The ε–δ definition of a limit

What does it mean for f(x) to get close to L as x gets close to a?

However small a band someone sets around L, you can find a window around a that keeps the graph inside the band. The point x = a itself is left out.

Before you start · 3 quick checks on absolute value, about 2 min

Definition

For \(f:D\to\mathbb{R}\), \(\lim_{x\to a} f(x)=L\) means: for every \(\varepsilon>0\) there is a \(\delta>0\) such that for all \(x\in D\),

\[0<\lvert x-a\rvert<\delta \;\Longrightarrow\; \lvert f(x)-L\rvert<\varepsilon.\]

In words: ε is the half-height of the band around L, and δ is the half-width of the window around a.

The domain, and why x = a is left out

Let \(f:D\to\mathbb{R}\), and let \(a\) be a limit point of \(D\), so that every interval around \(a\) contains points of \(D\) other than \(a\).

The condition \(0<\lvert x-a\rvert\) excludes \(x=a\), so the limit depends only on values of f near a and never on f(a), which may differ from L or be undefined.

1 See

Defend the limit against the page

The page names a band half-height ε. You answer with a window half-width δ and press Defend. The page then looks for an x inside the window whose value leaves the band.

Free play: set both ε and δ yourself

Inside the window the curve is drawn thick, in one color where it stays in the band and another where it escapes. The ring marks the sampled escaping point nearest to a. When δ fails, the caption prints a point x that satisfies 0 < |x − a| < δ and |f(x) − L| ≥ ε.

What to try

(A) Shrink ε and watch how small δ must become. Auto-find δ estimates the largest passing δ within the slider range.
(B) f is undefined at 1, and the limit is still 2. Off x = 1 this function equals x + 1, so δ = ε works.
(C) Try each claimed L. For ε ≤ 1/2, auto-find δ finds no passing δ for any of them.
(D) Turn on the zoom and shrink δ. The oscillation gets faster near 0, so a smaller window still reaches 1 and −1.
(E) Set δ = ε. Since |x·sin(1/x)| ≤ |x|, every point in the window is within ε of 0.

Auto-find δ checks several thousand points in the window, packed more tightly near a, and bisects on δ. A sampled check can miss an escaping point between samples. The sampled result requires an analytic justification.

Key point One point inside the window that leaves the band is enough to defeat a δ. For x² a small enough δ defends every ε. For the jump and for sin(1/x), no δ defends once ε is small.

2 Try

Predictions about δ and ε

Click an answer to see the result. Predictions 1 and 3 also show the case in the figure below them.

Goals: tune δ until the goal turns green

Each goal sets the function and ε. You move δ, and the page checks each value as you go.

Compute the largest δ for x² exactly

Key point ε comes first, and δ may depend on it. Shrinking δ cannot help when the values just beside a stay far from L.

3 Prove

A proof that x² tends to 1

The idea in three steps

Someone gives you ε. You need |x² − 1| = |x − 1|·|x + 1| to be below it.
Keep x within 1 of 1. Then |x + 1| is less than 3.
Choose δ = min(1, ε/3). Then |x − 1| is below ε/3, so the product is below 3 · ε/3 = ε.

Theorem

\(\lim_{x\to1} x^2 = 1\).

Here \(f(x)=x^2\) with domain \(D=\mathbb{R}\), \(a=1\) and \(L=1\). The claim is: for every \(\varepsilon>0\) there is \(\delta>0\) such that \(0<\lvert x-1\rvert<\delta\) implies \(\lvert x^2-1\rvert<\varepsilon\).

Why sin(1/x) has no limit at 0

Proposition

\(\lim_{x\to0}\sin(1/x)\) does not exist.

Here \(f(x)=\sin(1/x)\) with domain \(D=\mathbb{R}\setminus\{0\}\) and \(a=0\), a limit point of \(D\). The claim is that no real number \(L\) satisfies the definition.

Key point The proof chooses δ = min(1, ε/3). The cap at 1 keeps |x + 1| below 3, and ε/3 then makes |x − 1|·|x + 1| less than ε.

4 Apply

Problems

Each problem has hints and a full solution.