Continuity Describes Nearby Function Values
Connectedness results earlier in this course used continuity to guarantee intermediate values and preserve connectedness. We now express what continuity means locally: when inputs in the domain approach a point, their function values approach the value assigned at that point. The connection is precise. At an accumulation point of the domain, continuity holds exactly when the limit of the function equals its value there.
A limit concerns values of the function at inputs near a point, but not necessarily at the point itself. Continuity adds the requirement that the function’s assigned value agrees with that nearby behavior. This distinction will let us recognize when changing a function at one point repairs a discontinuity—and when no choice of value can do so.
Limits Relative to a Domain
Let \(E\subseteq\mathbb{R}\), let \(f:E\to\mathbb{R}\), and let \(a\) be an accumulation point of \(E\). Being an accumulation point means every open interval around \(a\) contains a point of \(E\) other than \(a\). This condition ensures there are domain inputs, distinct from \(a\), that can approach \(a\). A limit at \(a\) is defined using those inputs.
The condition \(0<|x-a|\) excludes the input \(a\). Consequently, the limit depends on function values near \(a\), not on the value \(f(a)\). The domain matters too: the inputs approaching \(a\) must belong to \(E\). For example, if \(E=[0,\infty)\), a limit as \(x\) tends to \(0\) through \(E\) uses only nonnegative inputs.
Unlike the definition of a limit, the definition of continuity includes \(x=a\). When \(x=a\), the inequality on function values is satisfied because \(|f(a)-f(a)|=0\). The definitions therefore differ only in the target value: a limit asks whether nearby values approach some \(L\), while continuity asks whether they approach \(f(a)\).
The Limit Criterion for Continuity
Proof. Suppose first that \(f\) is continuous at \(a\). Let \(\varepsilon>0\). By continuity, there is a \(\delta>0\) such that for every \(x\in E\), if \(|x-a|<\delta\), then \(|f(x)-f(a)|<\varepsilon\). This implication applies in particular to inputs satisfying \(0<|x-a|<\delta\). Thus the definition of limit holds with \(L=f(a)\), and \(\lim_{x\to a,\ x\in E}f(x)=f(a)\).
Conversely, suppose \(\lim_{x\to a,\ x\in E}f(x)=f(a)\). Given \(\varepsilon>0\), the definition of limit supplies a \(\delta>0\) such that \(x\in E\) and \(0<|x-a|<\delta\) imply \(|f(x)-f(a)|<\varepsilon\). For an input \(x\in E\) with \(|x-a|<\delta\), either \(x\ne a\), in which case this implication applies, or \(x=a\), in which case \(|f(x)-f(a)|=0<\varepsilon\). This is exactly continuity at \(a\). \(\square\)
The accumulation-point hypothesis is essential to the statement about a limit: our definition of limit applies only when there are domain points other than \(a\) arbitrarily close to \(a\). At an isolated point \(a\in E\), continuity still makes sense and holds automatically. Indeed, some \(\delta>0\) has no points \(x\in E\) with \(0<|x-a|<\delta\), so the continuity condition can only need to be checked at \(x=a\), where it is satisfied.
Worked Example: Continuity of the Square Function at Any Point
Let \(f:\mathbb{R}\to\mathbb{R}\) be \(f(x)=x^2\), and fix \(a\in\mathbb{R}\). We verify that \(\lim_{x\to a}x^2=a^2\). For any \(x\), \[ |x^2-a^2|=|x-a||x+a|. \] If \(|x-a|<1\), then the triangle inequality gives \(|x+a|\leq |x-a|+2|a|<1+2|a|\). Given \(\varepsilon>0\), choose \[ \delta=\min\left\{1,\frac{\varepsilon}{1+2|a|}\right\}. \] Whenever \(|x-a|<\delta\), we have \[ |x^2-a^2|=|x-a||x+a|<\delta(1+2|a|)\leq\varepsilon. \] In fact the final inequality is strict: if \(\delta=\varepsilon/(1+2|a|)\), the strict bound on \(|x+a|\) gives a strict result; if \(\delta=1\), then \(\varepsilon\geq 1+2|a|\) and \(|x-a|(1+2|a|)<1+2|a|\leq\varepsilon\). Therefore the limit is \(a^2=f(a)\), and the limit criterion proves continuity at \(a\).
Changing a Value Can Repair a Discontinuity
Since the limit does not depend on the value at \(a\), it is possible to change the function’s value there without changing the limit. If the limit exists, choosing the value to equal that limit makes the function continuous at \(a\). This is called a removable discontinuity when the original value either differs from the limit or is not assigned.
Worked Example: Choosing the Value at a Removable Discontinuity
Define \(g:\mathbb{R}\to\mathbb{R}\) by \[ g(x)= \begin{cases} \dfrac{x^2-4}{x-2}, & x\ne 2,\\ 4, & x=2. \end{cases} \] For \(x\ne2\), factoring gives \(x^2-4=(x-2)(x+2)\), so \(g(x)=x+2\). The factorization is valid on this part of the definition because \(x-2\ne0\). Hence for \(x\ne2\), \[ |g(x)-4|=|x+2-4|=|x-2|. \] Given \(\varepsilon>0\), choosing \(\delta=\varepsilon\) ensures that \(0<|x-2|<\delta\) implies \(|g(x)-4|<\varepsilon\). Thus \(\lim_{x\to2}g(x)=4=g(2)\), so \(g\) is continuous at \(2\). If instead the function had been assigned \(g(2)=7\), the limit would still be \(4\), but it would not equal the assigned value; the function would not be continuous at \(2\).
A limit can also fail to exist, in which case no value assigned at the point can make the function continuous there. A useful way to detect such failure is to approach the point along two different sequences of domain inputs. If the function values approach different numbers along those sequences, they cannot both approach one common limit.
A Sequential Test for Limits
Proof. Suppose first that \(\lim_{x\to a,\ x\in E}f(x)=L\), and let \((x_n)\) be any sequence in \(E\setminus\{a\}\) converging to \(a\). Given \(\varepsilon>0\), choose \(\delta>0\) from the definition of the limit so that \(x\in E\) and \(0<|x-a|<\delta\) imply \(|f(x)-L|<\varepsilon\). Since \(x_n\to a\), there is an index \(N\) such that \(n\geq N\) implies \(|x_n-a|<\delta\). Also \(x_n\ne a\) for every \(n\), so for \(n\geq N\) we obtain \(|f(x_n)-L|<\varepsilon\). Therefore \(f(x_n)\to L\).
For the reverse implication, suppose the limit is not \(L\). Negating the definition of limit gives an \(\varepsilon_0>0\) such that for every \(\delta>0\), there is an \(x\in E\) with \(0<|x-a|<\delta\) and \(|f(x)-L|\geq\varepsilon_0\). For each positive integer \(n\), use this statement with \(\delta=1/n\) to choose \(x_n\in E\) such that \[ 0<|x_n-a|<\frac{1}{n} \quad\hbox{and}\quad |f(x_n)-L|\geq\varepsilon_0. \] The bound \(|x_n-a|<1/n\) implies \(x_n\to a\), but the function values do not converge to \(L\), because they remain at least \(\varepsilon_0\) from \(L\). This contradicts the assumed sequential property. Hence the limit must equal \(L\). \(\square\)
Combining this result with the Limit Criterion for Continuity gives a sequential test for continuity: at an accumulation point \(a\in E\), \(f\) is continuous if and only if every sequence in \(E\) converging to \(a\) has function values converging to \(f(a)\). Sequences that equal \(a\) cause no difficulty, since their function values equal \(f(a)\). The test is especially useful for disproving continuity: one sequence whose function values fail to approach \(f(a)\) is enough.
Worked Example: Oscillation Prevents a Limit
Define \(h(x)=\sin(1/x)\) for \(x\ne0\). Consider the two sequences \[ x_n=\frac{1}{\pi/2+2\pi n}, \qquad y_n=\frac{1}{3\pi/2+2\pi n}, \] for positive integers \(n\). Both sequences consist of nonzero real numbers and converge to \(0\), since their positive denominators tend to infinity. Substitution gives \[ h(x_n)=\sin(\pi/2+2\pi n)=1 \quad\hbox{and}\quad h(y_n)=\sin(3\pi/2+2\pi n)=-1. \] The first sequence of function values converges to \(1\), while the second converges to \(-1\). If \(h(x)\) had a limit as \(x\) tends to \(0\), the sequential test would require both sequences of function values to converge to that same limit. Since \(1\ne-1\), the limit does not exist. Assigning any value to \(h(0)\) cannot make the function continuous at \(0\).
What the Limit Viewpoint Clarifies
The limit criterion separates two questions that are easy to confuse. First, what do the function values do near \(a\)? This is the limit question. Second, does the assigned value \(f(a)\) match that behavior? This is the continuity question. A function may have a limit at a point and still fail to be continuous there, as changing the assigned value in the removable-discontinuity example demonstrated.
There is also a useful domain-related caution. Limits and continuity are always understood relative to the function’s domain. When the domain omits nearby inputs, the limit tests only the inputs that remain. At an endpoint of an interval, for example, the limit through the domain is one-sided in effect; it does not require function values on the other side of the endpoint. And at an isolated domain point, continuity is automatic even though the limit, under the definition used here, is not defined there.
Check Your Understanding
Use the definitions and the sequential test to distinguish nearby behavior from the value assigned at the point.
- Why does the definition of a limit exclude the input \(x=a\), and why does the continuity definition include it?
- State the exact condition that makes continuity at an accumulation point equivalent to a limit statement.
- Suppose \(\lim_{x\to a}f(x)=L\), but \(f(a)\ne L\). What does this tell you about continuity at \(a\)?
- How can two sequences approaching the same point show that a function has no limit there?
- Why is every function continuous at an isolated point of its domain, even though the limit there may not be defined?