The Limit of a Sequence

Tell an engineer a part must be 10 mm and the first thing they ask is "within what tolerance?" A machine whose output creeps ever closer to 10 mm — landing inside any tolerance you demand, and staying there — is doing exactly what mathematicians mean when they say a sequence has a limit. This page makes that idea precise.

Look at the run of numbers 0.9,\ 0.99,\ 0.999,\ 0.9999,\ \dots Everyone can feel where it is heading: to 1. But feel is not a proof, and "gets closer and closer" turns out to be a treacherously slippery phrase — the terms 0.9, 0.99, 0.999, \dots also get closer and closer to 2 (each one is nearer to 2 than the one before!), yet nobody thinks the sequence tends to 2. For roughly 150 years after Newton, calculus ran on this kind of vague talk. The fix — the first genuinely rigorous definition most mathematics students ever meet — is a surprisingly playable idea: a challenge-and-response game.

Here are the rules. Your opponent names a tolerance — any positive number \varepsilon, as brutally small as they like: 0.01, 10^{-6}, one over a googol. You must answer with a cutoff N and guarantee that from the N-th term onward, every single term of the sequence sits within \varepsilon of the claimed limit L. Not "gets within \varepsilon at some point" — gets within \varepsilon and stays there forever. If you have a winning strategy — an answer for every possible \varepsilon — the sequence converges to L. If even one \varepsilon stumps you, it doesn't.

This page teaches you to play that game: what the rules say precisely, how to win it for classic sequences, and how the limit laws let you bank old wins instead of replaying the game from scratch.

The rules, in symbols

We say the terms of a sequence (a_n) converge to a number L — written

\lim_{n \to \infty} a_n = L \qquad\text{or}\qquad a_n \to L

— when the terms get, and stay, arbitrarily close to L. The informal idea of a limit is the same as for functions; what is new is that the "approach" happens along the discrete ticks n = 1, 2, 3, \dots going to infinity. The game above, written out in quantifiers, reads:

\forall\, \varepsilon > 0 \ \ \exists\, N \in \mathbb{N} \ \ \text{such that}\ \ n \ge N \implies |a_n - L| < \varepsilon.

Each piece of notation is one of the game's moves:

\forall\, \varepsilon > 0 — "for all \varepsilon": the opponent moves first and may pick any tolerance whatsoever. You don't get to veto small ones.
\exists\, N — "there exists N": you move second, and your cutoff is allowed to depend on the \varepsilon you were handed. Order matters enormously: \varepsilon first, then N.
n \ge N \implies |a_n - L| < \varepsilon — the win condition: every term from index N onward lies inside the open band (L - \varepsilon,\, L + \varepsilon). Equivalently: at most finitely many terms are allowed to sit outside the band.

A concrete round of the game

Take a_n = 1/n, claimed limit L = 0. Your opponent opens with \varepsilon = 0.01. You need every term from some cutoff on to satisfy \left|\tfrac1n - 0\right| < 0.01, i.e. \tfrac1n < \tfrac{1}{100}, i.e. n > 100. So you answer with any cutoff past 100 — say N = 101 — and check: a_{101} = \tfrac{1}{101} \approx 0.0099, a_{102}, a_{103}, \dots smaller still. Every term from N on is trapped in the band, and none ever escapes. Round won. Opponent tries \varepsilon = 10^{-6}? You answer N = 10^6 + 1. Whatever they throw, you have a formula that answers it — and that formula is the proof.

A sequence (a_n) converges to L \in \mathbb{R}, written \lim_{n\to\infty} a_n = L, if

\forall\, \varepsilon > 0 \ \ \exists\, N \in \mathbb{N} : \ \ n \ge N \implies |a_n - L| < \varepsilon.

The limit, when it exists, is unique — a sequence cannot converge to two different numbers.
N is allowed to depend on \varepsilon (smaller tolerance, later cutoff); only finitely many terms may sit outside the band.
A sequence with no such L is said to diverge.

The limit need not be a term of the sequence. 1/n converges to 0, yet no term equals 0 — and that's fine. The definition only asks the terms to enter every band around L, never to touch L itself. (Touching is allowed too: the constant sequence 5, 5, 5, \dots converges to 5 perfectly happily — every band contains all the terms, so any N works.)
Getting close once is worthless — you must stay. The sequence 1, 0, 1, 0, 1, 0, \dots lands exactly on 0 infinitely often, but it does not converge to 0: past any cutoff there is always another 1 waiting to jump back out of the band \varepsilon = \tfrac12. The win condition is n \ge N \implies inside — every term after the cutoff, with no later escapes, ever.
The game is asymmetric. If a cutoff N wins against some \varepsilon, the same N automatically wins against every bigger \varepsilon' (a wider band captures everything the narrow one did). So a bigger \varepsilon needs a (weakly) smaller N, and all the difficulty lives at tiny \varepsilon. That is why proofs may assume, say, \varepsilon < 1 without loss of generality — the big tolerances come free.

Worked example 1: proving \tfrac1n \to 0 from the definition

The harmonic sequence a_n = 1/n obviously "wants" to head to 0. Let us earn that claim with the \varepsilon–N definition, with L = 0. The whole proof is just answering the adversary's \varepsilon with a concrete N — the single round we played above, upgraded to a strategy for all rounds at once.

Step 1 — Let the adversary fix \varepsilon. Take any \varepsilon > 0, however small. Our job is to produce an N that works for this \varepsilon.

\frac{1}{n} < \varepsilon.

Step 3 — Solve it for n. Both sides are positive, so we may take reciprocals (which flips the inequality):

\frac{1}{n} < \varepsilon \iff n > \frac{1}{\varepsilon}.

Step 4 — Choose N. We need a natural number bigger than 1/\varepsilon. The Archimedean property guarantees one exists; an explicit choice is

N = \left\lfloor \frac{1}{\varepsilon} \right\rfloor + 1, \qquad\text{so}\qquad N > \frac{1}{\varepsilon}.

Step 5 — Verify the cutoff works. Suppose n \ge N. Then n \ge N > 1/\varepsilon, and running Steps 2–3 backwards (reciprocate again):

n > \frac{1}{\varepsilon} \implies \frac{1}{n} < \varepsilon \implies \left|\frac1n - 0\right| < \varepsilon.

Step 6 — Collect. For every \varepsilon > 0 we exhibited an N (namely \lfloor 1/\varepsilon\rfloor + 1) past which every term lies within \varepsilon of 0. That is exactly the definition, so

\lim_{n \to \infty} \frac{1}{n} = 0. \qquad \blacksquare

Concretely: hand me \varepsilon = 0.01 and I answer N = 101; hand me \varepsilon = 10^{-6} and I answer N = 1000001. There is always an answer, which is the whole point. Notice the shape of the proof — it is always this shape: unpack |a_n - L|, solve the inequality for n, read off N, then verify forwards.

Worked example 2: \tfrac{n}{n+1} \to 1

The terms \tfrac12, \tfrac23, \tfrac34, \tfrac45, \dots creep up towards 1 from below. Same game, one extra flourish of algebra: this time the interesting work is simplifying the distance to the limit before solving for n.

Unpack the distance. Put the two terms over a common denominator:

\left| \frac{n}{n+1} - 1 \right| = \left| \frac{n - (n+1)}{n+1} \right| = \left| \frac{-1}{n+1} \right| = \frac{1}{n+1}.

A messy-looking distance has collapsed to something monotone and friendly. This step — wrestling |a_n - L| into a simple decreasing expression — is where most of the skill in \varepsilon–N proofs lives.

Solve the win condition for n. Given \varepsilon > 0, we need

\frac{1}{n+1} < \varepsilon \iff n + 1 > \frac{1}{\varepsilon} \iff n > \frac{1}{\varepsilon} - 1.

Choose the cutoff and verify. Take N = \left\lfloor \tfrac{1}{\varepsilon} \right\rfloor + 1; then N > \tfrac{1}{\varepsilon} > \tfrac{1}{\varepsilon} - 1, so for all n \ge N:

n > \frac{1}{\varepsilon} - 1 \implies n + 1 > \frac{1}{\varepsilon} \implies \left| \frac{n}{n+1} - 1 \right| = \frac{1}{n+1} < \varepsilon. \qquad \blacksquare

Play a round: \varepsilon = 0.01 demands n > 99, so the smallest winning cutoff is N = 100 — indeed a_{100} = \tfrac{100}{101}, which misses 1 by \tfrac{1}{101} < 0.01. (Our formula's N = 101 also wins, just less thriftily: any cutoff at or past the smallest one is a valid answer. The definition asks for an N, not the best one.)

Worked example 3: banking wins with the limit laws

Nobody runs the duel by hand for every sequence. Once a stock of limits is proven, the limit laws carry over verbatim to sequences: if a_n \to A and b_n \to B, then

a_n + b_n \to A + B, \qquad a_n b_n \to AB, \qquad \frac{a_n}{b_n} \to \frac{A}{B}\ \ (B \ne 0).

Armed with just 1/n \to 0 and the laws, whole families of limits come free: 3 + \tfrac{5}{n} \to 3, \tfrac{1}{n^2} = \tfrac1n \cdot \tfrac1n \to 0, and so on — no fresh \varepsilon-juggling required.

The classic application is a rational-function sequence. What is

\lim_{n \to \infty} \frac{3n^2 + 5n}{2n^2 - 7}\ ?

Top and bottom both blow up, so the quotient law does not apply directly (\tfrac{\infty}{\infty} is not a number). The trick: divide top and bottom by the highest power of n — here n^2:

\frac{3n^2 + 5n}{2n^2 - 7} = \frac{3 + \dfrac{5}{n}}{2 - \dfrac{7}{n^2}}.

Now every piece converges: \tfrac5n \to 0 and \tfrac{7}{n^2} \to 0, so the numerator tends to 3, the denominator to 2 \ne 0, and the quotient law finishes the job:

\lim_{n \to \infty} \frac{3n^2 + 5n}{2n^2 - 7} = \frac{3 + 0}{2 - 0} = \frac{3}{2}.

The general moral: when numerator and denominator have equal degree, the limit is the ratio of leading coefficients; when the denominator's degree is higher, the limit is 0; when the numerator's is higher, the sequence diverges to \pm\infty.

The other workhorse is the squeeze theorem: if b_n \le a_n \le c_n for all large n, and both flanks converge to the same limit b_n, c_n \to L, then the trapped sequence has no choice but

a_n \to L.

For example -\tfrac1n \le \tfrac{\sin n}{n} \le \tfrac1n, and both edges go to 0, so \tfrac{\sin n}{n} \to 0 despite the wobbling numerator — a limit you would struggle to prove head-on, since \sin n never settles down. The squeeze converts "I can't compute it" into "I don't need to".

Divergence: when no answer exists

The definition also tells us when a limit fails — compare two sequences side by side. The point is that a limit can fail to exist even when the terms stay perfectly bounded.

Convergent. a_n = 1/n settles into every band, as we proved: whatever \varepsilon the opponent names, we have an answer.

Divergent by oscillation. a_n = (-1)^n reads -1, 1, -1, 1, \dots forever. No single L can work: pick the band \varepsilon = \tfrac12. To capture the +1 terms L would need |1 - L| < \tfrac12, and to capture the -1 terms it would need |{-1} - L| < \tfrac12 — but those two intervals are disjoint, so no L sits in both. Past any cutoff N there are still +1's and -1's, so the band is never entered for good. Hence (-1)^n diverges, though it stays bounded. Here the opponent wins the game — one well-chosen \varepsilon that you can never answer is all it takes.

Divergent to infinity. a_n = n escapes every band around every candidate L; we write a_n \to \infty as useful shorthand, but strictly the sequence diverges — \infty is not a number and no N ever pins the terms into a finite band.

Newton and Leibniz built calculus in the 1660s–80s on "infinitely small" quantities that were nonzero when you divided by them and zero when you discarded them. It worked — planets obeyed it — but the logic was mush, and in 1734 the philosopher-bishop George Berkeley skewered it in a pamphlet, The Analyst: these vanished increments, he jeered, were "neither finite quantities, nor quantities infinitely small, nor yet nothing. May we not call them the ghosts of departed quantities?" The mathematicians had no comeback. For over a century calculus ran on brilliant instinct and shaky foundations.

The cleanup came in two waves. In the 1820s Augustin-Louis Cauchy re-founded analysis on limits, defining a limit as a value that a variable "approaches indefinitely" — nearly rigorous, still worded in motion. Then in the 1860s Karl Weierstrass, lecturing in Berlin, replaced the motion with pure logic: the static \varepsilon–N (and \varepsilon–\delta) inequalities you have just learned. No flowing, no vanishing, no ghosts — only "for every tolerance there is a cutoff". Berkeley's mockery was finally, definitively answered: about 150 years late, and worth the wait. The definition on this page is the exact one Weierstrass's students copied down.

Watch the band close in

Here is the whole game as a picture. The dots are a_n = 1/n, the dashed line is the limit L = 0, and the shaded strip is the opponent's challenge: a band of half-width \varepsilon around L. Your answer is the vertical line at N = \lfloor 1/\varepsilon\rfloor + 1.

Now play the opponent: drag \varepsilon smaller and watch the band tighten. The cutoff slides right — dots before it (in the warning colour) may still poke outside the band, and that is allowed: the definition tolerates any finite amount of early misbehaviour. But every dot from N onward is captured, and stays captured. However far you shrink the band, the cutoff always catches up. That sliding line is Step 4 of the proof, drawn.