Prove a Basic Epsilon-N Limit

Turn the Definition into a Proof Plan

Proof Strategy Mastery emphasized fixing the exact target and identifying the bridge between the assumptions and the conclusion. For a basic epsilon-\(N\) limit proof, that bridge is an estimate: we bound the distance between the sequence term and its proposed limit, then choose an integer threshold that makes the bound smaller than a prescribed tolerance.

Let \((a_n)\) be a real sequence and let \(L\in\mathbb R\). The statement \(a_n\to L\) says that, no matter how small a positive tolerance is requested, every sufficiently late term lies within that tolerance of \(L\). The threshold may depend on the tolerance, but it must work for every index from that threshold onward.

The quantifiers determine the proof’s order. First, take an arbitrary \(\varepsilon>0\). Next, choose \(N\), possibly using \(\varepsilon\). Finally, prove the required inequality for an arbitrary integer \(n\geq N\). A proof that selects one \(N\) before knowing \(\varepsilon\), or checks only a single term, does not establish the definition.

An Error-Bound Theorem

Often the difficult algebra is not the choice of \(N\), but obtaining a useful estimate for the error. The following theorem isolates the common pattern. It also allows the estimate to begin only after some initial index, a detail that should not be hidden in a convergence proof.

Proof. Let \(\varepsilon>0\) be arbitrary. If \(C=0\), the assumed inequality gives \(|a_n-L|\leq0\) for every \(n\geq n_0\). Since an absolute value is nonnegative, \(|a_n-L|=0<\varepsilon\). Thus \(N=n_0\) works.

Suppose instead that \(C>0\). Choose

This is a positive integer, and \(N> C/\varepsilon\). Take any integer \(n\geq N\). Then \(n\geq n_0\), so the assumed error estimate applies. Also \(n\geq N> C/\varepsilon\), which implies \(C/n<\varepsilon\). Consequently,

This verifies the definition for the arbitrary \(\varepsilon>0\). Therefore \(a_n\to L\). \(\square\)

The floor in the choice of \(N\) is a convenient way to guarantee an integer threshold strictly greater than \(C/\varepsilon\). Any positive integer with that property and with \(N\geq n_0\) would also work. The theorem does not say that this choice is the smallest possible threshold; convergence requires a suitable threshold, not an optimal one.

Use a Bound That Matches the Target

The estimate in the theorem is a reusable technique: once the error is bounded by \(C/n\), convergence follows with an explicit threshold. The estimate itself must still be justified. In other problems, rationalizing a difference or using the triangle inequality may reveal a bound that tends to zero. The next example shows how to control a square-root expression without relying on a limit law.

This proof has two distinct jobs. Rationalization creates an error bound, and the threshold converts that bound into the exact strict inequality in the definition. Choosing \(N\) before finding an estimate would leave no reason to believe that all later terms are close enough.

When an Estimate Holds Only Eventually

A formula or inequality may be convenient only for indices beyond some starting point. That is not an obstacle: the definition concerns all \(n\geq N\), and \(N\) can be chosen beyond the starting point of the estimate. More generally, changing finitely many terms does not affect whether a sequence converges. This principle makes it legitimate to focus on a tail, provided the proof says where that tail begins.

Proof. First suppose \(a_n\to L\). Let \(\varepsilon>0\). By convergence, there is a positive integer \(N_a\) such that \(|a_n-L|<\varepsilon\) for every \(n\geq N_a\). Set \(N=\max\{N_a,m\}\). For any \(n\geq N\), we have both \(n\geq N_a\) and \(n\geq m\). Thus \(|a_n-L|<\varepsilon\), and \(b_n=a_n\). Hence \(|b_n-L|<\varepsilon\). Since this holds for every \(\varepsilon>0\), \(b_n\to L\).

Conversely, suppose \(b_n\to L\). Given \(\varepsilon>0\), choose \(N_b\) such that \(|b_n-L|<\varepsilon\) whenever \(n\geq N_b\). Set \(N=\max\{N_b,m\}\). For every \(n\geq N\), \(a_n=b_n\), so \(|a_n-L|=|b_n-L|<\varepsilon\). Therefore \(a_n\to L\). The two implications show that either sequence converges exactly when the other does, and the argument gives the same limit \(L\) in both directions. \(\square\)

This theorem also explains why an estimate that holds for \(n\geq n_0\) is enough. The finitely many terms before \(n_0\) need not satisfy that estimate: choose the convergence threshold at or beyond \(n_0\). The threshold is allowed to skip an initial segment, but not to skip any terms after it.

Common Gaps in Epsilon-\(N\) Proofs

A complete proof should make the threshold and the strict inequality visible. Several tempting shortcuts fail precisely at these points:

• Using a real cutoff without choosing an integer. A threshold in the definition is a positive integer. If an estimate suggests \(n>R\), specify an integer such as \(\lfloor R\rfloor+1\).
• Proving only that the error is at most \(\varepsilon\). The definition asks for \(|a_n-L|<\varepsilon\). Arrange a strict bound, often by choosing \(N> C/\varepsilon\), rather than merely \(N=C/\varepsilon\).
• Forgetting the quantifiers. The proof must work for every positive \(\varepsilon\), and then for every \(n\geq N\). Testing a few terms is not a substitute.
• Ignoring where an estimate applies. If an inequality is valid only for \(n\geq n_0\), make sure the chosen \(N\) is at least \(n_0\).
• Choosing a threshold before understanding the error. A plausible-looking \(N\) is not enough; the proof must show how \(n\geq N\) forces the required error bound.

The essential pattern is concise: fix \(\varepsilon>0\), estimate \(|a_n-L|\), choose a positive integer \(N\) that makes the estimate strictly smaller than \(\varepsilon\), and verify the claim for arbitrary \(n\geq N\). In the next tutorial, the same care with quantifiers will support a proof that a sequence cannot have two different limits.