Limit and continuity · Topos Uranos

Summary

This lesson carries limits to their first structural consequence, continuity, which the following lesson extends to all elementary functions. A function is continuous at a point when its value there is the one announced by the nearby values; with the limits of the preceding lessons, this idea is written as the usual challenge and response, with a single difference: the point itself is no longer excluded. It follows that continuity at an accumulation point amounts to the limit existing and being equal to the value, and that almost all its theorems are inherited from the algebra of limits without any new proof. Discontinuities are classified as removable, jump and essential, and the removable ones are repaired by continuous extension. The only new argument is the one for composition, where continuity does more than the limit: it removes the hypothesis that the theorem on the limit of a composition required, and it allows the limit to be brought inside a continuous function.

Learning objectives

Define continuity at a point, on one side and on an interval, and relate it to the limit.
Classify discontinuities as removable, jump and essential, and extend by continuity a function with a removable discontinuity.
Prove the continuity of sums, products, quotients, roots and compositions of continuous functions, and deduce that of polynomials, rational functions and trigonometric functions.
Compute limits by bringing them inside a continuous function, and study functions that are continuous at a single point or at none.

Definition

In the lesson on the limit of functions of a real variable, the definition deliberately excluded the point $a$ : the limit describes what happens near $a$ , not at $a$ , and the function may not be defined there. Continuity compares both things: what happens nearby and what happens at the point.

DefinitionFunction continuous at a point

Let $f$ be a function defined on a set $A$ and $a$ an element of $A$ . We say that $f$ is continuous at $a$ if for every $\varepsilon > 0$ there exists a $\delta > 0$ such that every $x$ in $A$ that satisfies $\abs{x - a} < \delta$ also satisfies $\abs{f(x) - f(a)} < \varepsilon$ :

(\forall \varepsilon > 0)(\exists \delta > 0)(\forall x \in A)\left(\abs{x - a} < \delta \rightarrow \abs{f(x) - f(a)} < \varepsilon\right)

If $f$ is not continuous at $a$ , we say that it is discontinuous at $a$ . If it is continuous at all the points of a set $S$ contained in $A$ , we say that it is continuous on $S$ ; and if it is so on its whole domain, simply that it is continuous.

The definition differs from that of the limit in three respects. The number that the values must approach is not an arbitrary $L$ , but $f(a)$ ; that is why $a$ must belong to the domain. The response $\abs{x - a} < \delta$ does not exclude $x = a$ , although including it costs nothing, because $\abs{f(a) - f(a)} = 0$ . And $a$ is not required to be an accumulation point of $A$ . The negation is obtained as in the lesson on the limit: $f$ is discontinuous at $a$ if there exists an $\varepsilon > 0$ such that, for every $\delta > 0$ , there is an $x$ in $A$ with $\abs{x - a} < \delta$ and $\abs{f(x) - f(a)} \geq \varepsilon$ .

PropositionContinuity and limit

Let $f$ be a function defined on $A$ and $a$ an element of $A$ . If $a$ is an accumulation point of $A$ , then $f$ is continuous at $a$ if and only if $\lim_{x \to a} f(x) = f(a)$ . If $a$ is an isolated point of $A$ , then $f$ is continuous at $a$ . The two definitions ask for the same thing except at the point $a$ , where continuity holds by itself. In the figure, the band surrounds the value $f(a)$ , which is now a filled dot on the graph; at the end, an isolated point of the domain appears.

Proof

$0 < \abs{x - a} < \delta \rightarrow \abs{x - a} < \delta \rightarrow \resaltar{\abs{f(x) - f(a)} < \varepsilon}$
If $f$ is continuous at $a$ , the response $\delta$ to each challenge $\varepsilon$ holds in particular for the $x$ with $0 < \abs{x - a} < \delta$ : it is a response for the limit, with $L = f(a)$ .
$x = a \rightarrow \abs{f(a) - f(a)} = \resaltar{0} < \varepsilon$
Conversely, if the limit is $f(a)$ , the response of the limit covers the $x$ different from $a$ , and at $x = a$ the condition always holds: it is a response for continuity.
$\abs{x - a} < \delta \wedge x \in A \rightarrow x = a \rightarrow \abs{f(x) - f(a)} = 0 < \varepsilon$
If $a$ is an isolated point, there exists a $\delta > 0$ such that the only element of $A$ with $\abs{x - a} < \delta$ is $a$ itself. That $\delta$ responds to all the challenges at once.

Therefore, at an accumulation point of the domain, continuity consists of three things: that $f$ be defined at $a$ , that it have a limit at $a$ , and that this limit be $f(a)$ . Each one can fail separately, and the classification of discontinuities comes from this. For example, the function that equals $1$ at every $x \neq 0$ and $0$ at $x = 0$ , from the counterexample on composition in the lesson on the algebra of limits, has limit $1$ at $0$ , different from its value: it is discontinuous at $0$ .

In the rest of this lesson

One-sided continuity and continuity on an interval
Classification of discontinuities
Operations with continuous functions
Composition
Local properties
Solved problems
Two pieces that join
Three different discontinuities
A damped oscillation and one that is not
Limits inside continuous functions
A function continuous at a single point
Proposed problems
Three pieces
The absolute value of a continuous function
The maximum of two continuous functions
An additive function
Thomae’s function