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The three conditions, the three kinds of break, and the two existence theorems a closed interval buys.
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 test continuity at a point against all three of its conditions, classify a discontinuity as removable, jump or infinite from the one-sided limits, repair a removable one, choose a constant that makes a piecewise function continuous at its junction, and state the intermediate and extreme value theorems with every hypothesis — and say, for each hypothesis, the counterexample that shows it cannot be dropped.
$f$ is continuous at $a$ when all three of these hold:
Stating them separately is the point. The last lesson spent its time on functions where (2) held and (1) did not, or where both held and (3) failed, and each of those is a different repair job.
The three kinds of break.
| Kind | What the limits do | Repairable? |
|---|---|---|
| removable | the two-sided limit exists | yes — assign it as the value |
| jump | both one-sided limits exist and differ | no — no single value joins them |
| infinite | the function is unbounded near $a$ | no — there is no limit to match |
Continuity on an interval means continuity at each of its points, with one-sided continuity at a closed endpoint.
Everything built from continuous pieces is continuous: sums, products, quotients (where the denominator does not vanish) and composites, $\lim_{x \to a} f(g(x)) = f\bigl(\lim_{x \to a} g(x)\bigr)$ when $f$ is continuous at that limit. That last one is what lets a limit be pushed inside a continuous function, and it is used constantly without being named.
What continuity buys. Two existence theorems, both for $f$ continuous on a closed bounded interval $[a, b]$:
A theorem is its hypotheses. Quoting the conclusion of one whose hypotheses you have not checked is not a proof of anything — and in this course it is the single most common way a correct-looking argument turns out to be wrong.
Another way: picture
Continuity is the graph you can draw without lifting the pen — accurate enough for the three breaks. A removable break is a single hole: the pen was lifted for an instant and put down in the same place. A jump is a step: lifted and put down somewhere else. An infinite break is the pen running off the page. And the intermediate value theorem is the observation that an unlifted pen going from below a line to above it must have crossed it.
Another way: steps
To classify a break at $a$:
The IVT is an existence theorem, and reading more into it is the standard error. It says a $c$ exists. It does not say how many, where, or how to find one.
Used properly it is the licence behind root-finding: to show $x^3 - x - 1 = 0$ has a solution, note $f(1) = -1 < 0$ and $f(2) = 5 > 0$, observe that a polynomial is continuous on $[1,2]$, and conclude a root lies between. Bisection then finds it — by repeatedly applying the same theorem to half the interval — and Newton's method later finds it faster. The theorem supplies the guarantee that there is something to find, and nothing else.
Used improperly it is quoted with a hypothesis missing. The function must be continuous on the closed interval, and $N$ must lie between the two endpoint values. $f(x) = 1/x$ on $[-1, 1]$ has $f(-1) = -1$ and $f(1) = 1$ and never takes the value $0$: the theorem does not apply, because $f$ is not continuous at $0$, and the failure is exactly where the hypothesis is.
Checking only the limit. All three conditions are required. A function can have a perfectly good limit at a point where it is undefined, and it is not continuous there.
Reading the value to classify the break. The kind of discontinuity is decided by the limits. Whether $f(a)$ exists, and what it is, decides whether there is a break, not which one.
Applying the IVT on an open interval. $(0,1)$ will not do; the theorem needs the endpoints, because it needs $f(a)$ and $f(b)$ to exist.
Expecting the EVT on an open or unbounded interval. $1/x$ on $(0,1]$ has no maximum; $x$ on $[0,\infty)$ has none either. Both hypotheses are load-bearing and each has its own counterexample.
Assuming a function with the IVT property is continuous. The converse is false, and interestingly so: $\sin(1/x)$ extended by $0$ takes every intermediate value on any interval round $0$ and is not continuous there.
Two confusions travel together here. The first is that a continuous function must be nice: $|x|$ is continuous everywhere and has a corner; $x\sin(1/x)$ extended by $0$ is continuous at $0$ and oscillates infinitely often in every neighbourhood of it. Continuity forbids breaks, not misbehaviour.
The second is the converse of the IVT. A function that takes every intermediate value is not thereby continuous — $\sin(1/x)$ with $f(0) = 0$ is the standard counterexample, and it takes every value in $[-1,1]$ on every interval round the origin. The pen-lifting picture is a good first intuition and is not the definition, and the two part company at exactly the examples worth knowing.
$f(x) = \dfrac{(x-1)(x-2)}{(x-1)(x-3)}$. At $x = 1$ both factors cancel: the limit is $\dfrac{1-2}{1-3} = \tfrac12$, so the break is removable and $f(1) = \tfrac12$ repairs it.
Cancelling factor, existing limit.
At $x = 3$ the denominator vanishes and the numerator does not, so $|f| \to \infty$: infinite, and nothing can repair it.
Surviving denominator zero.
Everywhere else $f$ is a quotient of polynomials with a non-vanishing denominator, hence continuous. Two different breaks in one formula, told apart by cancelling or not.
A continuous $f$ on $[0, 10]$ attains a maximum somewhere, by the EVT. That is a promise, made before any calculation.
Existence first.
The later method — check the critical points and the two endpoints, take the largest — is a search, and a search is only guaranteed to succeed because the thing being searched for is known to be there.
The theorem permits the method.
Drop closedness and the method fails: on $(0, 10)$ the largest value among the critical points may not be a maximum, because the supremum can sit at an endpoint the interval does not contain.
From the right the quotient is $1$; from the left it is $-1$.
Two one-sided limits.
Both exist and differ, so there is no two-sided limit: a jump.
No value at $4$ can repair it — no single number is both $1$ and $-1$.
Match each local behavior to the discontinuity type.
| Removable discontinuity | Jump discontinuity | Infinite discontinuity | Continuous at the point | |
|---|---|---|---|---|
| Both sides approach 2, but the point is missing | ||||
| Sides approach 2 and 5 | ||||
| Values grow without bound near the point | ||||
| Both sides and the value equal 2 |
Match each condition to the conclusion it warrants.
| Not continuous at $a$ | No two-sided limit exists | Continuous at $a$ | Some $c$ in $[a,b]$ has $f(c)=N$ | |
|---|---|---|---|---|
| $\lim_{x\to a}f(x)=3$, but $f(a)$ is undefined | ||||
| The one-sided limits at $a$ differ | ||||
| $f(a)=L=\lim_{x\to a}f(x)$ | ||||
| $f$ is continuous on $[a,b]$ and $N$ lies between $f(a)$ and $f(b)$ |
Match each condition set to the conclusion it justifies.
| $f$ is continuous at $a$ | Some $c\in[a,b]$ has $f(c)=r$ | $f$ attains a maximum and minimum on the interval | No two-sided limit exists, so continuity fails | |
|---|---|---|---|---|
| $\lim_{x\to a}f(x)=f(a)=L$ | ||||
| $f$ continuous on $[a,b]$; $r$ lies between $f(a)$ and $f(b)$ | ||||
| $f$ continuous on closed, bounded $[a,b]$ | ||||
| The two one-sided limits at $a$ are different |
$f(x) = \dfrac{x^2 - 4}{x - 2}$ for $x \ne 2$. What value at $x = 2$ makes $f$ continuous there?
Answer:
What kind of discontinuity does $f(x) = \lfloor x \rfloor$ have at $x = 2$?
To show $\cos x = 2x - 4$ has a solution, the IVT is applied to some $h$. If $h(x) = \cos x - 2x + 4$, what is $h(0)$?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
$f(x) = 4x$ for $x \le 3$ and $f(x) = 5x + k$ for $x > 3$. For which $k$ is $f$ continuous at $3$?
Answer:
You can classify and repair discontinuities and state both existence theorems with their hypotheses. Say in your own words what the intermediate value theorem does not promise. Next: limits at infinity, and the asymptotes they describe.
8. Your turn: is $f(x) = \dfrac{|x - 4|}{x - 4}$ continuous at $4$, and what kind of break is it?, step 3