Pointwise Limits Need Not Preserve Continuity
A sequence of continuous functions can converge at every point of its domain and still have a discontinuous limit. The reason is that pointwise convergence controls the values at each fixed input separately; it does not by itself control values across a whole neighborhood. We will compute examples, then give a sequence-based test that identifies exactly what a discontinuity looks like.
Unless another domain is specified, \(\mathbb{R}\) has its standard topology and intervals have the subspace topology. A function is continuous at \(x_0\) if for every \(\varepsilon>0\) there is a \(\delta>0\) such that \(x\) in its domain and \(|x-x_0|<\delta\) imply \(|f(x)-f(x_0)|<\varepsilon\).
Even when every \(f_n\) is continuous, pointwise convergence only gives a bound at a fixed point \(x\). It does not promise that the same \(N\) works for every \(x\) in a neighborhood. That distinction is central: continuity asks for control as the input varies near \(x_0\), while pointwise convergence fixes the input before choosing \(N\).
The Power Sequence on the Unit Interval
Worked Example: \(x^n\) on \([0,1]\)
Let \(f_n:[0,1]\to\mathbb{R}\) be \(f_n(x)=x^n\). Each \(f_n\) is continuous because it is a polynomial restricted to \([0,1]\). We find its pointwise limit by fixing \(x\) first.
If \(0\le x<1\), then \(x^n\to0\): for \(x=0\), every term is \(0\); for \(0<x<1\), the powers of a fixed number strictly between \(0\) and \(1\) tend to \(0\). At \(x=1\), every term is \(1^n=1\), so the limit is \(1\). Thus the pointwise limit \(f\) is
For example, at \(x=\tfrac12\), the first three values are \((\tfrac12)^1=\tfrac12\), \((\tfrac12)^2=\tfrac14\), and \((\tfrac12)^3=\tfrac18\), and these tend to \(0\), as the formula for \(f\) predicts. At the endpoint \(x=1\), the values are \(1,1,1,\ldots\), so \(f(1)=1\).
The function \(f\) is continuous at every \(x_0<1\): choose \(\delta=(1-x_0)/2>0\). If \(x\in[0,1]\) and \(|x-x_0|<\delta\), then \(x<x_0+\delta=(1+x_0)/2<1\). Hence \(f(x)=f(x_0)=0\). At \(x_0=1\), take \(x_k=1-\tfrac1k\) for \(k\ge2\). Then \(x_k\in[0,1)\), \(x_k\to1\), and \(f(x_k)=0\) for every \(k\), whereas \(f(1)=1\). So \(f(x_k)\) does not tend to \(f(1)\), and \(f\) is discontinuous at \(1\).
Check: the pointwise formula gives \(f(x_k)=0\) because \(1-\tfrac1k<1\), and it gives \(f(1)=1\). The sequence therefore confirms, rather than merely suggests, the discontinuity.
The failure here happens at the right endpoint. The next example places a discontinuity at the left endpoint instead, with a different pointwise formula.
Worked Example: A Limit That Jumps at Zero
On \([0,1]\), define \(g_m(x)=\dfrac{mx}{1+mx}\) for \(m\in\mathbb{N}\). The denominator satisfies \(1+mx\ge1\), so it never vanishes on the domain; each \(g_m\) is continuous there.
At \(x=0\), \(g_m(0)=0\) for every \(m\), so the limit is \(g(0)=0\). If \(x>0\), divide numerator and denominator by \(mx\) to get \(g_m(x)=\dfrac{1}{1+1/(mx)}\). Since \(1/(mx)\to0\) for this fixed \(x\), \(g_m(x)\to1\). Therefore
At \(x=\tfrac12\), direct substitution gives \(g_m(\tfrac12)=\dfrac{m/2}{1+m/2}=\dfrac{m}{m+2}\). For \(m=1\) this is \(\tfrac13\), for \(m=2\) it is \(\tfrac12\), and as \(m\to\infty\) it tends to \(1=g(\tfrac12)\). To check the discontinuity, take \(x_k=\tfrac1k\) for \(k\in\mathbb{N}\). These points lie in \((0,1]\) and tend to \(0\); the limit function has \(g(x_k)=1\) for every \(k\), while \(g(0)=0\). Thus \(g\) is not continuous at \(0\).
On \((0,1]\), \(g\) is constant, so it is continuous at every point in that subspace. The problem is specifically the value at \(0\): values of \(g\) at positive inputs remain \(1\), however close those inputs are to \(0\).
A Sequence Test for Continuity
The examples use sequences of inputs to expose a mismatch between the limiting values and the value at the point. The following criterion says this method detects every failure of continuity for maps from metric spaces to \(\mathbb{R}\).
Proof. Suppose first that \(f\) is continuous at \(x_0\), and let \(x_k\to x_0\). For any \(\varepsilon>0\), continuity gives \(\delta>0\) such that \(d(x,x_0)<\delta\) implies \(|f(x)-f(x_0)|<\varepsilon\). Since \(x_k\to x_0\), there is \(K\) such that \(k\ge K\) implies \(d(x_k,x_0)<\delta\). Therefore \(k\ge K\) implies \(|f(x_k)-f(x_0)|<\varepsilon\), proving \(f(x_k)\to f(x_0)\).
Conversely, suppose every sequence converging to \(x_0\) has images converging to \(f(x_0)\). We prove continuity by contradiction. If \(f\) is not continuous at \(x_0\), the negation of the \(\varepsilon\)-\(\delta\) definition gives an \(\varepsilon_0>0\) such that for every \(\delta>0\) there is \(x\in X\) with \(d(x,x_0)<\delta\) but \(|f(x)-f(x_0)|\ge\varepsilon_0\). For each \(k\in\mathbb{N}\), apply this statement with \(\delta=1/k\) to choose \(x_k\) satisfying \(d(x_k,x_0)<1/k\) and \(|f(x_k)-f(x_0)|\ge\varepsilon_0\). The inequalities \(d(x_k,x_0)<1/k\) imply \(x_k\to x_0\), but the image values never get within \(\varepsilon_0\) of \(f(x_0)\). This contradicts the assumed sequence property. Hence \(f\) is continuous at \(x_0\). \(\square\)
In particular, a function on a metric space is discontinuous at \(x_0\) exactly when there is some sequence \(x_k\to x_0\) for which \(f(x_k)\not\to f(x_0)\). The sequences in the two examples verify the failure directly. The metric hypothesis matters for the proof: it lets us choose inputs within distance \(1/k\) when continuity fails.
When a Pointwise Limit Does Stay Continuous
Pointwise limits do not always lose continuity. One useful sufficient condition is a shared Lipschitz bound. A function \(h:X\to\mathbb{R}\) on a metric space \((X,d)\) is \(L\)-Lipschitz, for a constant \(L\ge0\), if \(|h(x)-h(y)|\le Ld(x,y)\) for all \(x,y\in X\).
Proof. Fix any \(x,y\in X\). Pointwise convergence gives \(f_n(x)\to f(x)\) and \(f_n(y)\to f(y)\). Thus \(|f_n(x)-f_n(y)|\to|f(x)-f(y)|\), because subtraction and absolute value preserve limits of real sequences. For every \(n\), the Lipschitz assumption gives \(|f_n(x)-f_n(y)|\le Ld(x,y)\). Taking limits in this inequality yields \(|f(x)-f(y)|\le Ld(x,y)\). This holds for every \(x,y\), so \(f\) is \(L\)-Lipschitz. If \(L>0\), given \(\varepsilon>0\) choose \(\delta=\varepsilon/L\); then \(d(x,y)<\delta\) implies \(|f(x)-f(y)|\le Ld(x,y)<\varepsilon\). If \(L=0\), the inequality gives \(|f(x)-f(y)|=0\) for all \(x,y\), so \(f\) is constant and continuous. If \(X\) is empty, continuity is vacuous. \(\square\)
Worked Example: A Continuous Pointwise Limit
On \([0,1]\), let \(h_n(x)=x/n\). Each \(h_n\) is continuous. For every fixed \(x\in[0,1]\), \(0\le x/n\le1/n\), and \(1/n\to0\), so \(h_n(x)\to0\). The pointwise limit is the constant function \(h(x)=0\), which is continuous.
This also illustrates the proposition: for \(x,y\in[0,1]\), \(|h_n(x)-h_n(y)|=|x-y|/n\le |x-y|\), so every \(h_n\) is \(1\)-Lipschitz with the same bound. The limit is \(0\)-Lipschitz, since \(|h(x)-h(y)|=0\) for every pair \(x,y\).
Check: at \(x=\tfrac12\), \(h_n(\tfrac12)=\tfrac{1}{2n}\), which tends to \(0=h(\tfrac12)\). At \(x=1\), \(h_n(1)=1/n\), which also tends to \(0=h(1)\). These substitutions agree with the pointwise formula.
The Quantifier Pitfall
Common mistake. “Every \(f_n\) is continuous and \(f_n\to f\) pointwise, so \(f\) is continuous.” This inference is false. For the power example, every \(f_n(x)=x^n\) is continuous on \([0,1]\), but the limit has \(f(1)=1\) and \(f(1-\tfrac1k)=0\) for all \(k\ge2\), even though \(1-\tfrac1k\to1\).
The quantifiers explain the error. Pointwise convergence says that for each fixed \(x\), an index \(N\) can be chosen to make \(f_n(x)\) close to \(f(x)\). Continuity at \(x_0\), by contrast, must control \(f(x)\) for inputs \(x\) throughout a sufficiently small neighborhood of \(x_0\). The allowed dependence of \(N\) on the fixed input supplies no such neighborhood-wide control on its own.
Check Your Understanding
For each question, state the domain topology when it is relevant and distinguish the value at a point from values approaching that point.
- Find the pointwise limit of \(x^n\) on \([0,1)\). Is that limit continuous on this domain?
- For the limit of \(x^n\) on \([0,1]\), give a sequence witnessing discontinuity at \(1\), and compute both the sequence of inputs and the sequence of function values.
- For \(g_m(x)=\dfrac{mx}{1+mx}\) on \([0,1]\), compute the pointwise limit at \(x=0\) and at a fixed \(x>0\).
- State the sequential criterion for continuity at a point of a metric space.
- Why does a common Lipschitz constant for a pointwise-convergent sequence prevent its limit from being discontinuous?