The precise definition of a limit
Within tolerance
The statement \lim_{x\to a} f(x) = L says that f(x) can be made as close to L as we like by taking x close enough to a, with x \ne a. The ε-δ definition gives "as close as we like" and "close enough" an exact meaning.
In the graph below, the ochre band contains every y within \varepsilon of L, and the indigo band contains every x within \delta of a. Choose \varepsilon. How narrow must the indigo band be before the graph, at every point of it other than x=a, stays inside the ochre band?
Drag the ochre or indigo edges to change ε and δ. Point at the graph, or tap it, to read off x, f(x) and |f(x) − L|.
Reading the definition
The formal definition:
- For every \varepsilon > 0
- \varepsilon is given to you and can be any positive number, however small. The \delta you choose has to work for that \varepsilon, so \delta is usually written as a formula in \varepsilon.
- there is a \delta > 0
- You need one \delta for each \varepsilon. If a \delta works, every smaller positive \delta works too, so there is no single correct answer. Turn on "Show the largest δ that works" to see the cutoff.
- 0 < |x-a| < \delta
- x is within \delta of a, and x \ne a. The value f(a) is never used and does not need to exist. See the hole function in the graph above.
- |f(x)-L| < \varepsilon
- f(x) is within \varepsilon of L, so the point (x, f(x)) is inside the ochre band.
The order of the quantifiers
Could one \delta work for every \varepsilon at once? The definition does not ask for that. It says that for every \varepsilon there is a \delta, so \delta may change when \varepsilon changes. Swapping the quantifiers gives a different, stronger statement. For 2x+1 at x=3 that statement is false, because a fixed \delta fails as soon as \varepsilon < 2\delta.
The button below fixes \delta = 0.2. Press "Divide ε by 4" twice and the line leaves the ochre band.
Finding δ and writing the proof
To find \delta, start with |f(x)-L| and simplify it until a factor of |x-a| appears. This scratch work is not part of the proof. The proof starts by choosing \delta and then shows that |f(x)-L|<\varepsilon.
A line: \lim_{x\to 3}(2x+1)=7
We need 2|x-3| < \varepsilon, that is, |x-3| < \varepsilon/2. Choose \delta = \varepsilon/2.
Let \varepsilon > 0 and set \delta = \varepsilon/2. If 0<|x-3|<\delta, then
Since \varepsilon > 0 was arbitrary, the limit is 7. \blacksquare
A parabola: \lim_{x\to 2}x^2=4
Here |x^2-4| = |x-2|\cdot|x+2|, and the factor |x+2| depends on x. The usual fix is to first require \delta\le c for some number c, which bounds |x+2|. Textbooks usually take c=1, but any c between 0 and 4 works. What changes in the proof when you choose a different c?
A common error is \delta=\varepsilon/4, based on |x+2|\approx 4 near x=2. It fails for every \varepsilon, because |x+2|>4 when x>2. You can check this in the tester below.
Test a δ
Enter \delta as a formula in \varepsilon. The page checks it for \varepsilon = 10, 1, 0.1, 0.01, and 0.001. Type \varepsilon as e. You can also use min, max, sqrt, abs, and ^ for powers.
| ε | δ(ε) | Largest |f(x) − L| | Result |
|---|
The δ game
Six rounds. Each round gives a limit and a value of \varepsilon. Enter a \delta that works, either as a number or as a formula in e. You score a point when your \delta works. Answers within a factor of 2 of the largest working \delta also earn a mark for precision.
Solve it step by step
For each limit, find a δ that works for every ε, then use it for one particular ε. Decimals are fine. The score counts steps you answer correctly on the first try.
What if no δ works?
Let H(x)=0 for x<0 and H(x)=1 for x\ge 0. Every interval around 0 contains points where H(x)=0 and points where H(x)=1. If \varepsilon\le\tfrac12, the interval (L-\varepsilon, L+\varepsilon) is open and has length at most 1, so it cannot contain both 0 and 1, for any L.
To show that a limit does not exist, it is enough to find one \varepsilon for which no \delta works. Here \varepsilon=\tfrac12 is such a value, whatever L is. In the graph, drag L up and down and turn on "Show the largest δ that works".