolerance, the tighter the radius may need to be. The limit exists if this challenge can always be met.

The sequential characterization

There is an equivalent formulation in terms of sequences, which is often easier to use in proofs.

Theorem. $\lim_{x \to a} f(x) = L$ if and only if for every sequence $(x_n)$ in $D \setminus \{a\}$ with $x_n \to a$, one has $f(x_n) \to L$.

Proof.

($\Rightarrow$) Suppose the ε–δ condition holds. Let $(x_n)$ be any sequence in $D \setminus \{a\}$ with $x_n \to a$. Given $\varepsilon > 0$, choose $\delta$ from the ε–δ definition. Since $x_n \to a$, there exists $N$ such that $|x_n - a| < \delta$ for all $n \geq N$. Since also $x_n \neq a$, the ε–δ condition gives $|f(x_n) - L| < \varepsilon$ for all $n \geq N$. So $f(x_n) \to L$.

($\Leftarrow$) Suppose the ε–δ condition fails. Then there exists $\varepsilon > 0$ such that for every $\delta > 0$ there is some $x \in D \setminus \{a\}$ with $|x - a| < \delta$ but $|f(x) - L| \geq \varepsilon$. Choosing $\delta = 1/n$ at each step produces a sequence $(x_n)$ in $D \setminus \{a\}$ with $x_n \to a$ but $|f(x_n) - L| \geq \varepsilon$ for all $n$, so $f(x_n) \not\to L$. $\square$

The sequential form is particularly useful for proving a limit does not exist: find two sequences approaching $a$ along which $f$ tends to different values.

Uniqueness

If the limit of $f$ at $a$ exists, it is unique. The proof is immediate from the sequential characterization: if $f(x) \to L$ and $f(x) \to L'$, then for any sequence $x_n \to a$ with $x_n \neq a$, both $f(x_n) \to L$ and $f(x_n) \to L'$, so $L = L'$ by uniqueness of sequence limits.

One-sided limits

Sometimes $f$ approaches different values depending on the direction of approach. The left-hand limit is

$$\lim_{x \to a^-} f(x) \coloneqq L$$

if for every $\varepsilon > 0$ there exists $\delta > 0$ such that $-\delta < x - a < 0$ implies $|f(x) - L| < \varepsilon$. The right-hand limit $\lim_{x \to a^+} f(x)$ is defined symmetrically.

The two-sided limit exists and equals $L$ if and only if both one-sided limits exist and equal $L$.

Computing limits

Polynomials and rational functions

For any polynomial $p$ and any $a \in \mathbb{R}$:

$$\lim_{x \to a} p(x) = p(a).$$

This follows from $\lim_{x \to a} x = a$ and the arithmetic rules for limits proved in Local Properties of Limits. For a rational function $r = p/q$ with $q(a) \neq 0$, the quotient rule gives $\lim_{x \to a} r(x) = r(a)$.

When $q(a) = 0$ but $p(a) = 0$ as well, you may be able to cancel a common factor before taking the limit.

A fundamental trigonometric limit

$$\lim_{x \to 0} \frac{\sin x}{x} = 1.$$

The standard geometric proof shows $\cos x < \dfrac{\sin x}{x} < 1$ for $0 < |x| < \pi/2$. Since $\cos x \to 1$ as $x \to 0$, the squeeze theorem forces the middle expression to $1$ as well.

Summary

• The ε–δ definition: $\lim_{x \to a} f(x) = L$ means for every $\varepsilon > 0$ there exists $\delta > 0$ such that $0 < |x - a| < \delta$ implies $|f(x) - L| < \varepsilon$. The value (or existence) of $f$ at $a$ itself is irrelevant.
• Sequential characterization: $\lim_{x \to a} f(x) = L$ if and only if $f(x_n) \to L$ for every sequence $x_n \to a$ with $x_n \neq a$. Both characterizations are equivalent.
• The two-sided limit exists iff both one-sided limits exist and agree.
• Limits of polynomials and rational functions (with nonzero denominator) equal the function value at the point; the squeeze theorem handles many other cases.