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Limit of a sequence

Limits of sequences lie at the foundation of mathematical analysis: they make the intuitive notion of "getting arbitrarily close" rigorous and underpin almost every later construction, from continuity and derivatives to integrals, infinite series, and convergence in more general spaces such as metric and normed spaces.
Geometric meaning of the ε\varepsilon--NN definition: for every horizontal band of half-width ε\varepsilon around bb, there is an index NN past which every term xnx_n stays inside the band.
LIMIT OF A SEQUENCE
Let {xn}n=1∞\{x_n\}_{n=1}^{\infty} be a sequence of real numbers. We say that b∈Rb \in \mathbb{R} is the limit of the sequence and write
lim⁡n→∞xn=b\begin{equation*}\lim_{n \to \infty} x_n = b\end{equation*}
if and only if
∀ε>0,  ∃N∈N,  ∀n>N:∣xn−b∣<ε.\begin{equation*}\forall \varepsilon > 0, \; \exists N \in \mathbb{N}, \; \forall n > N : |x_n - b| < \varepsilon .\end{equation*}
A sequence that admits such a limit is called convergent.
The modern ε\varepsilon-NN formulation was developed in the nineteenth century by Bolzano, Cauchy and Weierstrass. It replaces vague phrases like "tends to" or "becomes infinitely close" with a precise quantifier statement: no matter how small a tolerance ε\varepsilon we choose, eventually all terms of the sequence stay within that tolerance of the limit. This formulation is essential not only in real analysis but also in numerical analysis, probability theory and many branches of applied mathematics, where one constantly needs to control how fast an approximation approaches its target value.

The uniqueness theorem follows Abbott2015, Theorem 2.2.7.

Proof of uniqueness of the limit using the triangle inequality.
Proof of uniqueness of the limit using the triangle inequality.
UNIQUENESS OF LIMITS
If a sequence {xn}n=1∞\{x_n\}_{n=1}^{\infty} has a limit, then this limit is unique.
\quad Proof. Suppose that xn→L1x_n \to L_1 and xn→L2x_n \to L_2. Let ε>0\varepsilon>0 be arbitrary.
Since xn→L1x_n \to L_1, there exists N1∈NN_1 \in \mathbb{N} such that
n>N1  ⟹  ∣xn−L1∣<ε2.(1)\begin{equation*}n > N_1 \implies |x_n - L_1| < \frac{\varepsilon}{2}. \qquad (1)\end{equation*}
Since xn→L2x_n \to L_2, there exists N2∈NN_2 \in \mathbb{N} such that
n>N2  ⟹  ∣xn−L2∣<ε2.(2)\begin{equation*}n > N_2 \implies |x_n - L_2| < \frac{\varepsilon}{2}. \qquad (2)\end{equation*}
Choose n>max⁡(N1,N2)n > \max(N_1,N_2). Then both inequalities hold at the same time. By the triangle inequality,
∣L1−L2∣≤Triangle inequality∣L1−xn∣+∣xn−L2∣<(1), (2)ε2+ε2=ε.\begin{equation*}|L_1 - L_2|\overset{\text{Triangle inequality}}{\leq}|L_1 - x_n| + |x_n - L_2|\overset{\text{(1), (2)}}{<}\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.\end{equation*}
Since ε>0\varepsilon>0 was arbitrary, this forces ∣L1−L2∣=0|L_1-L_2|=0. Therefore L1=L2L_1=L_2, and the limit is unique. □\Box

The boundedness theorem follows Abbott2015, Theorem 2.3.2.

All terms of a convergent sequence lie below the level M=max⁡(∣x1∣,…,∣L+1∣)M = \max(|x_1|, \dots, |L+1|).
BOUNDEDNESS OF CONVERGENT SEQUENCES
If a sequence {xn}n=1∞\{x_n\}_{n=1}^{\infty} converges to a finite limit L∈RL \in \mathbb{R}, then it is bounded; that is, there exists a constant M>0M > 0 such that
∣xn∣≤Mfor all n∈N.\begin{equation*}|x_n| \leq M \qquad \text{for all } n \in \mathbb{N}.\end{equation*}
\quad Proof. Apply the definition of the limit with ε=1\varepsilon = 1: there exists N∈NN \in \mathbb{N} such that
∀n>N  ⟹  ∣xn−L∣<1.\begin{equation*}\forall n > N \implies |x_n - L| < 1.\end{equation*}
By the triangle inequality, for every n>Nn > N
∣xn∣=∣xn−L+L∣≤∣xn−L∣+∣L∣<1+∣L∣=∣L+1∣.\begin{equation*}|x_n| = |x_n - L + L| \leq |x_n - L| + |L| < 1 + |L| = |L + 1|.\end{equation*}
Hence every term of the sequence with index n>Nn > N lies in the band of half-width 11 around LL, and in particular ∣xn∣<∣L+1∣|x_n| < |L + 1|.
It remains to control the finite list of initial terms x1,x2,…,xNx_1, x_2, \dots, x_N. Set
M=max⁡(∣x1∣,∣x2∣,…,∣xN∣,∣L+1∣).\begin{equation*}M = \max\big(|x_1|, |x_2|, \dots, |x_N|, |L + 1|\big).\end{equation*}
Then for every n∈Nn \in \mathbb{N} we have ∣xn∣≤M|x_n| \leq M: if n≤Nn \leq N this holds because ∣xn∣|x_n| is one of the numbers in the maximum, and if n>Nn > N this holds because ∣xn∣<∣L+1∣≤M|x_n| < |L + 1| \leq M. Therefore the sequence is bounded. □\Box

References

  • [Abbott2015]Abbott, Stephen. Understanding Analysis. Second edition. Springer, 2015.