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A limit taken one point at a time, the moving bump that defeats it, and the interchange of limits that fails.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to compute the pointwise limit of a sequence of functions, find the set on which it exists, and produce the standard examples showing that continuity, the value of the integral and boundedness are all lost. You will also be able to state the failure as an interchange of two limits, which is the form every theorem in the rest of this unit answers.
The limit of a sequence of numbers, continuity, and the integral. This unit puts a whole function in the place of a number, and the first question is which of those three survive the change.
$(f_n)$ is a sequence of functions on a set $S$. It converges pointwise to $f$ when, for every $x \in S$, the numbers $f_n(x)$ converge to $f(x)$. Written out: for every $x$ and every $\varepsilon > 0$ there is $N$ with $|f_n(x) - f(x)| < \varepsilon$ for all $n \ge N$ — and $N$ may depend on $x$ as well as on $\varepsilon$.
Pointwise convergence is the weakest reasonable notion: the domain is examined one point at a time, and the points never have to agree about anything.
That weakness has a cost, and it is not small.
Continuity is lost. $f_n(x) = x^n$ on $[0,1]$: each term is a polynomial, and the limit is $0$ below $1$ and $1$ at $1$.
The integral is lost. $f_n(x) = n$ on $(0, 1/n)$ and $0$ elsewhere: the limit is identically $0$, so $\int f = 0$, while $\int f_n = 1$ for every $n$. The limit of the integrals is $1$ and the integral of the limit is $0$.
The derivative is lost. $f_n(x) = \frac{\sin(nx)}{\sqrt{n}}$ converges uniformly to $0$, and $f_n'(x) = \sqrt{n}\cos(nx)$ converges to nothing.
What the failure is. Continuity of the limit at $c$ says $$\lim_{x \to c}\lim_{n \to \infty} f_n(x) = \lim_{n \to \infty}\lim_{x \to c} f_n(x),$$ an exchange of two limits. Pointwise convergence gives no licence to exchange them, and the examples above are that licence being absent.
The rest of this unit is about a hypothesis strong enough to grant it.
Another way: picture
Watch the graphs of $x^n$ on the unit interval. Each is a smooth curve from the origin to the corner at height one. As $n$ grows the curve is pressed flat against the axis and then leaps up at the very last moment. Every individual point ends up where it should; the shape of the graph does not settle at all.
Another way: steps
To find a pointwise limit:
Step 4 is a new question, not a continuation of step 3.
Let $f_n(x) = \dfrac{nx}{1 + n^2x^2}$ on $[0,1]$.
Pointwise. For fixed $x > 0$, divide top and bottom by $n^2$: the limit is $0$. At $x = 0$ every term is $0$. So $f_n \to 0$ pointwise, and the limit is as well behaved as anything could be.
And yet. At $x = 1/n$ the value is $\dfrac{1}{2}$, for every $n$. So $\sup_x |f_n(x) - 0| = \dfrac{1}{2}$ always: the largest gap between a term and the limit never shrinks at all.
Nothing contradicts anything. Each point is eventually close to the limit, and the place where the terms are still far away moves steadily to the left, escaping any fixed point but never disappearing. That moving bump is the mechanism behind every failure in this lesson, and measuring it — with the supremum of the gap — is exactly what the next lesson does.
| Property of the terms | Kept by a pointwise limit | Kept by a uniform limit |
|---|---|---|
| continuity | no | yes |
| integrability, with the value | no | yes |
| differentiability, with the derivative | no | no |
| boundedness | no | yes |
The third row is the awkward one and will need a separate theorem with a hypothesis on the derivatives rather than on the functions.
The fourth is worth a moment: $f_n(x) = \min(n, 1/x)$ on $(0,1]$ is bounded for each $n$ and converges pointwise to $1/x$, which is not. Even boundedness, the least demanding property in the table, is not inherited.
That the limit inherits the terms' properties. It inherits none of them. Every entry in the first column of the table above is a counterexample away.
That a discontinuous limit means something went wrong. Nothing went wrong. The limit of $x^n$ is genuinely discontinuous, and the sequence genuinely converges at every point.
That the index can be chosen once. $N$ depends on $x$ as well as on $\varepsilon$, and in the interesting examples it must: for $x^n$ and $\varepsilon = 1/2$, the index needed grows without bound as $x$ approaches $1$.
That the two iterated limits are the same question. Exchanging them is a claim, and the whole unit is about when it is a true one.
Let $f_n(x) = x^n$ on $[0,1]$. Fix $x < 1$: the powers of a number below one tend to $0$.
Fix the point, then take the limit.
At $x = 1$ every term is $1$, so the limit there is $1$.
The end point is its own separate sequence.
The limit function is $0$ below $1$ and $1$ at $1$: a jump, from a sequence of polynomials.
Continuity lost, with nothing gone wrong.
Let $f_n$ be $n$ on the interval from $0$ to $1/n$ and $0$ elsewhere on $[0,1]$.
A tall thin spike, moving and narrowing.
For fixed $x > 0$, once $n > 1/x$ the value is $0$, so the pointwise limit is $0$ everywhere.
Every point is eventually outside the spike.
But $\int_0^1 f_n = 1$ for every $n$, while $\int_0^1 f = 0$. The limit of the integrals is not the integral of the limit.
Area that escapes the limit entirely.
For $0 \le x < 1$ the numerator tends to $0$ and the denominator to $1$, so the limit is $0$.
Split the domain by the size of the point.
At $x = 1$ every term is $1/2$; for $x > 1$ divide top and bottom by $x^n$ to get a limit of $1$.
Two more cases, each a separate sequence.
So the limit is $0$, then $1/2$, then $1$: a function with a jump and an isolated middle value, from a sequence of continuous functions. Three cases, three limits, and the limit function's shape was not visible in any single one of them.
Each sequence below is taken on the interval from $0$ to $1$. Match it to what its pointwise limit does.
| the limit is discontinuous, though every term is continuous | the limit is zero, and the largest gap shrinks to zero | the limit is zero, and the largest gap never shrinks | |
|---|---|---|---|
| $f_n(x) = x^n$ | |||
| $f_n(x) = x/n$ | |||
| $f_n(x) = \dfrac{nx}{1 + n^2x^2}$ | |||
| $f_n(x) = \dfrac{1}{6n}$ |
Let $f_n(x) = \dfrac{nx}{1 + nx^2}$. What is the pointwise limit at $x = 9$?
Answer:
For which real $x$ does the sequence $f_n(x) = \left(\dfrac{x}{5}\right)^{\!n}$ converge as $n$ grows?
This task has no paper form; do it on a device.
Put the steps of finding a pointwise limit and deciding whether it inherits continuity into order.
Number the steps in order (write the number in the box):
Here is an argument about the powers $f_n(x) = x^n$ on the interval from $0$ to $1$. Mark the one step that does not follow.
This task has no paper form; do it on a device.
For $f_n(x) = x^n$ on the interval from $0$ to $1$, compare $\lim_{x \to 1^-}\lim_{n} f_n(x)$ with $\lim_{n}\lim_{x \to 1^-} f_n(x)$. What does the comparison show?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Each sequence below is taken on the interval from $0$ to $1$. Match it to what its pointwise limit does.
| the limit is discontinuous, though every term is continuous | the limit is zero, and the largest gap shrinks to zero | the limit is zero, and the largest gap never shrinks | |
|---|---|---|---|
| $f_n(x) = x^n$ | |||
| $f_n(x) = x/n$ | |||
| $f_n(x) = \dfrac{nx}{1 + n^2x^2}$ | |||
| $f_n(x) = \dfrac{1}{5n}$ |
You can find a pointwise limit and its domain, and give an example in which continuity or the integral is lost. Say in your own words why the index must be allowed to depend on the point. Next: the stronger hypothesis that repairs most of this.
10. Your turn: the pointwise limit of $f_n(x) = \dfrac{x^n}{1 + x^n}$ on $[0, \infty)$, step 3