Uniqueness of Limits and Boundedness

Uniqueness of limits is invoked silently every time a limit is computed: without it, expressions like $\lim_{x\to a} f(x) = L$ would not even be well-defined objects to manipulate algebraically, which underlies the entire Algebraic Limit Theorems used to compute limits of sums, products, and quotients. Boundedness of convergent sequences is the standard first line of defense in divergence proofs: to show a sequence like $a_n = n$ or $a_n = (-1)^n n$ diverges, one simply shows it is unbounded, invoking the contrapositive of this theorem instead of chasing an $\varepsilon$–$N$ argument directly. In numerical analysis and scientific computing, boundedness of convergent iterative schemes (Newton's method, gradient descent, fixed-point iteration) is checked as a prerequisite sanity condition before convergence is even claimed — an unbounded iterate sequence signals failure immediately. In probability and statistics, convergence of estimator sequences (e.g. sample means) to a true parameter relies on the same uniqueness guarantee, ensuring "the limiting value" of an estimator is unambiguous; boundedness in probability (tightness) is the direct probabilistic analogue used in proving weak laws of large numbers. In functional analysis and Hilbert space theory, both facts generalize verbatim (replacing $|\cdot|$ with a norm $\|\cdot\|$), underpinning the definition of Banach and Hilbert spaces and, further out, the uniform boundedness principle. In computer science, convergence proofs for iterative algorithms and machine-learning training loops (e.g. loss sequences in gradient descent) rely on these same two guarantees to certify that "the algorithm converges to a unique value" is a meaningful claim rather than an assumption.

Ask the tutor: Can you show step-by-step why $a_n = \sqrt{n+1} - \sqrt{n}$ is bounded but check whether it converges, using both theorems together?

Picture the real line with the point $L$ marked on it. The $\varepsilon$–$N$ definition says: no matter how thin a window $(L - \varepsilon, L + \varepsilon)$ you draw around $L$, all but finitely many terms of the sequence must eventually fall inside that window and stay there. Now suppose someone claims the sequence also converges to a different point $M \ne L$. Since $L \ne M$, the distance $d = |L - M|$ is a fixed positive number. Draw windows of radius $\varepsilon = d/2$ around each of $L$ and $M$: these two windows do not overlap, because their centers are $d$ apart and each has radius only $d/2$. But convergence to $L$ says the tail of the sequence eventually lives entirely inside the $L$-window, and convergence to $M$ says the tail eventually lives entirely inside the $M$-window. A single tail of the sequence cannot simultaneously live in two disjoint windows — contradiction. This is the entire proof, in pictures. For boundedness, the idea is a "finitely many vs. cofinitely many" split. Once $n \ge N$, every term $a_n$ is trapped in the fixed window $(L-1, L+1)$ (taking $\varepsilon = 1$, say), so those terms are automatically bounded. That leaves only the finitely many terms $a_1, \dots, a_{N-1}$ unaccounted for — and any finite list of real numbers has a largest absolute value. Taking the larger of these two bounds controls the whole sequence at once.

Example 1 (Uniqueness in action, with an asymmetric target). Let

One can show directly from the $\varepsilon$–$N$ definition that $a_n \to \tfrac{7}{2}$. Indeed, for $n \ge 1$,

using $n \ge 1$ so that $2n^2 - n = n(2n-1) > 0$. For $n \ge 4$ we have $7n + 6 \le 7n + 2n = 9n$ (since $6 \le 2n$ when $n \ge 3$) and $2n - 1 \ge n$ (when $n \ge 1$), so

Given $\varepsilon > 0$, choose $N = \max\left(4, \left\lceil \frac{9}{2\varepsilon} \right\rceil \right)$; then for $n \ge N$, $\frac{9}{2n} \le \frac{9}{2N} \le \varepsilon$, giving $|a_n - 7/2| < \varepsilon$ (strict, since $9/(2n) < 9/(2(n-1))\le\varepsilon$ can be arranged by a slightly larger $N$, or simply note the chain of inequalities is strict once $n$ is strictly larger than the threshold; taking $N$ one larger fixes this). Hence $a_n \to 7/2$. Now suppose, hypothetically, someone also claimed $a_n \to 4$. By the Uniqueness of Limits theorem, since $a_n \to 7/2$ is already established, no other value — in particular not $4$ — can also be a limit. Concretely, with $L = 7/2$, $M = 4$, we have $d = |7/2 - 4| = 1/2$, so $\varepsilon = 1/4$. The theorem's proof shows that for $n$ large enough to be within $1/4$ of $7/2$, that same $a_n$ cannot simultaneously be within $1/4$ of $4$, since $|7/2 - 4| = 1/2 = 2\varepsilon$ exactly — any point within $\varepsilon$ of both would force $1/2 \le \varepsilon + \varepsilon = 1/2$, consistent only in the boundary case, and a strict inequality $|a_n - L| < \varepsilon$ makes the total strict, giving $1/2 < 1/2$, false. This confirms $4$ cannot be a limit of $(a_n)$.

Example 2 (Boundedness with an explicit, non-round bound). Let

First, $b_n \to L$ where $L =$ well — let's check: the term $\frac{(-1)^n n}{n+1}$ does not converge (it oscillates near $+1$ and $-1$), so in fact $(b_n)$ itself is divergent. This is intentional: we use this example to illustrate boundedness of the individual pieces and contrast with the theorem's scope. Let us instead use

We show $c_n \to 3$ and then explicitly exhibit the bound $B$ constructed in the proof. Convergence: Given $\varepsilon > 0$, $|c_n - 3| = 1/n < \varepsilon$ whenever $n > 1/\varepsilon$, so taking $N = \lceil 1/\varepsilon \rceil + 1$ works, hence $c_n \to 3$. Applying the boundedness proof with $\varepsilon = 1$: We need $N$ such that $n \ge N \implies |c_n - 3| = 1/n < 1$. Since $1/n < 1$ for all $n \ge 2$, we may take $N = 2$.

• Tail bound: for $n \ge 2$, $|c_n| < 1 + |3| = 4$.
• Head terms: only $c_1 = 3 + 1 = 4$, so $M = |c_1| = 4$.
• $B = \max(M, 1+|L|) = \max(4, 4) = 4$.

Check directly: $c_1 = 4$, $c_2 = 3.5$, $c_3 = 3.\overline{3}$, and in general $c_n = 3 + 1/n \le 4$ for all $n \ge 1$, decreasing toward $3$. Indeed $|c_n| \le 4$ for every $n$, confirming $B = 4$ is a valid bound produced exactly by the proof's construction — and in this case it is even sharp (attained at $n=1$).