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The formal definition of a limit

Infinity and strange math · level 8, lesson 1 of 7

A question to start

In level 5 you met limits in words: a sequence has limit \(L\) if its terms "get as close as you like to \(L\) and stay there". Everyone agrees that \(1, \tfrac12, \tfrac13, \tfrac14, \ldots\) has limit 0.

But "as close as you like" is slippery. Close enough for whom? And how could you ever prove that a sequence gets close to 0 forever, when you can only check finitely many terms?

Here is a sharper version of the question. The sequence \(0.9, 0.99, 0.999, \ldots\) gets closer and closer to \(2\): every term is nearer to 2 than the one before. Is 2 its limit? Of course not, the limit is 1. So "getting closer and closer" is not the right definition. What is?

It took mathematicians well over a century after the invention of calculus to find the answer: a two-line definition, polished in the mid-1800s by Karl Weierstrass and his students, building on Augustin-Louis Cauchy and Bernard Bolzano. It is built from two Greek letters, \(\varepsilon\) (epsilon) and \(\delta\) (delta), and it turned the infinite into something that can be handled with finite, checkable steps.

The idea

The limit of a sequence as a challenge game

Imagine a game between a challenger and a prover, about a sequence \(a_1, a_2, a_3, \ldots\) and a number \(L\).

  1. The challenger picks a tolerance \(\varepsilon > 0\), as small as they like: \(0.1\), or \(0.000001\).
  2. The prover must answer with a position \(N\) such that every term after position \(N\) is within \(\varepsilon\) of \(L\).

If the prover can always answer, whatever \(\varepsilon\) the challenger chooses, then \(L\) is the limit. That is the whole definition, written in symbols:

\[\lim_{n \to \infty} a_n = L \quad\text{means}\quad \text{for every } \varepsilon > 0 \text{ there is an } N \text{ such that } |a_n - L| < \varepsilon \text{ for all } n > N.\]

Read \(|a_n - L|\) as "the distance between \(a_n\) and \(L\)". Each part does a job:

  • "for every \(\varepsilon > 0\)" is "as close as you like": every tolerance, not just some;
  • "there is an \(N\)" is "eventually": we may ignore any finite number of early terms;
  • "for all \(n > N\)" is "and stays there": not just one term inside, but all of them from then on.

In a picture: draw a band from \(L - \varepsilon\) to \(L + \varepsilon\). Whatever the width of the band, from some point \(N\) on, all the terms are inside it.

L L+ε L−ε n N

A smaller \(\varepsilon\) (a thinner band) usually needs a larger \(N\). That is fine: \(N\) is allowed to depend on \(\varepsilon\).

Now the sequence \(0.9, 0.99, 0.999, \ldots\) and the number 2: challenge with \(\varepsilon = 0.5\). Every term is less than 1, so its distance from 2 is more than 1, bigger than \(0.5\). No \(N\) works, so 2 is not the limit. Getting closer is not enough; you must get within every tolerance.

How a proof goes: work backwards, then write forwards

To prove \(\lim a_n = L\) you must give a recipe for \(N\) that works for every \(\varepsilon\). The usual method:

  1. Scratch work. Write \(|a_n - L|\), simplify it, and ask: how large must \(n\) be to make this less than \(\varepsilon\)? Solve for \(n\).
  2. The proof. Let \(\varepsilon > 0\). Choose \(N\) (the answer from the scratch work). Then show that every \(n > N\) gives \(|a_n - L| < \varepsilon\).

For \(a_n = \tfrac1n\) and \(L = 0\): \(|a_n - 0| = \tfrac1n\), and \(\tfrac1n < \varepsilon\) exactly when \(n > \tfrac1\varepsilon\). So the proof is: given \(\varepsilon > 0\), let \(N = \tfrac1\varepsilon\). For every \(n > N\), \(\tfrac1n < \tfrac1N = \varepsilon\). ∎

Saying "no limit" precisely

To say that \(L\) is not the limit, turn every "for every" into "there is" and the reverse (this is how negation works in logic):

there is some \(\varepsilon > 0\) such that for every \(N\) there is an \(n > N\) with \(|a_n - L| \ge \varepsilon\).

In game terms: the challenger has one winning tolerance, and however late the prover starts, some term after it falls outside the band.

Limits of functions: \(\varepsilon\) and \(\delta\)

For a function, \(\lim_{x \to a} f(x) = L\) says that \(f(x)\) is close to \(L\) whenever \(x\) is close to \(a\). The game is the same, except that "far along the sequence" becomes "close to \(a\)":

  1. The challenger picks \(\varepsilon > 0\), a tolerance around \(L\) on the \(y\)-axis.
  2. The prover answers with \(\delta > 0\), a distance around \(a\) on the \(x\)-axis, such that every \(x\) within \(\delta\) of \(a\) (but not equal to \(a\)) has \(f(x)\) within \(\varepsilon\) of \(L\).
\[\lim_{x \to a} f(x) = L \quad\text{means}\quad \text{for every } \varepsilon > 0 \text{ there is a } \delta > 0 \text{ such that } 0 < |x - a| < \delta \implies |f(x) - L| < \varepsilon.\]

The condition \(0 < |x - a|\) says that we never look at \(x = a\) itself. The limit is about what \(f\) does near \(a\), not at \(a\); \(f(a)\) may be different, or not even defined.

The example: \(\lim_{x \to 3} (2x + 1) = 7\)

Scratch work. \(|f(x) - 7| = |2x + 1 - 7| = |2x - 6| = 2|x - 3|\). We want \(2|x - 3| < \varepsilon\), that is \(|x - 3| < \tfrac{\varepsilon}{2}\). So \(\delta = \tfrac\varepsilon2\) should work.

Proof. Let \(\varepsilon > 0\), and choose \(\delta = \tfrac\varepsilon2\). If \(0 < |x - 3| < \delta\), then

\[|(2x + 1) - 7| = 2|x - 3| < 2\delta = \varepsilon.\]

So for every \(\varepsilon\) we have found a \(\delta\) that works, and \(\lim_{x \to 3}(2x + 1) = 7\). ∎

In the picture, \(\varepsilon = 1\) and \(\delta = 0.5\): over the orange strip \(2.5 < x < 3.5\), the line stays inside the blue band \(6 < y < 8\).

3−δ 3 3+δ 7+ε 7 7−ε y = 2x + 1

The line has slope 2, so it magnifies distances by 2. That is why \(\delta\) is half of \(\varepsilon\): the \(x\)-window must be half the height of the \(y\)-window.

Why bother?

With this definition, every statement about limits becomes something you can prove, with no talk of "infinitely small" amounts or "the last term". The infinite process is replaced by a finite promise: "tell me your \(\varepsilon\), and I will tell you my \(N\) (or \(\delta\))". All of calculus, from derivatives to integrals to infinite series, rests on it.

Worked examples

1. A number for \(N\). For \(a_n = \tfrac1n\) and \(\varepsilon = 0.01\), which \(N\) works? \(\tfrac1n < 0.01\) when \(n > 100\), so \(N = 100\). Check: \(a_{101} = \tfrac1{101} \approx 0.0099 < 0.01\). ✓

2. Prove that \(\lim \tfrac{3n}{n + 2} = 3\). Scratch: \(\left| \tfrac{3n}{n+2} - 3 \right| = \left| \tfrac{3n - 3n - 6}{n + 2} \right| = \tfrac{6}{n + 2}\). This is less than \(\varepsilon\) when \(n + 2 > \tfrac6\varepsilon\), that is \(n > \tfrac6\varepsilon - 2\).

Proof: let \(\varepsilon > 0\) and \(N = \tfrac6\varepsilon\). For \(n > N\), \(n + 2 > n > \tfrac6\varepsilon\), so \(\left|\tfrac{3n}{n+2} - 3\right| = \tfrac6{n+2} < \varepsilon\). ∎ (Choosing a slightly bigger \(N\) than needed is fine; it keeps the proof tidy.)

3. A sequence with no limit. Show that \(a_n = (-1)^n\) (that is \(-1, 1, -1, 1, \ldots\)) has no limit.

Suppose its limit were some \(L\). Challenge with \(\varepsilon = 1\). Consecutive terms are \(-1\) and \(1\), which are 2 apart. If both were within 1 of \(L\), then by the triangle inequality \(2 = |1 - (-1)| \le |1 - L| + |L - (-1)| < 1 + 1 = 2\), which is false. So after any \(N\) there is always a term at distance at least 1 from \(L\). No \(L\) works.

4. A \(\delta\) for a given \(\varepsilon\). For \(\lim_{x \to 3}(2x + 1) = 7\) with \(\varepsilon = 0.1\): \(\delta = 0.05\). Test \(x = 3.04\): \(f(x) = 7.08\), within \(0.1\) of 7. ✓

5. A curved function: \(\lim_{x \to 3} x^2 = 9\). Scratch: \(|x^2 - 9| = |x - 3|\,|x + 3|\). The factor \(|x + 3|\) is not constant, so first keep \(x\) near 3: if \(|x - 3| < 1\), then \(2 < x < 4\), so \(|x + 3| < 7\). Then \(|x^2 - 9| < 7|x - 3|\), which is less than \(\varepsilon\) if \(|x - 3| < \tfrac\varepsilon7\).

Proof: let \(\varepsilon > 0\) and \(\delta = \min\left(1, \tfrac\varepsilon7\right)\), the smaller of the two. If \(0 < |x - 3| < \delta\), then \(|x + 3| < 7\) (since \(\delta \le 1\)) and \(|x - 3| < \tfrac\varepsilon7\), so \(|x^2 - 9| < 7 \cdot \tfrac\varepsilon7 = \varepsilon\). ∎

6. The value at \(a\) does not matter. \(f(x) = \dfrac{x^2 - 9}{x - 3}\) is not defined at \(x = 3\). For \(x \ne 3\) it equals \(x + 3\), so \(|f(x) - 6| = |x - 3|\), and \(\delta = \varepsilon\) works: \(\lim_{x \to 3} f(x) = 6\).

Try it

  1. For \(a_n = \tfrac1n\), find an \(N\) that works for \(\varepsilon = 0.001\).
  2. Prove that \(\lim_{n \to \infty} \tfrac{1}{n^2} = 0\).
  3. Prove that \(\lim_{n \to \infty} \tfrac{2n - 1}{n} = 2\).
  4. Prove that \(\lim_{x \to 2} (5x - 4) = 6\). What is \(\delta\) when \(\varepsilon = 0.1\)?
  5. For \(\lim_{x \to 3} (2x + 1) = 7\) and \(\varepsilon = 0.3\), what is the largest \(\delta\) that works?
  6. Prove that \(\lim_{x \to 1} x^2 = 1\). (Hint: first insist \(|x - 1| < 1\).)
  7. Let \(f(x) = 1\) for \(x > 0\) and \(f(x) = -1\) for \(x < 0\). Show that \(\lim_{x \to 0} f(x)\) does not exist.
  8. Write in symbols what it means for a sequence to have no limit at all (for every \(L\), it does not converge to \(L\)).
Answers
  1. \(\tfrac1n < 0.001\) when \(n > 1000\), so \(N = 1000\).
  2. Let \(\varepsilon > 0\) and \(N = \tfrac{1}{\sqrt\varepsilon}\). For \(n > N\), \(n^2 > N^2 = \tfrac1\varepsilon\), so \(\left|\tfrac1{n^2} - 0\right| = \tfrac1{n^2} < \varepsilon\). ∎
  3. \(\left|\tfrac{2n-1}{n} - 2\right| = \left|\tfrac{-1}{n}\right| = \tfrac1n\). Let \(\varepsilon > 0\) and \(N = \tfrac1\varepsilon\); for \(n > N\), \(\tfrac1n < \varepsilon\). ∎
  4. \(|(5x - 4) - 6| = |5x - 10| = 5|x - 2|\). Let \(\delta = \tfrac\varepsilon5\). If \(0 < |x - 2| < \delta\) then \(|(5x - 4) - 6| < 5\delta = \varepsilon\). ∎ For \(\varepsilon = 0.1\), \(\delta = 0.02\).
  5. We need \(2|x - 3| < 0.3\), that is \(|x - 3| < 0.15\). So \(\delta = 0.15\) is the largest. (Check: at \(x = 3.15\), \(f(x) = 7.3\), exactly at the edge; anything closer is inside.)
  6. \(|x^2 - 1| = |x - 1|\,|x + 1|\). If \(|x - 1| < 1\) then \(0 < x < 2\), so \(|x + 1| < 3\). Let \(\delta = \min\left(1, \tfrac\varepsilon3\right)\). If \(0 < |x - 1| < \delta\) then \(|x^2 - 1| < 3 \cdot \tfrac\varepsilon3 = \varepsilon\). ∎
  7. Suppose the limit were \(L\). Take \(\varepsilon = 1\). Any \(\delta > 0\) allows both \(x = \tfrac\delta2\) (where \(f = 1\)) and \(x = -\tfrac\delta2\) (where \(f = -1\)). Both values within 1 of \(L\) would give \(2 = |1 - (-1)| < 1 + 1 = 2\), which is false. So no \(L\) works, and the limit does not exist.
  8. For every \(L\), there is \(\varepsilon > 0\) such that for every \(N\) there is \(n > N\) with \(|a_n - L| \ge \varepsilon\). (Such a sequence is said to diverge.)

Common mistake

Choosing \(\varepsilon\) yourself. In a proof that a limit is \(L\), you do not get to pick \(\varepsilon\); you must handle every \(\varepsilon\) at once, so your \(N\) or \(\delta\) is a formula in \(\varepsilon\). (In a proof that a limit is not \(L\), you do get to pick one bad \(\varepsilon\).)

Letting \(\delta\) depend on \(x\). \(\delta\) may depend on \(\varepsilon\) (and on \(a\)), but it must be chosen before \(x\): one window that works for every \(x\) in it. Writing "\(\delta = \tfrac{\varepsilon}{|x+3|}\)" is not allowed; that is why worked example 5 first trapped \(|x + 3|\) below 7.

Where it leads