Real function/Continuous/Rules/Section

Lemma

Let ${}D\subseteq \mathbb {R}$ and ${}E\subseteq \mathbb {R}$ be subsets and let

$f\colon D\longrightarrow \mathbb {R}$

and

$g\colon E\longrightarrow \mathbb {R}$

denote functions with

${}f(D)\subseteq E$. Then the following statements hold.

1. If ${}f$ is continuous in ${}x\in D$ and ${}g$ is continuous in ${}f(x)$, then also the composition ${}g\circ f$ is continuous in ${}x$.
2. If ${}f$ and ${}g$ are continuous, so is ${}g\circ f$.

Proof

The first statement follows from fact. This implies also the second statement.

$\Box$

Lemma

Let ${}D\subseteq \mathbb {R}$ be a subset and let

$f,g\colon D\longrightarrow \mathbb {R}$

be continuous functions. Then also the functions

$f+g\colon D\longrightarrow \mathbb {R} ,x\longmapsto f(x)+g(x),$

$f-g\colon D\longrightarrow \mathbb {R} ,x\longmapsto f(x)-g(x),$

$f\cdot g\colon D\longrightarrow \mathbb {R} ,x\longmapsto f(x)\cdot g(x),$

are continuous. For a subset ${}U\subseteq D$ such that ${}g$ has no zero in ${}U$, also

$f/g\colon U\longrightarrow \mathbb {R} ,x\longmapsto f(x)/g(x),$

is continuous.

Proof

This follows from fact and fact.

$\Box$

Corollary

Polynomial functions

$P\colon \mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto P(x),$

are

continuous.

Proof

Due to example and fact, the powers

$\mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto x^{n},$

are continuous for every ${}n\in \mathbb {N}$. Hence, also the functions

$\mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto ax^{n},$

are continuous for every ${}a\in \mathbb {R}$ and therefore, again due to fact, the functions

$\mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{1}x+a_{0},$

are continuous.

$\Box$

Rational functions are continuous on their domain.

Corollary

Let ${}P,Q\in \mathbb {R} [X]$ be polynomials and let ${}U:={\left\{x\in \mathbb {R} \mid Q(x)\neq 0\right\}}$. Then the rational function

$U\longrightarrow \mathbb {R} ,x\longmapsto {\frac {P(x)}{Q(x)}},$

is continuous.

Proof

This follows from fact and fact.

$\Box$