Continuity

MATH1051

5.1 Definition of Continuity

We say that a function $f$ is continuous at $a$ if $f(a)$ is defined, ${\lim }_{x\to a}f(x)$ exists and ${\lim }_{x\to a}f(x)=f(a)$.

If $f$ is not continuous at $a$, we say that $f$ is discontinuous at $a$, or $f$ has a discontinuity at $a$.

A function may not be continuous at $x=a$ for a number of reasons.

5.1.1 Example

Let

$$f(x)=\{\begin{array}{ll}x,& x\ne 0\\ 1,& x=0.\end{array}$$

Then $f$ has a discontinuity at $x=0$. This is because $f(0)=1$ while ${\lim }_{x\to 0}f(x)=0$. Hence the third condition does not hold.

5.2 Continuity of Intervals

We say that $f$ is continuous on the open interval $(a,b)$ if $f$ is continuous at $c$, for all $c\in (a,b)$.

If $f$ is continuous on the closed interval $[a,b]$, then $f$ is continuous on $(a,b)$ and

$$\underset{x\to a^{+}}{\lim }f(x)=f(a),\underset{x\to b^{-}}{\lim }f(x)=f(b).$$

You could think of a continuous function being one that on an interval can be drawn without lifting your pen.

5.3 Properties of Continuous Functions

If $f(x)$ and $g(x)$ are continuous at $x=a$ and $c$ is a constant, then the following functions are all continuous at $x=a$.

$$\begin{array}{rlr}& 1.& f(x)\pm g(x)\\ & 2.& c\times f(x)\\ & 3.& f(x)\times g(x)\\ & 4.& \frac{f(x)}{g(x)},s.t.g(a)\ne 0.\end{array}$$

Since the function $f(x)=x$ is continuous, this proves that any polynomial is continuous, and any ratio of polynomials is continuous, provided the denominator is not zero.

5.4 The Intermediate Value Theorem (IVT)

Suppose that $f$ is continuous on the closed interval $[a,b]$ and let $N$ be any number between $f(a)$ and $f(b)$, where $f(a)\ne f(b)$. Then there exists a number $c\in (a,b)$ such that $f(c)=N$.

5.4.1 Example

You want to show that there is at least one real solution for a function. Say, for example, show that the equation $3x+2\cos x+5=0$ has one real solution for $x$.

First, you can just test out any two numbers in between the function that would make sense. Since we have a cosine ($\cos x$), we will use $-\pi$ or $\pi$ for one of our $x$.

$$f(0)=0+2+\cos (0)+5$$ $$f(0)=7,7>0$$

Now that we shown that this $x$ bound is too big, we go for a smaller number, say $-\pi$.

$$f(-\pi )=-3\pi +2\cos (-\pi )+5$$ $$f(-\pi )=-3\pi -2+5$$ $$f(-\pi )=3(-\pi +1),3(-\pi +1)<0$$

Since the higher bound of $x$ and the lower bound of $x$ are between 0, that means that there will be a real solution for $x$ at some point in between.

5.5 Application of the IVT (Bisection Method)

The bisection method is a procedure for approximating the zeros of a continuous function. It first cuts the interval $[a,b]$ in half (say, at a point $c$) and then decides in which of the smaller intervals ($[a,c]$, or $[c,b]$) the zero lies. This process is repeated until the interval is small enough to give a significant approximation of the zero itself.

We can present this bisection method as an algorithm:

1. Given $[a,b]$ such that $f(a)f(b)<0$, let $c=\frac{a+b}{2}$.
2. If $f(c)=0$ then quit; $c$ is a zero of $f$
3. If $f(c)\ne 0$ then:
   1. If $f(a)f(c)<0$, a zero lies in the interval $[a,c)$. So replace $b$ by $c$.
   2. If $f(a)f(c)>0$, replace $a$ by $c$.
4. If the interval $f[a,b]$ is small enough to give a precise enough approximation then quit. Otherwise, go to step 1.