We provide all necessary properties of functions of one variable such as limit at a given point, continuity, and differentiability. Such functions are defined over a domain of the set of real numbers.

## Continuity of functions of one variable

We denote by $\mathbb{R}$ the set of real numbers. If $f$ is a real function, then we denote by $\mathscr{D}_f$ the domain of the definition of $f$. That is a part of $\mathbb{R}$ in which $f$ is well defined. For some function we can have $\mathscr{D}_f=\mathbb{R}$.

To represent the function, we write $f:\mathscr{D}_f\to \mathbb{R}$ and the graph of $f$ is a subset of $\mathbb{R}^2$ defined by $$G(f):=\{(x,f(x)):x\in \mathscr{D}\}.$$

### Algebraic properties of functions

A function $f:\mathscr{D}_f\to \mathbb{R}$ is said to be injective if $f(x)=f(y)$ implies that $x=y$. Injectivity is related to the uniqueness of the solution of algebraic equations. As an example the function $f(x)=2^x$ for $x\in\mathscr{D}_f=\mathbb{R}$ is injective. In fact, if $f(x)=f(y),$ the $2^x=2^y$. By applying in both sides the logarithmic function we obtain $x\ln(2)=y\ln(2),$ so that $x=y$.

We say that $f$ is surjective if for any $y\in \mathbb{R},$ there exists a real number $x\in \mathscr{D}_f$ such that $y=f(x)$. We note that the surjectivity is related to the existence of the solution of algebraic equations.

The function $f$ is said to be bijective if it is both injective and surjective. This means that for every $y\in \mathbb{R},$ there exists a unique $x\in \mathscr{D}_f$ such that $y=f(x)$. We note that bijectivity is related to the existence and uniqueness of the solutions of algebraic equations.

If $f:\mathscr{D}_f\to\mathbb{R}$ is bijective, we denote by $f^{-1}:\mathbb{R}\to \mathscr{D}_f$ the inverse of $f$. This inverse it satisfies $f\circ f^{-1}=f^{-1}\circ f=id$.

### Limit and continuity of a function

Let $f:\mathscr{D}_f\to\mathbb{R}$ be a function of one variable. Assume that there exists $x_0\in \mathbb{R}\setminus \mathscr{D}_f$ and that for any $\varepsilon>0,$ $(x_0-\varepsilon,x_0+\varepsilon)\cap \mathscr{D}_f\neq \emptyset$.

We say that $f$ admits a real number $\ell$ at $x_0$ is for any $\varepsilon>0,$ there exists $\alpha>0$ such that for any $x\in \mathbb{D_f},$ we have $$|x-x_0|<\alpha \Rightarrow |f(x)-\ell|<\varepsilon.$$

We say that $f$ admit $+\infty$ a a limit at $x_0$ if for any $A>0,$ $\alpha>0$ such that for any $x\in \mathbb{D_f}$, $$|x-x_0|<\alpha \Rightarrow f(x)>A.$$

The limit of the function $f$ at $x_0$ is $-\infty,$ if for any $A>0,$ $\alpha>0$ such that for any $x\in \mathbb{D_f}$, $$|x-x_0|<\alpha \Rightarrow f(x)<-A.$$

For more details on limits, you may also consult: on how to find the limit of a function.

Let us now discuss the continuity of functions of one variable.

A function $f:\mathscr{D}_f\to\mathbb{R}$ is continuous at a point $a\in \mathscr{D}_f$ if the limit of $f$ at $a$ is exactly $f(a)$. That is, for any $\varepsilon>0,$ there exists $\alpha>0$ such that for any $x\in \mathscr{D}_f,$ we have $$|x-a|<\alpha \Rightarrow |f(x)-f(a)|<\varepsilon.$$

Sometimes it is more practical to verifie $\lim_{h\to 0}f(a+h)=f(a)$ to prove that the function $f$ is continuous at $a$.

We also say that $f$ is continuous on $\mathscr{D}_f$ if the function $f$ is continuous at any point of $\mathscr{D}_f$.

### Examples of continuous functions of one variable

The function $x\mapsto \sin(x)$ is continuous on $\mathbb{R}$. In fact, let $a,h\in\mathbb{R}$. By using the following trigonometric identity $$\sin(a+h)=\sin(a)\cos(h)+\cos(a)\sin(h),$$ we have \begin{align*}\sin(a+h)-\sin(a)=\sin(a)(\cos(h)-1)+\cos(a)\sin(h).\end{align*} We know that $$\lim_{h\to 0}\cos(h)=1.\quad \lim_{h\to 0}\sin(h)=0.$$ Then $\lim_{h\to 0}\sin(a+h)=\sin(a)$. This ends the proof.

You may also consult a post on continuous functions of one variable for more details and exercises with detailed answers.

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