Continuity Proof Mastery I

Planning a Continuity Proof

The Open-Set Criterion for Continuity gives a useful way to recognize continuous functions, but many proofs still come down to controlling the change in a function’s value near a chosen input. The difficult part is often not writing the epsilon-delta definition; it is finding an estimate that turns a desired output bound into a manageable condition on the input. This tutorial develops a proof strategy for that task and applies it to two operations that arise frequently in formulas: taking a square root and taking a maximum or minimum.

A reliable proof begins by fixing a point \(a\) in the domain and an output tolerance \(\varepsilon>0\). Next, compare the quantity \(|f(x)-f(a)|\) with an expression involving \(|x-a|\). The estimate should make the final choice of \(\delta\) visible. Sometimes a proof also needs separate cases: for example, a formula involving a square root behaves differently at the endpoint zero than it does at a positive point. Accounting for that distinction is part of the proof, not an optional refinement.

The estimate is the central part of the argument. A choice of \(\delta\) written down without showing why it works is not a complete proof. Conversely, once a valid estimate has been found, the choice often becomes straightforward. We will see both patterns below.

Continuity of the Square-Root Function

The square root is defined only for nonnegative inputs, and the point zero requires care. At a positive input, a difference of square roots can be rewritten by multiplying by the sum of the roots. At zero, that denominator argument is unavailable: the sum of the roots has no fixed positive lower bound as the input approaches zero. We therefore prove continuity at positive points and at zero separately.

Proof. Fix \(a\in[0,\infty)\) and \(\varepsilon>0\). First suppose \(a>0\). For any \(x\geq0\),

because \(\sqrt{x}+\sqrt{a}\geq\sqrt{a}>0\). Choose \(\delta=\varepsilon\sqrt{a}\), which is positive. If \(x\in[0,\infty)\) and \(|x-a|<\delta\), the estimate gives

Thus \(s\) is continuous at every positive \(a\).

Now suppose \(a=0\). Choose \(\delta=\varepsilon^2>0\). If \(x\in[0,\infty)\) and \(|x-0|<\delta\), then \(x\geq0\) and \(x<\varepsilon^2\). Since both sides are nonnegative, taking square roots gives \(\sqrt{x}<\varepsilon\). Hence

Therefore \(s\) is continuous at zero as well, completing the proof. \(\square\)

The endpoint case illustrates a useful proof habit: do not use an algebraic estimate outside the range where its denominator is safely controlled. At \(a>0\), the denominator \(\sqrt{x}+\sqrt{a}\) is at least \(\sqrt{a}\). At \(a=0\), that lower bound would be zero, so it cannot yield a useful estimate. The separate choice \(\delta=\varepsilon^2\) handles the endpoint directly.

Maximum and Minimum of Continuous Functions

A formula involving a maximum or minimum may look piecewise: one expression is selected in one region, and another expression is selected elsewhere. It can be awkward to prove continuity by dividing the domain into regions and then checking every point where the selected expression changes. A better approach is to express maximum and minimum using absolute value. Since the absolute-value function is continuous and the algebra of continuous functions is already established, this gives a short proof that works even at the switching points.

Proof. For any real numbers \(u\) and \(v\),

To verify the first identity, if \(u\geq v\), then \(|u-v|=u-v\), so \((u+v+|u-v|)/2=(u+v+u-v)/2=u\). If \(u<v\), then \(|u-v|=v-u\), so \((u+v+|u-v|)/2=(u+v+v-u)/2=v\). In each case the expression equals the larger number. For the second identity, if \(u\geq v\), then \((u+v-|u-v|)/2=(u+v-u+v)/2=v\); if \(u<v\), then \((u+v-|u-v|)/2=(u+v-v+u)/2=u\). It equals the smaller number in either case.

Applying these identities with \(u=f(x)\) and \(v=g(x)\) gives

Because \(f\) and \(g\) are continuous, their difference \(f-g\) is continuous. The absolute value of a continuous function is continuous, and sums and constant multiples of continuous functions are continuous. Therefore both displayed expressions define continuous functions on \(E\). These are exactly \(h\) and \(k\), so the maximum and minimum functions are continuous. \(\square\)

What Makes These Proofs Reliable?

These examples use two complementary methods. For the square root, an explicit estimate reveals a valid \(\delta\). The estimate differs at positive inputs and at zero, so the proof treats those cases separately. For maxima and minima, an identity replaces a piecewise-looking operation with sums, differences, and absolute values. Previously established continuity results then do the remaining work.

A common pitfall is to assume that a piecewise description automatically proves continuity because each formula is continuous on its own region. That only checks behavior away from the points where the selected formula changes. A proof must also account for each joining point. The maximum-and-minimum theorem avoids this gap: it applies at every point, including ties. Alternatively, a direct piecewise proof must show that the expressions agree at each join and verify continuity there.

Another pitfall is choosing \(\delta\) before deriving an estimate. In the square-root example, the denominator after rationalization is at least \(3\), which suggests that \(|x-4|<3\varepsilon\) will suffice. The resulting inequality then verifies the choice. In other problems the estimate may require first restricting \(x\) to a neighborhood of \(a\); if so, the final \(\delta\) can be chosen as the smaller of the radii that enforce the needed conditions. What matters is that every restriction is stated and that the final choice guarantees all of them.