Difficulty: Intermediate to Advanced | Prerequisites: Limits (§1.5), Limit Laws (§1.6). You should be comfortable evaluating limits before tackling the epsilon-delta definition.
§1.7 gives the formal, rigorous definition of a limit using epsilon (ε) and delta (δ). This is where calculus stops being intuitive and starts being precise. §1.8 then defines continuity in terms of limits and introduces the Intermediate Value Theorem (IVT), one of the most powerful existence results in calculus. If you skipped three weeks: §1.7 is proof-heavy and may not appear on every exam, but understanding the ε-δ idea pays off throughout the course. §1.8 is more broadly tested.
The epsilon-delta definition makes the phrase "arbitrarily close" precise: for every output tolerance ε > 0, there exists an input accuracy δ > 0 such that |x − a| < δ forces |f(x) − L| < ε. A function is continuous at a point when its limit equals its value there. The IVT says a continuous function on a closed interval hits every value between f(a) and f(b).
Epsilon (ε)
The output error tolerance. A positive number representing how close you need f(x) to be to L. Think of ε as "how accurate does the answer need to be?"
Delta (δ)
The input accuracy. A positive number representing how close x must be to a to guarantee the output stays within ε of L. Think of δ as "how precise does the input need to be?"
Formal definition of a limit
lim(x→a) f(x) = L means: for every ε > 0, there exists a δ > 0 such that 0 < |x − a| < δ implies |f(x) − L| < ε.
Continuous (at a point)
A function f is continuous at x = a when lim(x→a) f(x) = f(a). Three conditions must all hold: f(a) is defined, the limit exists, and the limit equals f(a).
Removable discontinuity
A point where the limit exists but either f(a) is undefined or f(a) ≠ lim(x→a) f(x). The "hole" could be "filled" by redefining f(a).
Jump discontinuity
A point where the left and right limits both exist but are unequal. The graph jumps from one value to another.
Vertical asymptote discontinuity
A point where at least one one-sided limit is ±∞. The function blows up.
Oscillation discontinuity
A point where the function oscillates so wildly that no one-sided limit exists. Example: sin(1/x) near x = 0.
Intermediate Value Theorem (IVT)
If f is continuous on [a, b] and r is between f(a) and f(b), then there exists some c in [a, b] with f(c) = r. In simple terms, a continuous function cannot skip over any value between its endpoints.
Why the formal definition matters
Limits underpin every definition in calculus. The ε-δ definition makes the informal idea ("f(x) gets close to L") mathematically precise. Think of it as an error-control problem: given any output tolerance ε, find an input accuracy δ that guarantees the output stays within bounds.
The definition
lim(x→a) f(x) = L means: for every ε > 0, there exists δ > 0 such that 0 < |x − a| < δ implies |f(x) − L| < ε.
In words: no matter how tight the tolerance on the output, you can always find an input window that keeps f(x) within that tolerance.
Worked example 1: proving lim(x→5) (3x − 2) = 13
Start with the output requirement |f(x) − L| < ε:
|(3x − 2) − 13| < ε ⟹ |3x − 15| < ε ⟹ 3|x − 5| < ε ⟹ |x − 5| < ε/3
So choose δ = ε/3. Then 0 < |x − 5| < δ forces |(3x−2) − 13| < ε. Done.
Worked example 2: proving lim(x→3) √x = √3
The algebra is harder. Starting from |√x − √3| < ε and squaring (carefully), you work out that δ = 2ε√3 − ε² guarantees the bound. The key takeaway: for any ε you specify, a suitable δ exists.
Proving general theorems
The ε-δ framework lets you prove the Limit Laws rigorously. The Sum Law proof, for instance, splits the tolerance in half: require each of f(x) and g(x) to be within ε/2 of their respective limits, so their sum is within ε of L + M.
The Squeeze Theorem is also proved this way: if f(x) and h(x) are both within ε of L, and g(x) is sandwiched between them, then g(x) is automatically within ε of L.
The Substitution Theorem (for composition of limits) chains two ε-δ arguments: the outer function's δ becomes the target for the inner function's limit.
Definition
f(x) is continuous at x = a when lim(x→a) f(x) = f(a). This requires three things at once: f(a) exists, the limit exists, and they are equal.
Types of discontinuity
Removable (type i): f(a) is undefined, but the limit exists. The graph has a hole. You could fix it by defining f(a) = L.
Removable (type ii): f(a) is defined but equals the "wrong" value, not the limit. The graph has a filled dot in the wrong place.
Jump (type iii): The left and right limits exist but disagree. The graph jumps. Example: the floor function, piecewise-defined functions.
Vertical asymptote (type iv): One or both one-sided limits are ±∞. Example: 1/(x − 3) at x = 3.
Oscillation (type v): One or both one-sided limits fail to exist because of wild oscillation. Example: sin(1/x) at x = 0.
Continuity on an interval
f is continuous on [a, b] when it is continuous at every interior point and the appropriate one-sided limits match at the endpoints: lim(x→a⁺) f(x) = f(a) and lim(x→b⁻) f(x) = f(b).
Domain of continuity
Almost all functions defined by formulas are continuous wherever they are defined. This follows from the Limit Laws. So the practical question is: where is the function undefined?
Example: g(x) = (x² − 3x + 1)√(x+1) / (x − 3). The domain requires x + 1 ≥ 0 (so x ≥ −1) and x ≠ 3. The function is continuous on [−1, 3) ∪ (3, ∞).
Composing continuous functions
If g is continuous at a and f is continuous at g(a), then the composite f(g(x)) is continuous at a. This is the Composition Law:
lim(x→a) f(g(x)) = f(lim(x→a) g(x))
Statement
If f is continuous on the closed interval [a, b], and r is any value between f(a) and f(b), then there exists at least one c in [a, b] such that f(c) = r.
In plain terms: a continuous function that starts below a target value and ends above it (or vice versa) must cross that value somewhere in between. No jumping allowed.
What the IVT does and does not tell you
The theorem guarantees existence ("there is a c") but does not tell you what c is or how to find it. It is a pure existence result.
The theorem fails for discontinuous functions. For example, g(x) = (x² − 3x + 1)√(x+1) / (x − 3) has g(2) ≈ 1.7 and g(4) ≈ 11.2, with 7 between them. But g has a vertical asymptote at x = 3, so there is no c in [2, 4] with g(c) = 7.
Classic application: showing a root exists
To prove cos(x) = x has a solution, define f(x) = cos(x) − x. Then f(0) = 1 > 0 and f(π) = −1 − π < 0. Since f is continuous and changes sign, the IVT guarantees some c in [0, π] with f(c) = 0, i.e. cos(c) = c.
This technique works whenever you can find two inputs where a continuous function takes opposite signs. The IVT then certifies a zero between them.
Formal limit definition
lim(x→a) f(x) = L ⟺ for every ε > 0, there exists δ > 0 such that 0 < |x − a| < δ ⟹ |f(x) − L| < ε
One-sided limit (from the right)
lim(x→a⁺) f(x) = L ⟺ for every ε > 0, there exists δ > 0 such that 0 < x − a < δ ⟹ |f(x) − L| < ε
Infinite limit
lim(x→a) f(x) = ∞ ⟺ for every bound B, there exists δ > 0 such that 0 < |x − a| < δ ⟹ f(x) > B
Continuity condition
f continuous at a ⟺ lim(x→a) f(x) = f(a)
Composition Law
lim(x→a) f(g(x)) = f(lim(x→a) g(x)), provided lim(x→a) g(x) = b and f is continuous at b
Students often think the ε-δ definition means "ε and δ are small." They can be any positive number. The point is that for every ε (no matter how small), a suitable δ exists.
A common mistake is choosing δ first, then ε. The proof structure is: let ε be given (arbitrary), then produce δ in terms of ε.
Students confuse "removable discontinuity" with "the function is continuous there." A removable discontinuity is still a discontinuity. The function is not continuous at that point; it merely could be made continuous by redefining one value.
The IVT does not say the function crosses the value only once. There could be many values of c where f(c) = r.
⚠️ Some exams ask you to produce an ε-δ proof for a simple linear function. The standard method: solve |f(x) − L| < ε for |x − a| and read off δ.
⚠️ Identifying the type of discontinuity (removable, jump, asymptote, oscillation) is a common exam question. Know the classification and be able to read it from a graph.
⚠️ The IVT is tested in two ways: (1) show that an equation has a solution by finding a sign change, and (2) explain why the theorem requires continuity by giving a counterexample with a discontinuous function.
⚠️ "At which points is f(x) continuous?" questions reduce to finding where the function is undefined (denominator zero, negative under an even root, etc.).
True or False: In an ε-δ proof, you choose ε first and then find δ.
False. ε is given to you (arbitrary). You must produce δ in terms of ε.
Fill in the blank: A function is continuous at x = a when lim(x→a) f(x) = ______.
f(a).
True or False: The Intermediate Value Theorem can be used to find the exact value of a root.
False. The IVT guarantees a root exists but does not tell you its value.
Name the type of discontinuity where lim(x→a⁺) f(x) ≠ lim(x→a⁻) f(x).
Jump discontinuity.
True or False: f(x) = |x| is continuous at x = 0.
True. The corner is not a discontinuity; the graph has no break there.
Q: Give an ε-δ proof that lim(x→2) (5x + 1) = 11.
A: Need |(5x+1) − 11| < ε, i.e. |5x − 10| < ε, i.e. 5|x − 2| < ε, i.e. |x − 2| < ε/5. Choose δ = ε/5. Then 0 < |x − 2| < δ forces |(5x+1) − 11| = 5|x−2| < 5δ = ε.
Q: Classify the discontinuity of f(x) = (x² − 1)/(x − 1) at x = 1.
A: f(1) is undefined (0/0), but lim(x→1) f(x) = lim(x→1)(x+1) = 2, which exists. This is a removable discontinuity (type i).
Q: Use the IVT to show that x³ + x − 1 = 0 has a solution between 0 and 1.
A: Let f(x) = x³ + x − 1. Then f(0) = −1 < 0 and f(1) = 1 > 0. Since f is a polynomial (hence continuous) and changes sign on [0, 1], the IVT guarantees some c in (0, 1) with f(c) = 0.
Q: Where is g(x) = √(x − 2) / (x − 5) continuous?
A: The domain requires x − 2 ≥ 0 (so x ≥ 2) and x ≠ 5. So g is continuous on [2, 5) ∪ (5, ∞). At x = 5 there is a vertical asymptote.
Q: True or false: if f is continuous on [0, 3] and f(0) = −2, f(3) = 4, then there exists c in [0, 3] with f(c) = 0.
A: True. Since −2 < 0 < 4 and f is continuous on [0, 3], the IVT guarantees such a c.
The ε-δ definition from §1.7 will reappear when the derivative is formally defined in Chapter 2. Continuity is a prerequisite for the Extreme Value Theorem and Rolle's Theorem (Chapter 4). The IVT is the theoretical backbone of numerical root-finding methods like the bisection algorithm. Composition of continuous functions is used throughout when differentiating composite functions via the Chain Rule.
epsilon-delta definition, ε-δ proof, formal limit definition, error control, continuity, continuous function, discontinuity, removable discontinuity, jump discontinuity, vertical asymptote, oscillation discontinuity, Intermediate Value Theorem, IVT, existence theorem, bisection method, composition law, Stewart calculus §1.7 §1.8, MATH 132, MSU calculus, domain of continuity