From a Pattern to a Precise Claim
A table or graph can suggest that a sequence approaches a number, but it cannot by itself guarantee what happens to every term far out in the sequence. The epsilon-N definition makes “approaches” precise: no matter how small a positive distance is required, all terms after some index must lie within that distance of the proposed limit.
Write a sequence as \((a_n)\), where \(n\) is a positive integer, and let \(L\) be a real number. The definition describes what it means for the terms \(a_n\) to approach \(L\) as \(n\) increases without bound.
The absolute value \(|a_n-L|\) is the distance between the term \(a_n\) and the proposed limit \(L\). The condition requires that this distance be less than \(\varepsilon\) for every index \(n\) from \(N\) onward, not merely for selected terms.
What the Quantifiers Require
The order of the statements in the definition matters. First, someone may choose any positive tolerance \(\varepsilon\), however small. Then there must be an index \(N\) that works for that tolerance. Once \(N\) is chosen, every index \(n\ge N\) must satisfy the distance condition.
- \(\varepsilon\) is the allowed distance from the limit. It can be any positive real number.
- \(N\) is a positive-integer threshold. It may depend on \(\varepsilon\).
- \(n\ge N\) means the condition must hold for all terms from \(a_N\) onward.
- \(|a_n-L|<\varepsilon\) is the specific requirement each of those terms must meet.
There is no requirement that \(N\) be the smallest possible threshold. Any positive integer that makes the implication true is sufficient. In practice, the task is to use the expression for \(|a_n-L|\) to find a condition on \(n\) that guarantees it is less than \(\varepsilon\).
A Reliable Threshold-Finding Method
When the distance from the proposed limit has a simple formula, the work often follows four steps. First calculate \(|a_n-L|\). Next require that this expression be less than \(\varepsilon\). Then solve for a condition on \(n\) and choose a positive integer threshold that guarantees that condition. Finally, start with \(n\ge N\) and verify the original distance inequality.
Find \(|a_n-L|\) using the proposed limit \(L\).
Determine a condition on \(n\) that ensures \(|a_n-L|<\varepsilon\).
Pick a positive integer \(N\) that guarantees the condition whenever \(n\ge N\).
Use \(n\ge N\) to prove the required strict inequality for every later term.
Worked Examples
Worked Example: Proving that \(1/n\) tends to zero
We will prove that \(\lim_{n\to\infty}1/n=0\) directly from the definition. Let \(\varepsilon>0\) be arbitrary. The distance from \(1/n\) to \(0\) is
because \(n\) is a positive integer. We need \(1/n<\varepsilon\), which is equivalent, for positive \(n\) and \(\varepsilon\), to \(n>1/\varepsilon\). Choose
Here \(\lfloor x\rfloor\) is the greatest integer less than or equal to \(x\). Thus \(N\) is a positive integer and \(N>1/\varepsilon\). If \(n\ge N\), then \(n\ge N>1/\varepsilon\), so taking reciprocals of positive numbers gives \(1/n<\varepsilon\). Therefore \(|1/n-0|<\varepsilon\) for every \(n\ge N\). Since this works for every \(\varepsilon>0\), the definition proves that the limit is \(0\).
Check with a particular tolerance: If \(\varepsilon=0.1\), then \(N=\lfloor 10\rfloor+1=11\). For \(n\ge11\), \(1/n\le1/11<0.1\), as required.
Worked Example: Proving a rational sequence tends to three
Consider \(a_n=(3n-1)/(n+2)\). We will prove that its limit is \(3\). First find the distance:
The denominator is positive because \(n\ge1\). To make the distance less than \(\varepsilon\), it is enough to require \(n+2>7/\varepsilon\). Choose the positive integer
Since \(N>7/\varepsilon\), every \(n\ge N\) satisfies \(n+2\ge N+2>7/\varepsilon\). Hence \(7/(n+2)<\varepsilon\), and therefore \(|a_n-3|<\varepsilon\). This proves from the definition that \(\lim_{n\to\infty}(3n-1)/(n+2)=3\).
Check with a particular tolerance: If \(\varepsilon=0.2\), then \(7/\varepsilon=35\) and \(N=36\). At every \(n\ge36\), the distance is \(7/(n+2)\le7/38<0.2\). The chosen threshold works, though it need not be the smallest one.
Worked Example: Proving that \(n/(n+1)\) tends to one
Consider \(a_n=n/(n+1)\). Subtract the proposed limit and simplify before choosing a threshold:
For any \(\varepsilon>0\), choose \(N=\lfloor1/\varepsilon\rfloor+1\). Then \(N>1/\varepsilon\), so if \(n\ge N\), we have \(n+1\ge N+1>1/\varepsilon\). Taking reciprocals gives \(1/(n+1)<\varepsilon\). Thus \(|a_n-1|<\varepsilon\) for every \(n\ge N\), proving that the limit is \(1\).
Check with a particular tolerance: If \(\varepsilon=0.01\), then \(N=101\). For \(n\ge101\), the distance is \(1/(n+1)\le1/102<0.01\). The definition asks for a working threshold, not an optimal one.
Why the Method Works
The proof strategy follows directly from the definition, rather than from a separate approximation rule. For example, in the first worked example, the definition asks for a positive integer \(N\) such that \(n\ge N\) forces \(1/n<\varepsilon\). Solving \(1/n<\varepsilon\) for \(n\) gives \(n>1/\varepsilon\). Choosing an integer strictly greater than \(1/\varepsilon\) ensures that every larger index also meets the inequality. The threshold therefore turns an informal claim about terms becoming close into the exact “for every tolerance, there is a threshold” statement.
More generally, suppose algebra shows that a distance can be written as \(C/(n+d)\), where \(C>0\), and the denominator is positive for the indices under consideration. Requiring \(C/(n+d)<\varepsilon\) is equivalent to \(n+d>C/\varepsilon\). A sufficiently large integer \(N\) can enforce that inequality for all \(n\ge N\). The crucial part is not merely finding an index where the inequality holds; it is ensuring it continues to hold for every later index.
When a Proposed Limit Fails
The definition also tells us how to refute a proposed limit. Negating its quantifiers gives a precise failure condition: there is some positive tolerance \(\varepsilon_0\) such that, no matter which threshold \(N\) is chosen, at least one index \(n\ge N\) has \(|a_n-L|\ge\varepsilon_0\). In other words, terms that violate the chosen distance requirement keep appearing arbitrarily far out in the sequence.
This is useful when a proposed limit cannot be verified. A single early term outside the tolerance does not refute the limit, because the threshold could be chosen later. To use the negation, the failure must occur beyond every possible threshold.
Common Pitfall: Choosing an Index That Does Not Guarantee the Inequality
Suppose we want to prove \(1/n\to0\) for \(\varepsilon=0.1\). A tempting choice is \(N=10\), because \(1/10=0.1\). But the definition requires a strict inequality: \(|1/n|<0.1\). At \(n=10\), the distance equals \(0.1\), so \(N=10\) does not work.
The correction is to choose \(N=11\). Then for every \(n\ge11\), \(1/n\le1/11<0.1\). The threshold has to guarantee the strict inequality starting at \(n=N\), not merely at indices larger than \(N\).
A related error is to check only a few terms and conclude that all later ones work. Checking terms can help test a proposed threshold, but the proof must show why the inequality holds for every \(n\ge N\). The algebraic bound supplies that guarantee.
Why the Definition Matters
The epsilon-N definition separates a convincing visual pattern from a mathematical proof. A graph can make a sequence appear close to a value, while the definition requires a guarantee that applies to every term after a threshold. It also clarifies exactly what a successful proof must provide: for an arbitrary positive tolerance, give an index threshold and verify the distance condition for all later indices.
When proving a sequence limit, keep the logical structure visible: let \(\varepsilon>0\) be arbitrary, choose a positive integer \(N\), and show that every \(n\ge N\) satisfies \(|a_n-L|<\varepsilon\). When challenging a proposed limit, identify one fixed positive tolerance for which no threshold can work. These two approaches are the definition in action.
Check Your Understanding
For each question, focus on what the definition requires of every index after the threshold.
- In the epsilon-N definition, what does \(\varepsilon\) represent, and what does \(N\) represent?
- For \(a_n=2/n\), find a positive-integer threshold \(N\) in terms of \(\varepsilon\) that guarantees \(|a_n-0|<\varepsilon\) for every \(n\ge N\).
- For \(a_n=(n+2)/(n+1)\), calculate \(|a_n-1|\) and use it to prove the limit is \(1\).
- Why does choosing \(N=10\) fail to verify \(|1/n-0|<0.1\) for every \(n\ge N\)?
- State the quantifier pattern in the negation of the definition of \(\lim_{n\to\infty}a_n=L\).
