The Epsilon–Delta Definition

A numerical table can make a limit plausible. A proof must handle every input sufficiently near the target, including inputs that no table contains. We therefore need a statement that says exactly what “sufficiently near” must accomplish.

Let $f$ be defined throughout a punctured neighborhood of $a$. We say $\lim_{x\to a}f(x)=L$ if

$$\text{for every }\varepsilon>0\text{ there exists }\delta>0 \text{ such that }0<|x-a|<\delta \implies |f(x)-L|<\varepsilon.$$

Here $\varepsilon$ is the requested output tolerance and $\delta$ is the promised input tolerance. We receive an arbitrary positive $\varepsilon$, then choose a suitable $\delta$, then guarantee the implication for every permitted $x$. A choice of $\delta$ may depend on $\varepsilon$, $a$, and the fixed function. It must not depend on an input $x$ selected afterward.

A linear proof from start to finish

Prove $\lim_{x\to2}(3x-1)=5$. The output error is

$$|(3x-1)-5|=3|x-2|.$$

For a given $\varepsilon>0$, choose $\delta=\varepsilon/3$. If $0<|x-2|<\delta$, then $|(3x-1)-5|=3|x-2|<3\delta=\varepsilon$. Since this works for every positive tolerance, the proof is complete.

Notice the distinction between scratch work and the proof. We discovered $\delta$ by working backward from the desired inequality. We presented the proof forward, starting with an arbitrary tolerance and the explicit choice. A successful proof does not require the largest possible $\delta$. Any positive choice that ensures the implication works.

A quadratic needs a local bound

To prove $\lim_{x\to2}x^2=4$, factor the error:

$$|x^2-4|=|x-2||x+2|.$$

The second factor depends on $x$. We first arrange $|x-2|<1$, which gives $1<x<3$ and hence $|x+2|<5$. Now choose

$$\delta=\min\{1,\varepsilon/5\}.$$

If $0<|x-2|<\delta$, then $|x^2-4|<5|x-2|<5\delta\leq\varepsilon$. The final strict inequality comes from the strict bound on $|x-2|$. The minimum enforces two requirements simultaneously: remain in a neighborhood where the auxiliary factor is controlled, and make the final error small.

For a general fixed real $a$, $|x-a|<1$ implies $|x+a|<2|a|+1$. Thus $\delta=\min\{1,\varepsilon/(2|a|+1)\}$ proves $\lim_{x\to a}x^2=a^2$.

The largest symmetric tolerance for a square

The laboratory uses an exact bound to make the geometry visible. Put $x=a+h$. For $|h|<\delta$,

$$|x^2-a^2|=|2ah+h^2| \leq2|a||h|+h^2 <2|a|\delta+\delta^2.$$

The supremum of the output error on this open neighborhood is $2|a|\delta+\delta^2$, approached from the outward direction. A sufficient and, for the symmetric neighborhood, maximal choice is

$$\delta_* = \sqrt{a^2+\varepsilon}-|a| =\frac{\varepsilon}{\sqrt{a^2+\varepsilon}+|a|}.$$

The second form avoids subtracting nearly equal numbers in computation. At $a=0$, this becomes $\sqrt\varepsilon$. At $a=2$ and $\varepsilon=0.1$, it is approximately $0.0248457$. Setting $\delta=\delta_*$ is valid: the boundary supremum equals $\varepsilon$, but boundary inputs are excluded and every interior error is strictly smaller. This is a useful example of why strict and nonstrict inequalities cannot be interchanged casually.

How to show a candidate fails

Negating the definition gives: there exists some $\varepsilon_0>0$ such that for every $\delta>0$, there is a permitted $x$ with $0<|x-a|<\delta$ and $|f(x)-L|\geq\varepsilon_0$.

For $f(x)=|x|/x$, the candidate $L=1$ fails at zero. Choose $\varepsilon_0=1$. Whatever $\delta>0$ is offered, choose $x=-\delta/2$. Then $f(x)=-1$ and $|f(x)-1|=2\geq1$. To rule out every candidate, the one-sided disagreement established previously is more economical: any two-sided limit would equal both $-1$ and $1$.

A finite limit is unique

Suppose two distinct finite limits $L$ and $M$ existed. Let $\varepsilon=|L-M|/3$. There would be neighborhoods in which outputs are within $\varepsilon$ of each limit. Choose a noncentral input inside both neighborhoods. The triangle inequality would give

$$|L-M|\leq|L-f(x)|+|f(x)-M|<2|L-M|/3,$$

a contradiction. The existence of nearby domain points matters; our punctured-neighborhood assumption supplies them. This proof establishes uniqueness without relying on a picture.

Exercises

1. Prove $\lim_{x\to-1}(5x+2)=-3$ with an explicit $\delta$.
2. Prove $\lim_{x\to0}x^2=0$ using the largest symmetric tolerance.
3. Prove $\lim_{x\to3}x^2=9$ using a local bound of radius one.
4. Explain why choosing $\delta=\varepsilon/|x+2|$ is not an acceptable construction in the quadratic proof.
5. For $a=1$, $\varepsilon=0.25$, does $\delta=0.2$ guarantee the square-function tolerance? Find the supremum error.
6. Show that the candidate limit $L=0$ fails for the constant function $f(x)=2$ at zero.

Worked solutions

1. Choose $\delta=\varepsilon/5$. The output error is $5|x+1|<\varepsilon$.
2. Choose $\delta=\sqrt\varepsilon$. Then $0<|x|<\delta$ implies $x^2<\varepsilon$.
3. If $|x-3|<1$, then $|x+3|<7$. Choose $\min\{1,\varepsilon/7\}$ and factor $|x^2-9|$.
4. This choice depends on the later input $x$. We must provide a single radius that works for every input inside it; bounding $|x+2|$ by a fixed constant accomplishes that.
5. The supremum is $2(1)(0.2)+0.2^2=0.44$, greater than $0.25$, so the guarantee fails. For example, $x=1.19$ is allowed and produces error $0.4161$.
6. Choose $\varepsilon_0=1$. Every nonzero input within every proposed positive radius has error $2\geq1$.