The concept of continuity of a function is intuitively clear, however, very complex.

Less formal, if the domain of the function is an interval, then the graph of the function we can draw without lifting the pencil from the paper. Therefore, for instance, the $sign$ function sgn: $\mathbb{R} \to \mathbb{R}$ is not continuous at $x=0$, since $\lim_{x \to 0}$ sgn $x$ does not exist:

We can assume that the existence of the limit of a function at the point is necessary condition that a function is continuous at the same point. If we consider the following graph of the function,

we can see it is not a sufficient condition. From the graph we can see that the function has the limit $l$ at point $x_0$, however, the limit $l$ is not equal to the value of the function at point $x_0$ ($f(x_0)$).

Therefore, the formal definition follows.

Let $I \subseteq \mathbb{R}$ be an open interval and $x_o \in I$. The function $f: I \to \mathbb{R}$ is continuous at point $x_0$ if:

1.) $$\lim_{x \to x_0}f(x)$$

exists

and

2.) $$\lim_{x \to x_0}f(x)=f(x_0).$$

The function $f$ is the continuous function if it is continuous at every point from the domain of the function.

**Cauchy definition**

Let $I \subseteq \mathbb{R}$ be an open interval, $x_0 \in I$ and $f: I \to \mathbb{R}$ function. The function $f$ is continuous at point $x_0$ iff $\forall x \in I$:

$$(\forall \epsilon > 0) (\exists \delta > 0) ((|x-x_0| < \delta) \Rightarrow (|f(x) – f(x_0)|)).$$

For example, the constant function $f: \mathbb{R} \to \mathbb{R}$, $f(x) =x$, $\forall x \in \mathbb{R}$ is continuous at every point $x_0 \in \mathbb{R}$. Namely,

$$(\forall \epsilon > 0)(\exists \delta > 0)(\forall x \in \mathbb{R})((|x-x_0| < \delta) \Rightarrow (|c-c| = 0 < \epsilon)).$$

The sine function $\sin: \mathbb{R} \to [-1,1 ]$ is continuous on $\mathbb{R}$. For proof, we will use the trigonometric identity for the difference of sines and the following inequality:

$$| \sin x| \le |x|,$$

for $|x| < \frac{\pi}{2}$.

Therefore, for any $x_0 \in \mathbb{R}$ we have

$$|\sin x – \sin x_0| = \left | 2 \sin \left ( \frac{x – x_0}{2}\right ) \cdot \left( \cos \frac{x + x_0}{2} \right) \right | $$

$$ \le 2 \left | \sin \left ( \frac{x – x_0}{2} \right) \right | $$

$$ \le 2 \left | \frac{x-x_0}{2} \right | $$

$$ \le 2 \frac{|x – x_0|}{2} $$

$$ \le | x – x_0|$$,

when $|x – x_0| < \frac{\pi}{2}$.

Now we have:

$$\lim_{x \to x_0}|\sin x – \sin x_0| \le \lim_{x \to x_0} |x – x_0| = 0,$$

that is

$$\lim_{x \to x_0} \sin x = \lim_{x \to x_0} \sin x_0 = \sin x_0,$$

which means that the function sine is continuous at point $x_0$.

Trigonometric functions are continuous on theirs domains.

**Properties of continuous functions**

Let $I \subseteq \mathbb{R}$ be an open interval, $x_0 \in I$ and $f,g: I \to \mathbb{R}$ continuous functions at $x_0$. Then:

1.) $\forall \alpha, \beta \in \mathbb{R}$, the function $\alpha f + \beta g$ is continuous at $x_0$,

2.) the function $f \cdot g$ is continuous function at $x_0$,

3.) if $g(x) \neq 0$, $\forall x \in I$, then the function $\frac{f}{g}$ is continuous function at $x_0$.

The function $f: \mathbb{R} \to \mathbb{R}$, $f(x) = x^n$, $\forall n \in \mathbb{N}$ and $\forall x \in \mathbb{R}$ is continuous function on $\mathbb{R}$.

Every polynomial function $p: \mathbb{R} \to \mathbb{R}$ defined as

$$p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_2 x^2 + a_1 x +a_0,$$

$a_n \neq 0$, $ a_0, \ldots, a_n \in \mathbb{R}$, is continuous function on $\mathbb{R}$.

Every rational function is continuous on its domain. Rational function is a ratio of two polynomials $p$ and $q$, that is, a ratio of two continuous functions:

$$r(x) = \frac{p(x)}{q(x)}.$$

Therefore, the rational function $r(x)$ is a continuous function on $\mathbb{R}$ such that $q(x) \neq 0$.

**Continuity of composite functions**

Let $A, B \subseteq \mathbb{R}$ be open intervals and $f: A \to \mathbb{R}$, $g: B \to \mathbb{R}$ functions for which is valid $f(A) \subseteq B$, that is, the function $g \circ f: A \to \mathbb{R}$ is well defined. If the function $f$ is continuous at point $x_0 \in A$ and function $g$ continuous at point $f(x_o) \in B$, then the function $g \circ f$ is continuous at point $x_o$.

**Uniformly continuous functions**

Let $ I \subseteq \mathbb{R}$ be an open interval and $f: I \to \mathbb{R}$. Then the function $f$ is uniformly continuous on $I$ if:

$$(\forall \epsilon > 0)(\exists \delta > 0) ((|x-y| < \delta)(x, y \in I) \Longrightarrow (|f(x) – f(y)| < \epsilon)).$$

This means that $\delta$ depends on $\epsilon$, not on $x$ and $y$.