Mathematics · Glossary

What is Limit of a sequence?

Definition 11.1 University Mathematics — Year 1 · Chapter 11 — Sequences

A sequence (un)(u_n) of reals converges to R\ell \in \R when

ε>0, NN, nN,unε.\forall \varepsilon > 0,\ \exists N \in \N,\ \forall n \geq N, \qquad \abs{u_n - \ell} \leq \varepsilon .

One writes unu_n \to \ell or limun=\lim u_n = \ell. A sequence that does not converge (to any real) diverges. Divergence to ++\infty: M, N, nN, unM\forall M,\ \exists N,\ \forall n \geq N,\ u_n \geq M (similarly -\infty).

Examples

Example 11.2 (An ε\varepsilonNN proof, written out once)

Claim: un=n2+12n2312u_n = \dfrac{n^2 + 1}{2n^2 - 3} \to \dfrac12. First isolate the error:

un12=2(n2+1)(2n23)2(2n23)=522n23=52(2n23)(n2).\Bigl| u_n - \frac12 \Bigr| = \Bigl| \frac{2(n^2 + 1) - (2n^2 - 3)}{2(2n^2 - 3)} \Bigr| = \frac{5}{2\,\abs{2n^2 - 3}} = \frac{5}{2\,(2n^2 - 3)} \quad (n \geq 2).

Then dominate it by something simple: for n2n \geq 2, 2n23n22n^2 - 3 \geq n^2, so the error is 52n252n\leq \frac{5}{2n^2} \leq \frac 5{2n}. Given ε>0\varepsilon > 0, the Archimedean property provides Nmax(2,52ε)N \geq \max\bigl(2, \frac{5}{2\varepsilon}\bigr); for nNn \geq N the error is ε\leq \varepsilon. Done. The closing insight: an ε\varepsilonNN proof has exactly three moves — compute the error, bound it by a decreasing elementary expression, solve for the threshold — and after this chapter’s theorems (operations, squeeze) one almost never writes such a proof again: the theorems package the three moves once and for all.

Example 11.3 (Divergence to infinity, certified)

Claim: un=n2100n+u_n = n^2 - 100n \to +\infty. Factor the dominant term: un=n2(1100n)n22u_n = n^2\bigl(1 - \frac{100}{n}\bigr) \geq \frac{n^2}{2} for n200n \geq 200. Given MM, take N=max(200,2M)N = \max\bigl(200, \lceil\sqrt{2M}\rceil\bigr): for nNn \geq N, unn22Mu_n \geq \frac{n^2}{2} \geq M. Two habits are on display: dominant-term factoring converts a competition (n2n^2 against 100n-100n) into a single scale times a factor tending to 11; and the threshold may be huge (u100=0u_{100} = 0, the sequence is even negative before n=100n = 100) — divergence to ++\infty is a statement about the tail, indifferent to any finite amount of misbehavior.

Example 11.6 (Operations plus one algebraic trick)

Compute lim(n2+nn)\lim\,\bigl(\sqrt{n^2 + n} - n\bigr). The two pieces separately tend to ++\infty: the operations theorem says nothing about their difference (an indeterminate form). Multiply by the conjugate:

n2+nn=(n2+n)n2n2+n+n=nn2+n+n=11+1n+1.\sqrt{n^2 + n} - n = \frac{(n^2 + n) - n^2}{\sqrt{n^2 + n} + n} = \frac{n}{\sqrt{n^2+n} + n} = \frac{1}{\sqrt{1 + \frac1n} + 1} .

Now everything converges: 1+1n1\sqrt{1 + \frac1n} \to 1, because 01+h1=h1+h+1h0 \leq \sqrt{1 + h} - 1 = \frac{h}{\sqrt{1+h} + 1} \leq h (conjugate again, then squeeze with h=1nh = \frac1n); then the operations theorem gives the limit 11+1=12\frac{1}{1 + 1} = \frac12. The closing insight: the operations theorem is not a calculator for all limits — indeterminate forms (\infty - \infty, 00\frac00, 0×0 \times \infty, 11^\infty) must first be transformed by algebra (conjugates, factoring the dominant term) until every piece converges; the systematic machine for resistant cases is the asymptotic expansion of Chapter 16.

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