Jupiter2U School
Chapter 18

18. Limits & Continuity, Rigorously

College

Chapter 15 introduced the limit informally, as "what a function approaches." Here we make that precise with the epsilon–delta definition, define continuity rigorously, and prove two of the theorems that make continuous functions trustworthy.

At a glance
Core ideaThe ε–δ definition makes "approaches" precise, testable and provable.
Key termContinuity: f(a) is defined, the limit exists, and the two are equal.
You can…Prove a limit with ε–δ and apply the IVT and EVT.
Watch outA limit can exist where the function is undefined — that's a removable hole.
01 · Theory

The ε–δ definition of a limit

Formally, limx→a f(x) = L means:

ε–δ definitionFor every ε > 0 there exists δ > 0 such that, whenever 0 < |x − a| < δ, it follows that |f(x) − L| < ε.

In words: no matter how tight a tolerance ε you demand on the output, a tight enough window δ around the input guarantees it.

f is continuous at a when three things all hold: f(a) is defined, limx→a f(x) exists, and the two are equal. Discontinuities come in types: removable (a single missing or misplaced point), jump (the left and right limits disagree), and infinite (a vertical asymptote).

Two cornerstone theorems about continuous functions on a closed interval [a, b]:

Intermediate Value Theorem (IVT): if f is continuous on [a,b] and k is between f(a) and f(b), some c in [a,b] has f(c) = k.
Extreme Value Theorem (EVT): a continuous function on [a,b] attains both a maximum and a minimum value on that interval.
02 · Explanation

A tolerance game, played precisely

Think of ε–δ as a challenge-and-response game. A skeptic names how close to L they want f(x) to land — that's their ε. You must respond with a window δ around a that guarantees it, no matter how small their ε gets. If you can always win this game for every ε, the limit truly is L. This turns the vague phrase "gets arbitrarily close" into something you can actually verify or disprove.

Rigor matters because intuition alone breaks down on strange functions — some are continuous everywhere yet differentiable nowhere. The IVT is why bisection root-finding works (and why, somewhere on Earth, two antipodal points always have equal temperature): a continuous path from a negative value to a positive one cannot "jump over" zero.

03 · Practical

Worked example: a direct ε–δ proof

Problem

Prove, using the ε–δ definition, that limx→2 (3x + 1) = 7.

  1. Let ε > 0 be given. We need |(3x + 1) − 7| < ε whenever 0 < |x − 2| < δ.
  2. Simplify the target expression: |(3x + 1) − 7| = |3x − 6| = 3|x − 2|.
  3. We want 3|x − 2| < ε, i.e. |x − 2| < ε/3. So choose δ = ε/3.
  4. Verify: if 0 < |x − 2| < δ = ε/3, then |(3x+1) − 7| = 3|x − 2| < 3δ = ε. ∎

Answer: for any ε > 0, taking δ = ε/3 proves the limit is exactly 7.

04 · Q&A

Test your understanding

What's the difference between a limit existing at a point and the function being continuous there?

A limit can exist even if the function is undefined at that point, or defined to a different value — the limit only cares what happens as you approach. Continuity is stricter: it additionally requires f(a) to be defined and to equal that limit. A function with a "hole" at a point has a limit there but is not continuous there.

Give an example of a jump discontinuity.

A step function, e.g. f(x) = 0 for x < 0 and f(x) = 1 for x ≥ 0. Approaching 0 from the left gives limit 0; from the right gives limit 1. Since the one-sided limits disagree, the overall limit does not exist, and the function jumps.

Use the Intermediate Value Theorem to argue x³ − x − 1 = 0 has a root between 1 and 2.

Let f(x) = x³ − x − 1, continuous everywhere (polynomials always are). f(1) = 1 − 1 − 1 = −1 < 0 and f(2) = 8 − 2 − 1 = 5 > 0. Since f is continuous and changes sign, the IVT guarantees some c ∈ (1,2) with f(c) = 0 — a root must exist there, even without solving exactly.

Why must δ depend on ε, rather than being one fixed number?

Because a tighter output tolerance almost always demands a tighter input window. If someone demands ε = 0.001 instead of ε = 1, a δ that worked for the loose tolerance will usually be too generous for the strict one. The definition requires a valid δ for every ε, so in general δ is a function of ε (here, δ = ε/3).

Concept mind map

How the ideas connect

Every key idea in this chapter, branching from the core concept — use it to see the whole picture at a glance.

Formal limitEpsilon toleranceDelta boundFor all epsilonThere exists deltaDirect proofEpsilon-Delta Limits
Infographic

The process, step by step

Step 1State the targetGiven epsilon > 0, aim for |f(x) - L| < epsilon
Step 2Relate to |x - a|Bound |f(x) - L| by a multiple of |x - a|
Step 3Choose deltaPick delta so the bound stays under epsilon
Step 4Verify the implicationShow 0 < |x - a| < delta forces the goal
Solved examples

Worked problems, step by step

Follow each solution line by line, then try to reproduce it on paper before moving on.

Example 1Prove the limit of 2x as x approaches 3 is 6.

  1. Need |2x - 6| < epsilon
  2. |2x - 6| = 2|x - 3|
  3. Choose delta = epsilon/2 so 2|x-3| < epsilon

Example 2For limit of x + 1 at x = 2 equal to 3, find delta given epsilon.

  1. |(x + 1) - 3| = |x - 2|
  2. Require |x - 2| < epsilon
  3. Choose delta = epsilon
Practice problem set

Now you try

Work each one out first, then tap to reveal the worked answer.

1For limit of 5x at x = 1 equal to 5, choose delta.
|5x - 5| = 5|x-1|, so delta = epsilon/5.
2What does epsilon represent?
The allowed tolerance on the output distance from L.
3What does delta control?
How close x must be to a, the input tolerance.
4For limit of 3x - 1 at x = 2 equal to 5, find delta.
|3x - 1 - 5| = 3|x - 2|, so delta = epsilon/3.
5Why must the definition hold for every epsilon > 0?
The output can be forced arbitrarily close to L, no matter how small the tolerance.
6For constant function f(x) = 4, prove the limit is 4.
|4 - 4| = 0 < epsilon always, so any delta works.