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The limit agreeing with the value, the kinds of break, and moving a continuous function inside a limit.
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 decide whether a function is continuous at a point by checking all three conditions, name the kind of break when it is not, and repair a removable one. You will also be able to use the sequential criterion in both directions: with one sequence to disprove continuity, and forwards to move a continuous function inside a limit, which is the step every later interchange argument depends on.
The epsilon-delta definition of a limit and the sequential criterion. Continuity adds one requirement to a limit — that it equal the value — and that requirement is what makes every theorem of this unit possible.
A break is removable when the one-sided limits exist and agree but differ from the value, a jump when they exist and disagree, and essential when one of them does not exist. The sequential criterion restates continuity as the preservation of limits of sequences.
$f$ is continuous at $c$ when $\lim_{x \to c} f(x) = f(c)$. Written out: for every $\varepsilon > 0$ there is $\delta > 0$ with $|f(x) - f(c)| < \varepsilon$ whenever $|x - c| < \delta$.
Two differences from a limit are worth noticing. The condition $0 < |x - c|$ is gone, because at $x = c$ the conclusion holds anyway. And the target is not a free letter $L$ but the value $f(c)$, so three things are being asserted at once: the limit exists, the value exists, and they agree.
The sequential criterion. $f$ is continuous at $c$ exactly when $f(x_n) \to f(c)$ for every sequence $x_n \to c$. In the useful direction: a continuous function may be moved inside a limit.
What is continuous. Sums, products and quotients of continuous functions are continuous where defined, by the algebra of limits; compositions are continuous, by the sequential criterion; so polynomials, rational functions off their poles, roots, and the trigonometric and exponential functions are continuous. Almost nothing has to be checked from the definition once these are in hand.
Kinds of break. If the one-sided limits exist and agree but differ from the value, the break is removable — redefining one value repairs it. If they exist and disagree, it is a jump. If one fails to exist, the break is essential.
Another way: picture
The graph can be drawn near $c$ without lifting the pen. That picture is a good guide and a bad definition: it is a statement about a whole neighbourhood, whereas continuity is a statement at a point, and there are functions continuous at exactly one point that no pen could draw.
Another way: steps
To decide continuity at $c$:
All three yes means continuous. To disprove it, one sequence $x_n \to c$ with $f(x_n) \not\to f(c)$ is enough.
Suppose $f$ is continuous at $c$ and $x_n \to c$. Given $\varepsilon$, continuity gives $\delta$; convergence gives $N$ with $|x_n - c| < \delta$ for $n \ge N$; so $|f(x_n) - f(c)| < \varepsilon$ for $n \ge N$. Two definitions, composed.
The converse is proved by contraposition: if continuity fails, some $\varepsilon_0$ has no $\delta$, so for each $n$ there is $x_n$ within $1/n$ of $c$ with $|f(x_n) - f(c)| \ge \varepsilon_0$. That sequence converges to $c$ and its images do not converge to $f(c)$.
The criterion is what carries unit 2 across the bridge. The algebra of limits for functions is the algebra of limits for sequences plus this; the continuity of a composition is a one-line consequence; and identifying the limit of a recursive sequence by taking limits in the recursion is this theorem, used without comment.
| Function | Continuous at | Discontinuous at |
|---|---|---|
| $x^2$ | every point | nowhere |
| $\lfloor x \rfloor$ | every non-integer | every integer (jumps) |
| $\operatorname{sgn}(x)$ | every $x \ne 0$ | $0$ (jump) |
| $x \cdot \mathbf{1}_{\mathbb{Q}}(x)$ | $0$ only | every other point |
| $\mathbf{1}_{\mathbb{Q}}(x)$ | nowhere | everywhere |
The fourth row is worth dwelling on: a function continuous at exactly one point. Near $0$ the values are squeezed between $0$ and $|x|$, so the limit is $0$, which is the value; anywhere else the rationals and irrationals give sequences with different image limits. It is a reminder that continuity is defined pointwise and that the pen-and-paper picture is a picture of something else.
It is not 'the graph can be drawn without lifting the pen'. That is a statement about an interval, and the indicator of the rationals scaled by $x$ shows the two come apart.
It is not 'the limit exists'. The limit existing and the value existing is two conditions; continuity is three, and the third is the equality.
One sequence does not prove it. A single sequence with the right image limit is consistent with discontinuity — the criterion quantifies over all of them. One sequence can only disprove it.
It does not survive every operation. A quotient is continuous only where the denominator is non-zero, and a limit of continuous functions need not be continuous at all — which is the subject of the last unit of this course.
Let $f(x) = \dfrac{\sin x}{x}$ for $x \ne 0$, undefined at $0$.
The value is missing; the limit may not be.
The limit as $x \to 0$ is $1$, so the break is removable.
Limit exists, value does not.
Defining $f(0) = 1$ makes it continuous at $0$, and no other value does.
The repair is unique when it exists.
Let $f(x) = \operatorname{sgn}(x)$, with $f(0) = 0$. Is it continuous at $0$?
The value exists; the question is the limit.
Take $x_n = 1/n \to 0$. Then $f(x_n) = 1$ for every $n$, so $f(x_n) \to 1$.
One sequence, with computable images.
But $f(0) = 0 \ne 1$, so the criterion fails and $f$ is not continuous at $0$.
One sequence is enough to disprove.
At $0$: every value satisfies $|f(x)| \le |x|$, so the squeeze gives limit $0$, which is $f(0)$.
Bound the whole function, then squeeze.
At any $c \ne 0$: rationals near $c$ give images near $c$, and irrationals near $c$ give images $0$.
Two sequences with different image limits.
So it is continuous at $0$ and nowhere else. The graph is two dusts of points and could not be drawn by any pen, which is the clearest evidence that the pen picture is not the definition.
Match each function to its behaviour at the point $6$.
| continuous there | a removable break | a jump | an essential break | |
|---|---|---|---|---|
| $\dfrac{x^2 - 6^2}{x - 6}$, undefined at $6$ | ||||
| $\lfloor x \rfloor$ at $6$ | ||||
| $\sin\!\left(\dfrac{1}{x - 6}\right)$, with value $0$ at $6$ | ||||
| $x^2$ at $6$ |
Let $f(x) = \dfrac{x^2 - 3^2}{x - 3}$ for $x \ne 3$, and $f(3) = k$. For which $k$ is $f$ continuous at $3$?
Answer:
Put the steps of a proof that a function is discontinuous at $4$, by the sequential criterion, into order.
Number the steps in order (write the number in the box):
Build the proof that a function continuous at $c$ with $f(c) > 0$ is positive on some interval about $c$.
This task has no paper form; do it on a device.
Let $g(x) = \dfrac{x^2 - 3^2}{x - 3}$ for $x \ne 3$, with $g(3) = 0$. Here is an argument that $g$ is continuous at $3$. Mark the one step that does not follow.
This task has no paper form; do it on a device.
A sequence satisfies $x_n \to 3$, and $f$ is continuous at $3$. What follows?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Match each function to its behaviour at the point $8$.
| continuous there | a removable break | a jump | an essential break | |
|---|---|---|---|---|
| $\dfrac{x^2 - 8^2}{x - 8}$, undefined at $8$ | ||||
| $\lfloor x \rfloor$ at $8$ | ||||
| $\sin\!\left(\dfrac{1}{x - 8}\right)$, with value $0$ at $8$ | ||||
| $x^2$ at $8$ |
You can check continuity at a point, classify a break, and use the sequential criterion to disprove continuity or to exchange a limit with a function. Say in your own words why the limit existing is not enough. Next: the first theorem that needs continuity on a whole interval rather than at a point.
10. Your turn: where is $f(x) = x \cdot \mathbf{1}_{\mathbb{Q}}(x)$ continuous?, step 3