What "approaches" really means
Tonight's idea
Calculus runs on the phrase "\(f(x)\) approaches \(L\) as \(x\) approaches \(a\)". For over a century nobody could say exactly what it meant. This settled it.
\(f(x) \to L\) as \(x \to a\) means: for every \(\varepsilon > 0\) there is a \(\delta > 0\) such that \(0 \lt |x - a| \lt \delta\) implies \(|f(x) - L| \lt \varepsilon\).
Read it as a game. Your opponent names a tolerance \(\varepsilon\) (epsilon): how close to \(L\) outputs must be. You answer with a window \(\delta\) (delta) around \(a\). You win if every \(x\) in your window, other than \(a\) itself, sends \(f(x)\) inside the tolerance. The limit exists if you can beat every \(\varepsilon\); your \(\delta\) may depend on it. And \(f(a)\) never appears: \(f\) need not be defined at \(a\), so \(0/0\) is a question, not an answer.
To show there is no limit, swap roles: for each candidate \(L\), name one \(\varepsilon\) that no \(\delta\) can handle. Suppose \(1/x \to L\) as \(x \to 0\), and take \(\varepsilon = 1\). For any \(\delta\), \(x = \min\!\big(\delta/2,\ 1/(|L| + 2)\big)\) lies in the window, yet \(1/x \ge |L| + 2\), so \(|1/x - L| \ge 2 > 1\). Every \(L\) loses.
Watch one: prove that \(2x \to 6\) as \(x \to 3\)
Scratch work first. We want \(|2x - 6| \lt \varepsilon\). Since \(|2x - 6| = 2|x - 3|\), that is the same as \(|x - 3| \lt \varepsilon/2\). So the window should have half-width \(\varepsilon/2\).
Now the proof, written forwards. Let \(\varepsilon > 0\) be given. Choose \(\delta = \varepsilon/2\). If \(0 \lt |x - 3| \lt \delta\), then
\[|2x - 6| = 2|x - 3| \lt 2\delta = \varepsilon.\]
That handles every \(\varepsilon\) at once, which is the point. The pattern is always the same: work backwards from \(|f(x) - L| \lt \varepsilon\) to a condition on \(|x - a|\), then write it out forwards. Steeper functions need smaller windows: for \(2x\) the window is half the tolerance.
Your turn
- Prove that \(3x + 1 \to 7\) as \(x \to 2\). State your \(\delta\) in terms of \(\varepsilon\).
- Let \(f(x) = x^2\), \(a = 2\), \(L = 4\) and \(\varepsilon = 0.1\). Find a \(\delta\) that works. Hint: \(x^2 - 4 = (x - 2)(x + 2)\). First insist that \(\delta \le 1\), so that \(|x + 2| \lt 5\).
- Prove that \(x^2 \to 9\) as \(x \to 3\), for every \(\varepsilon\). Your \(\delta\) may be the smaller of two numbers, written \(\min(1, \ldots)\).
- Let \(s(x) = 1\) for \(x > 0\) and \(s(x) = -1\) for \(x \lt 0\). Prove that \(s\) has no limit at \(0\): show that for any \(L\), the choice \(\varepsilon = 1\) defeats every \(\delta\). Where does your argument use that \(1\) and \(-1\) are \(2\) apart?
Stretch: a denominator that moves
Prove that \(1/x \to 1/2\) as \(x \to 2\). The difficulty is that \(\tfrac1x - \tfrac12 = \tfrac{2 - x}{2x}\) has \(x\) in the denominator. Insist first that \(\delta \le 1\), so \(x > 1\), and bound the denominator from below. Then say in one sentence why the same plan fails at \(a = 0\).
Answers
- \(|(3x + 1) - 7| = 3|x - 2|\). Choose \(\delta = \varepsilon/3\). Then \(0 \lt |x - 2| \lt \delta\) gives \(3|x - 2| \lt 3\delta = \varepsilon\).
- If \(|x - 2| \lt 1\) then \(1 \lt x \lt 3\), so \(|x + 2| \lt 5\) and \(|x^2 - 4| = |x - 2|\,|x + 2| \lt 5|x - 2|\). To make this below \(0.1\) take \(|x - 2| \lt 0.02\). So \(\delta = 0.02\) works (anything smaller does too). Check: on that window \(|x^2 - 4|\) never exceeds about \(0.080\).
- If \(|x - 3| \lt 1\) then \(2 \lt x \lt 4\), so \(|x + 3| \lt 7\) and \(|x^2 - 9| = |x - 3|\,|x + 3| \lt 7|x - 3|\). Choose \(\delta = \min(1, \varepsilon/7)\). Then \(0 \lt |x - 3| \lt \delta\) gives both \(|x - 3| \lt 1\) (so the bound \(7\) applies) and \(7|x - 3| \lt \varepsilon\). So \(|x^2 - 9| \lt \varepsilon\).
- Suppose \(s(x) \to L\) and take \(\varepsilon = 1\). Any window \((-\delta, \delta)\) contains a positive \(x\), where \(s = 1\), and a negative \(x\), where \(s = -1\). Both would need \(|1 - L| \lt 1\) and \(|{-1} - L| \lt 1\). But then \(2 = |1 - (-1)| \le |1 - L| + |L + 1| \lt 2\), a contradiction. The gap of \(2\) is what makes \(\varepsilon = 1\) enough; with a gap of \(g\), any \(\varepsilon \le g/2\) would do.
Stretch. If \(|x - 2| \lt 1\) then \(x > 1\), so \(2x > 2\) and \(\left|\tfrac1x - \tfrac12\right| = \tfrac{|x - 2|}{2x} \lt \tfrac{|x - 2|}{2}\). Choose \(\delta = \min(1, 2\varepsilon)\): then \(\tfrac{|x - 2|}{2} \lt \tfrac{2\varepsilon}{2} = \varepsilon\). At \(a = 0\) the plan fails because no window around \(0\) keeps \(x\) away from \(0\), so the denominator cannot be bounded from below; indeed the idea section showed there is no limit there at all.