Target Audience: Anyone who can differentiate fluently and wants the definition the rules are actually derived from.
Prerequisites: Tier 0, especially Logic and Quantifiers — the whole of this page turns on the order of two quantifiers, and on negating them correctly.
Calculus is usually learned as a set of rules and only later justified. This page runs the other way: one definition is made precise in Section 2, and every result afterwards — the limit laws, continuity, the existence theorems, the derivative, and all the differentiation rules — is a consequence of it.
The explorer in Section 7 plays the definition as the game it is. A challenger names \(\varepsilon\); the explorer answers with the \(\delta\) the textbook derivation gives, verifies that it works, and then reports the largest \(\delta\) that works with an exact witness just outside it proving you cannot do better. One preset has no limit at all, and the explorer shows why no answer is possible.
It is exact rational arithmetic throughout. Each function supplies a closed form for the worst case on a neighbourhood, so “does this \(\delta\) work?” is a decided comparison rather than a numerical guess.
\(\lim_{x \to a} f(x) = L\) says the values of \(f\) can be forced as close to \(L\) as anyone demands, by keeping \(x\) close enough to \(a\).
That sentence is already almost precise. The only vague parts are “as close as anyone demands” and “close enough”, and the entire content of the next section is turning those two phrases into quantifiers — in the right order.
Every interesting limit is a \(0/0\) that cannot be evaluated by substitution. The derivative is the clearest case: the difference quotient is undefined at exactly the point of interest. A limit is the machinery that lets a function be studied near a point without being evaluated at it.
\(\lim_{x \to a} f(x) = L\) means: for every \(\varepsilon \gt 0\) there exists \(\delta \gt 0\) such that \[ 0 \lt |x - a| \lt \delta \quad\Longrightarrow\quad |f(x) - L| \lt \varepsilon. \]
A challenger names \(\varepsilon\) — how close to \(L\) the output must get. You answer with \(\delta\) — how close to \(a\) the input will be kept. The limit holds exactly when you can always answer. Losing a single round, one \(\varepsilon\) with no valid \(\delta\), disproves the limit outright.
\(\varepsilon\) comes first and \(\delta\) second, so \(\delta\) is allowed to depend on \(\varepsilon\) — and essentially always does. Swapping them states something entirely different and much stronger. This is the single place where the quantifier work from Tier 0 pays off most directly.
The condition is \(0 \lt |x - a|\), not \(|x - a| \lt \delta\) alone. The point \(a\) is deliberately excluded, because a limit describes the approach and not the arrival. The value \(f(a)\) — or whether \(f\) is defined at \(a\) at all — is irrelevant.
Bound \(|f(x) - L|\) by something involving \(|x - a|\), then solve for \(|x - a|\).
For \(f(x) = 3x + 1\) at \(a = 2\): \(|f(x) - 7| = 3|x - 2|\), so \(\delta = \varepsilon/3\) works — and nothing larger does, since at \(|x - 2| = \varepsilon/3\) the gap is exactly \(\varepsilon\).
When the bound still contains \(x\), restrict first. For \(x^2\) at \(a = 3\): \(|x^2 - 9| = |x - 3||x + 3|\), and insisting \(|x - 3| \lt 1\) makes \(|x + 3| \lt 7\), giving \(\delta = \min(1, \varepsilon/7)\).
The definition asks only that some \(\delta\) works, so a conservative one is not a worse answer — it is simply an answer. For \(1/x\) at \(a = 2\) the textbook \(\min(1, 2\varepsilon)\) is correct, while the exact largest is the more generous \(\frac{4\varepsilon}{1 + 2\varepsilon}\). The explorer reports both.
Uniqueness of Limits uses the definition directly: if a function approached two different values, an \(\varepsilon\) smaller than half their separation would force two disjoint neighbourhoods to overlap.
Negating the definition gives the test. A limit \(L\) fails when there is some \(\varepsilon \gt 0\) for which every \(\delta\) admits a point with \(|f(x) - L| \geq \varepsilon\).
Let \(f(x) = -1\) for \(x \lt 0\) and \(+1\) for \(x \gt 0\), at \(a = 0\).
Every punctured neighbourhood of \(0\), however small, contains points where \(f\) is \(-1\) and points where it is \(+1\). Whatever \(L\) is proposed, it is at least \(1\) away from one of those two values — so \(\varepsilon = 1\) can never be answered, and no limit exists. Shrinking \(\delta\) does not help, which is what makes this a genuine failure rather than a hard case.
Restricting the approach to one side gives \(\lim_{x \to a^{-}}\) and \(\lim_{x \to a^{+}}\). The two-sided limit exists exactly when both exist and agree. The jump above has both one-sided limits, \(-1\) and \(+1\), and no limit — which is precisely the distinction.
Returning to \(\varepsilon\) for every computation would be unbearable. The limit laws are what make routine limits routine: limits of sums, products and quotients are the sums, products and quotients of the limits — provided each limit exists, and a denominator’s limit is non-zero.
The Limit Laws (Sum and Product) proves them from the definition. The sum law splits \(\varepsilon\) in half and takes the smaller of the two deltas; the product law is the one that needs care, because a bound on one factor has to be established before it can be used.
If \(g(x) \leq f(x) \leq h(x)\) near \(a\), and \(g\) and \(h\) share the limit \(L\), then \(f\) has limit \(L\) too.
The Squeeze Theorem matters because some limits admit no algebraic simplification at all. The standard case is
proved on The Fundamental Limit sin(x)/x → 1 by squeezing between \(\cos x\) and \(1\) using a geometric area comparison. Section 9 shows that every trigonometric derivative rests on it.
\(f\) is continuous at \(a\) when \(\lim_{x \to a} f(x) = f(a)\).
That single equation asserts three things at once: the limit exists, \(f(a)\) is defined, and the two are equal. A discontinuity is a failure of any one of them.
\(f(x) = \dfrac{x^2 - 4}{x - 2}\) has limit \(4\) at \(x = 2\) and is not continuous there, because \(f(2)\) does not exist. Repairing it means defining \(f(2) = 4\) — which is why such discontinuities are called removable. The explorer carries this function, and \(\delta = \varepsilon\) answers every challenge exactly.
Proved on Differentiability Implies Continuity: write \(f(a + h) - f(a)\) as the difference quotient times \(h\), and the product tends to \(f^{\prime}(a) \cdot 0 = 0\). The converse fails, as Section 7 shows.
Continuity on an interval buys three theorems that guarantee something exists without producing it. That is worth noticing: each is an existence result, and none is constructive.
A function continuous on \([a, b]\) takes every value between \(f(a)\) and \(f(b)\).
Intermediate Value Theorem. It guarantees a root without locating it — and a bisection search is exactly the act of applying it repeatedly to close in.
A function continuous on a closed, bounded interval attains a maximum and a minimum.
\(1/x\) on \((0, 1]\) is continuous and unbounded; \(x\) on \((0, 1)\) attains neither a maximum nor a minimum. Neither contradicts The Extreme Value Theorem, because in each case the interval is not closed and bounded. A theorem whose hypotheses fail says nothing at all.
If \(f\) is continuous on \([a,b]\), differentiable on the interior, and \(f(a) = f(b)\), then \(f^{\prime}(c) = 0\) somewhere inside — Rolle’s Theorem. Dropping the equal-endpoints condition gives the Mean Value Theorem: some tangent matches the average slope, \(f^{\prime}(c) = \frac{f(b) - f(a)}{b - a}\). Both are proved on Rolle’s Theorem and the Mean Value Theorem, and the proof of Rolle leans on the Extreme Value Theorem.
This is not a new kind of object. It is the definition of Section 2 applied to a function built out of \(f\) — and the quotient is undefined at exactly \(h = 0\), which is precisely the situation a limit exists to handle.
For \(f(x) = x^2\) at \(a = 3\), the quotient is \(\dfrac{(3+h)^2 - 9}{h} = 6 + h\) exactly, for every \(h \neq 0\). The limit is then \(6\) by inspection.
That is the pattern behind every derivative proof on this site: the algebra removes the \(0/0\) first, and only then is a limit taken.
\(|x|\) is continuous at \(0\), but its difference quotient is \(-1\) from the left and \(+1\) from the right. The one-sided limits disagree, so the derivative does not exist. The implication runs one way only.
Choose a function and a challenge \(\varepsilon\). The explorer gives the \(\delta\) the textbook derivation produces and verifies it against the exact worst case on that neighbourhood; then it reports the largest \(\delta\) that works, with a witness point just outside it. The last preset has no limit, and the explorer shows why no \(\delta\) can be given.
Every delta, worst case and witness is computed live in exact rational arithmetic from a closed form for the supremum — nothing is sampled, and nothing is rounded except for display.
Each rule is proved once from the definition, and then never again. That is the return on Section 7.
| Rule | Statement | Proof |
|---|---|---|
| Linearity | \((af + bg)^{\prime} = af^{\prime} + bg^{\prime}\) | The limit laws applied to the quotient |
| Product | \((fg)^{\prime} = f^{\prime}g + fg^{\prime}\) | The Product Rule |
| Quotient | \(\left(\frac{f}{g}\right)^{\prime} = \frac{f^{\prime}g - fg^{\prime}}{g^2}\) | The Quotient Rule |
| Chain | \((f \circ g)^{\prime}(a) = f^{\prime}(g(a))\,g^{\prime}(a)\) | Derivative of the Chain Rule |
| Power | \(\frac{d}{dx}x^{n} = nx^{n-1}\) | The Power Rule by Induction |
The proof adds and subtracts \(f(a + h)g(a)\) to split one difference quotient into two, and the second summand is what that middle term becomes. Knowing where it comes from is what stops the rule being misremembered.
The outer derivative is evaluated at \(g(a)\), the inner one at \(a\). Writing \(f^{\prime}(a)g^{\prime}(a)\) is the most common error in the topic, and it is an error about evaluation points rather than about the shape of the formula.
Induction with the product rule gives positive integers; the quotient rule extends it to negative ones; Derivative of the Square Root Function and the inverse function theorem reach roots; and \(x^{a} = e^{a\ln x}\) with the chain rule reaches every real exponent. Logarithms follow the same route on Derivative of ln(x) and Derivative of ln(ax).
If \(f\) is differentiable and invertible near \(a\) with \(f^{\prime}(a) \neq 0\), then \[ (f^{-1})^{\prime}(f(a)) = \frac{1}{f^{\prime}(a)}. \]
Geometrically the graph of the inverse is the original reflected in the line \(y = x\), so slopes become reciprocals. The Inverse Function Derivative Theorem proves it, and the condition \(f^{\prime}(a) \neq 0\) is exactly what stops the reflected tangent being vertical.
Each is the theorem above plus a Pythagorean identity. Differentiating \(\sin(\arcsin x) = x\) puts \(\cos(\arcsin x)\) in the denominator, and that equals \(\sqrt{1 - x^2}\) — positive, because the range of \(\arcsin\) is restricted to where cosine is non-negative.
arcsin · arccos · arctan, with the arcsin case worked at length in the dedicated tutorial.
Derivatives of sin(x) and cos(x) reduces the difference quotient by the angle sum identity to two limits: \(\lim_{h \to 0}\frac{\sin h}{h} = 1\) and \(\lim_{h \to 0}\frac{\cos h - 1}{h} = 0\). The first is the fundamental trigonometric limit, which is why these derivatives are genuinely harder than a polynomial’s.
\(\lim_{h \to 0}\frac{\sin h}{h} = 1\) holds for radians only; in degrees the limit is \(\pi/180\). Every trigonometric derivative would pick up that factor. That is the real reason calculus uses radians — not tradition, but the derivative formulas.
Differentiating again gives \(f^{\prime\prime}\), which measures how the slope itself is changing.
If \(f^{\prime\prime} \gt 0\) on an interval then \(f\) is convex there: the first derivative is increasing, so the graph curves upward and lies above each of its tangent lines.
The Second-Derivative Test for Convexity proves it, and the Mean Value Theorem of Section 6 is what converts “the derivative is increasing” into a statement about the graph.
\(f^{\prime}(a) = 0\) is necessary for an interior extremum, not sufficient: \(x^3\) at \(0\) has a vanishing derivative and no extremum at all. That gap is why the second-derivative test exists — and why it is inconclusive when \(f^{\prime\prime}(a) = 0\) as well, which is exactly the \(x^3\) case.
Sequences and series apply the same \(\varepsilon\) machinery to infinite sums, and integration turns the limit on an area. Both are Tier 3; the definition on this page is what they are built from.
| Result | Statement | Proof on this site |
|---|---|---|
| Uniqueness of limits | A function cannot approach two different values | Uniqueness of Limits |
| Limit laws | Sums and products of limits are limits of sums and products | The Limit Laws |
| Squeeze | \(g \leq f \leq h\), \(\lim g = \lim h = L \Rightarrow \lim f = L\) | The Squeeze Theorem |
| \(\sin x / x\) | \(\lim_{x \to 0}\frac{\sin x}{x} = 1\), in radians | The Fundamental Limit |
| Differentiable \(\Rightarrow\) continuous | The converse is false | Differentiability Implies Continuity |
| IVT | Continuous on \([a,b]\) hits every intermediate value | Intermediate Value Theorem |
| EVT | Continuous on a closed bounded interval attains a max and a min | The Extreme Value Theorem |
| Rolle and MVT | \(f^{\prime}(c) = 0\), and \(f^{\prime}(c) = \frac{f(b)-f(a)}{b-a}\) | Rolle’s Theorem and the MVT |
| Inverse derivative | \((f^{-1})^{\prime}(f(a)) = 1/f^{\prime}(a)\) | The Inverse Function Derivative Theorem |
| Convexity | \(f^{\prime\prime} \gt 0 \Rightarrow\) convex | The Second-Derivative Test for Convexity |
| Largest \(\delta\) | Sufficient is enough, but the explorer reports the best possible | Explorer, Section 7 |