Limit of a Sequence¶

Definition¶

Let $L\in\mathbb{R}\text{.}$ A sequence $(a_n)_{n=1}^\infty$ converges to $L$ if for every $\varepsilon >0\text{,}$ there exists an $N \geq 1$ such that for every $n\geq N\text{,}$ we have $|a_n-L|<\varepsilon\text{.}$ In this case, the sequence $(a_n)_{n=1}^\infty$ is called convergent, and $L$ is called its limit. We write

L=\lim_{n\to\infty} a_n\quad\text{or}\quad a_n\to L\text{ as }n\to\infty.

If no such $L\in\mathbb{R}$ exists, the sequence $(a_n)_{n=1}^\infty$ is called divergent.

Interpretation¶

The limit of a sequence describes the value approached by its terms as the index increases. The definition formalizes "approached" as eventual closeness: however small a tolerance $\varepsilon>0$ is chosen, from some index $N$ on every term lies within $\varepsilon$ of $L$ .

The terms 1/n approaching 0 inside the band from -0.1 to 0.1 around 0; all terms from n = 11 on lie in the band.

$a_n=1/n\to0$ $0$ : for $\varepsilon=0.1$ $0.1$ , every term with $n\ge N=11$ (red) lies within $\varepsilon$ of the limit $0$ , since $1/n<0.1$ exactly when $n>10$ $10$ . A smaller $\varepsilon$ needs a larger $N$ : for $\varepsilon=0.01$ $0.01$ , $N=101$ works. The definition asks only that some $N$ exists for each $\varepsilon$ , not the smallest one.

Shrink $\varepsilon$ and watch $N$ grow. Then move the candidate $L$ away from the limit, or pick $(-1)^n$ : no $N$ keeps every later term in the band.

The order of the quantifiers carries the meaning: first $\varepsilon$ is given, then $N$ may be chosen depending on $\varepsilon$ , and the condition must hold for all $n\ge N$ $N$ . This last part rules out sequences that only come close now and then. The terms of $(-1)^n$ equal $1$ infinitely often but also return to $-1$ , so for $\varepsilon=1$ no $N$ keeps all later terms within $\varepsilon$ of any candidate $L$ . Only the tail matters: changing finitely many terms does not change whether the limit exists or what it is.

Two slips are common. "The terms get closer and closer to $L$ " is not the definition: the terms of $1/n$ also get closer to $-1$ at every step, but $-1$ is not the limit. And the terms need not approach from one side: $(-1)^n/n$ converges to $0$ while jumping from one side of $0$ to the other. The same $\varepsilon$ -idea returns later for limits of functions, as a limit at a point.

Key Results¶

Limits of sequences are unique when they exist.
Every convergent sequence is bounded.

Enrichment: non-assessable in 2026-27

Every convergent sequence is a Cauchy sequence, and over $\mathbb{R}$ the converse also holds. The Cauchy criterion is optional enrichment in Math I: Calculus and is not part of the assessable sequence-limit core.

Examples¶

$\displaystyle \lim_{n\to\infty}\frac{1}{n}=0\text{.}$
The sequence $(-1)^n$ does not converge.

Proving a sequence limit with $\varepsilon$ and $N$

Prove from the definition that $\displaystyle\lim_{n\to\infty}\frac{2n+1}{n+3}=2$ $2$ .

Simplify the distance to the limit

For every $n\ge1$ $1$ ,

\left|\frac{2n+1}{n+3}-2\right|=\left|\frac{2n+1-2n-6}{n+3}\right|=\frac{5}{n+3}.

Find a simpler bound and choose

N

Since $n+3>n$ $n$ , we have $\dfrac{5}{n+3}<\dfrac{5}{n}$ $\frac{5}{n}$ , and $\dfrac{5}{n}<\varepsilon$ as soon as $n>5/\varepsilon$ $5/ ε$ . So, given $\varepsilon>0$ $0$ , choose an integer $N\ge1$ with $N>5/\varepsilon$ $5/ ε$ ; it exists by the Archimedean property of the real numbers.

Verify the choice

For every $n\ge N$ $N$ ,

\left|\frac{2n+1}{n+3}-2\right|=\frac{5}{n+3}<\frac{5}{n}\le\frac{5}{N}<\varepsilon.

This holds for every $\varepsilon>0$ $0$ , so the sequence converges to $2$ .

Takeaway. Simplify $|a_n-L|$ first, then bound it by something simple that shrinks as $n$ grows, and read $N$ off that bound. $N$ may depend on $\varepsilon$ , and it does not have to be the smallest possible one.

Prerequisites¶

Sequence

Sources¶

Math I: Calculus · Lecture 3

Aliases: sequence limit Keywords: convergent sequence, convergence of a sequence Symbols: element of, epsilon, infinity, real numbers

Last updated: 27 October 2026 11:52