The Concept

Definition 3.1.1  
A function $f : \hbox {I \hskip -5.2pt {R}}\to \hbox {I \hskip -5.2pt {R}}: x \mapsto f(x)$ is continuous at a point $c$ iff

$$\lim_{x \to c} f(x) = f(c)$$ (3.1.1)

What this means is that if you approach a point $c$, then the function value will also approach $f(c)$. Intuitively, one could say that a function is continuous if it is possible to draw its graph without lifting the pen of the paper.

Remark 3.1.2  

• A function is called continuous if it is continuous for all points on the real line, i.e. if (3.1.1) is valid for all $c$ in $\hbox {I \hskip -5.2pt {R}}$. An example would be $f(x)=x^2+3$.
• A function is called ``continuous on its domain" if it is continuous for all points for which it is defined. A typical example is $f(x)=\frac{1}{x}$ which is continuous for all points except at 0, where $f$ is not defined.
• A function that is not continuous is called discontinuous.

Theorem 3.1.3  
The following statements are equivalent:

1. $f$ is continuous in $c$.
2. $$\lim_{h \to 0}f(c+h) = f(c)$$
3. $\forall \varepsilon > 0, \exists \delta >0 \hbox{ such that } \vert x - c \vert < \delta \implies \vert f(x) - f(c) \vert < \varepsilon$
4. For any sequence $(x_n)_n$ in the domain of $f$ such that $x_n \to c$, we have $f(x_n) \to f(c)$.

Examples 3.1.4 (Continuous Functions)  
Quite a few of our bread and butter functions happen to be continuous:

• All polynomials are continuous in every point $c$.
• Rational functions: Take $f$ and $g$ to be continuous, then $$\frac{f(x)}{g(x)}$$ are continuous in any $c$ as long as $g(c)\not=0$.
• Root functions: Take $f$ a continuous function, then $\sqrt[2n]{f(x)}$ is defined as long as $f(x)>0$ where it is also continuous. $\sqrt[2n+1]{f(x)}$ is defined over the entire real line and also continuous there-over.
• Trigonometric functions are continuous over their domain:
  • $\sin(x)$ and $\cos(x)$ are continuous for any $x$.
  • $\tan(x)$ and $\sec(x)$ are continuous as long as $x \not= \frac{\pi}{2} + k\pi$ with $k \in \hbox {{\helv Z} \hskip -6.5pt {\helv Z}}$.
  • $\hbox{cotan}(x)$ and $\hbox{cosec}(x)$ are continuous as long as $x \not= k\pi$ with $k \in \hbox {{\helv Z} \hskip -6.5pt {\helv Z}}$.
• Inverse trigonometric functions are continuous over their domain:
  • $\arctan(x)$ and $\hbox{arccot}(x)$ are continuous for any $x$.
  • $\arcsin(x)$ and $\arccos(x)$ are continuous over their domain $[-1,1]$.
• Exponential functions are continuous over the entire real line.
• The function $\log(x)$ is continuous for positive $x$.

In order to prove all the above we would have to go through an $\varepsilon\hbox{-}\delta$ procedure for all of them, but that would keep us here for too long.

Theorem 3.1.5  
Let $f$ and $g$ be continuous at a point $c$. Then the following holds:

(a)

$f + g$ is continuous at $c$.

(b)

$f \cdot g$ is continuous at $c$.

(a)

$f \circ g$ is continuous at $c$ if $g$ is defined at $f(c)$.

Proof.

(a)

$$\lim_{x \to c}f(x) + g(x) = \lim_{x \to c}f(x) + \lim_{x \to c}g(x)$$

(b)

$$\lim_{x \to c}f(x) \cdot g(x) = \lim_{x \to c}f(x) \cdot \lim_{x \to c}g(x)$$

(c)

$$\lim_{x \to c}g(f(x)) = g \left(\lim_{x \to c}f(x)\right) = g(f(c))$$

$\qedsymbol$

Remark 3.1.6  
The previous theorem allows us to notice that even a rather wacky function like

$$f : \hbox {I \hskip -5.2pt {R}}\to \hbox {I \hskip -5.2pt {R}}: x \mapsto \sqrt[3]{x}+e^{-x^2}\sin(4x+5)$$ (3.1.2)

is continuous because it is simply a composition of continuous functions out of the list.

Examples 3.1.7 (Discontinuous Functions)  
There are basically two reasons why a function would not be continuous at a certain point, because there are basically two things that can go wrong with a limit: either the limit goes to infinity or the limit does not exist (i.e. the left and right limit are different). Note that this is a rather analogous situation to the way sequences can diverge.

(a)

The function

$$f : \hbox {I \hskip -5.2pt {R}}_0 \to \hbox {I \hskip -5.2pt {R}}: x \mapsto \frac{1}{x^2}$$ (3.1.3)

clearly goes to infinity at 0, therefore the function is not continuous at 0.

200pt

(b)

The function defined as

$$f : \hbox {I \hskip -5.2pt {R}}\to \hbox {I \hskip -5.2pt {R}}: \... ...mapsto -1 & \text{if $x < 0$} \\ x \mapsto 1 & \text{if $x \geq 0$} \end{cases}$$ (3.1.4)

has another problem at 0. In fact, the limit in 0 depends on the fact if we come from the left or from the right. From the left, it is $-1$ while from the right, it is $1$

200pt

(c)

Sometimes, it is not so obvious to see if a function is continuous or not just by looking at a graph. Take e.g. the function

$$f : \hbox {I \hskip -5.2pt {R}}\to \hbox {I \hskip -5.2pt {R}}: \... ...\frac{1}{x}) & \text{if $x\not=0$} \\ x \mapsto 0 & \text{if $x=0$} \end{cases}$$ (3.1.5)

The graph doesn't really give you much information to guess if the function is continuous in 0 or not. The function certainly doesn't run of to infinity, but on the other hand, in oscillates wildly around zero.

200pt

This function is not continuous in 0. To see this, take the sequence

$$x_n := \frac{1}{\frac{\pi}{2}+2n\pi}$$ (3.1.6)

For this sequence, we have

$$f(x_n) = \sin\left(\frac{1}{x_n}\right) = \sin\left(\frac{\pi}{2}+2n\pi\right) = 1$$ (3.1.7)

This means that we have

$$\lim_{n \to \infty}f(x_n) = 1 \hbox{ but, } f \left( \lim_{n \to \infty}x_n \right) = 0$$ (3.1.8)

Therefore $f$ cannot be continuous at 0.

(d)

And then there are the functions with a reducible discontinuity. Take e.g.

$$f: \hbox {I \hskip -5.2pt {R}}\backslash \{1\} \to \hbox {I \hskip -5.2pt {R}}: x \mapsto \frac{x^2-1}{x-1}$$ (3.1.9)

This function is essentially the same as $x+1$ except for $x=1$ where $f$ is not defined. This type of discontinuity is called reducible because arbitrarily putting $f(1):=0$ makes $f$ a continuous function. With any of the previous functions such an action is impossible.