The Precise Definition of a Limit

The precise definition of a limit is simply a careful way of saying "as close as you like": you can make $f(x)$ as close to $L$ as anyone demands, just by keeping $x$ close enough to $a$. It's what mathematicians use to prove the limit laws are actually true. The definition looks scary, but it's only a few pieces: $\varepsilon$ (how close to $L$ the outputs have to be), $\delta$ (how close to $a$ you promise to keep the inputs), $0 < |x - a| < \delta$ ($x$ is within $\delta$ of $a$, but not $a$ itself) and $|f(x) - L| < \varepsilon$ (the output lands within $\varepsilon$ of $L$). Read in order, it's one sentence: whatever $\varepsilon$ you're given, there's a $\delta$ that does the job. It takes a little getting used to, so bear with it.

Since the very first lesson we've been saying things like "$f(x)$ gets as close to $L$ as we like." That's fine for intuition, but "close" is vague. How close? This definition answers that by turning it into a game. A challenger names a tolerance $\varepsilon$, as small as they like. You have to answer with a window $\delta$ so that every $x$ within $\delta$ of $a$ (other than $a$ itself) gives an $f(x)$ within $\varepsilon$ of $L$. If you can always answer, no matter how small $\varepsilon$ gets, the limit is $L$.

Linear functions are easy. For $f(x) = mx + b$, $|f(x) - L| = |m|\,|x - a|$. The output distance is just the input distance stretched by $|m|$, so to land within $\varepsilon$, keep the inputs within $\varepsilon/|m|$. In other words, $\delta = \varepsilon/|m|$ always works.

Nonlinear functions need a bound. The stretch factor isn't constant anymore, so you pin it down. For $\lim_{x\to3} x^2 = 9$, factor: $|x^2 - 9| = |x - 3|\,|x + 3|$. The $|x - 3|$ part is the one we control. The $|x + 3|$ part changes with $x$, so promise up front that $\delta \le 1$. Then $x$ is between 2 and 4, which keeps $|x + 3| < 7$. Now $|x^2 - 9| < 7|x - 3|$, and it's just like the linear case with a stretch of $7$: $\delta = \min\left(1, \frac{\varepsilon}{7}\right)$ works.

You'll rarely use $\varepsilon$–$\delta$ arguments to compute a limit. Their job is to make the foundations solid: every limit law in this unit is proved with them.

Manufacturing tolerances are $\varepsilon$–$\delta$ in disguise: to keep a part's output within spec ($\varepsilon$), how precisely must each input be controlled ($\delta$)?