# Continuity

Intuitively speaking, we say a function is continuous at a point if its graph has no separation, i.e. there is no hole or breakage, at that point. Such notion of continuity can be defined explicitly as follows.

Definition: A function $f(x)$ is said to be continuous at a point $x=a$ if $\lim_{x\to a}f(x)=f(a).$

Note that the above definition assumes the existence of both $\displaystyle\lim_{x\to a}f(x)$ and $f(a)$.

There are 3 different types of discontinuities.

• $f(a)$ is not defined.

For example, consider the function$f(x)=\frac{x^2-4}{x-2}.$ Clearly $f(2)$ is not defined. However the limit $\displaystyle\lim_{x\to 2}f(x)$ exists:\begin{eqnarray*}\lim_{x\to 2}\frac{x^2-4}{x-2}&=&\lim_{x\to 2}\frac{(x+2)(x-2)}{x-2}\\&=&\lim_{x\to 2}(x+2)=4.\end{eqnarray*} As a result the graph has a hole.

This kind of discontinuity is called a removable discontinuity, meaning that we can extend $f(x)$ to a function which is continuous at $x=a$ in the following sense: Define $g(x)$ by$g(x)=\left\{\begin{array}{ccc}f(x)\ \mbox{if}\ x\ne a,\\\lim_{x\to a}f(x)\ \mbox{if}\ x=a.\end{array}\right.$Then $g(x)$ is a continuous at $x=a$. The function $g(x)$ is called the continuous extension of $f(x)$. What we just did is basically filling the hole and the filling is the limit $\displaystyle\lim_{x\to a}f(x)$. For the above example, we define$g(x)=\left\{\begin{array}{ccc}\frac{x^2-4}{x-2} &\mbox{if}& x\ne 2,\\4 &\mbox{if}& x=2.\end{array}\right.$ Then $g(x)$ is continuous at $x=2$ and in fact, it is identical to $x+2$.

• $\displaystyle\lim_{x\to a}f(x)$ deos not exist.

Example. Let $f(x)=\left\{\begin{array}{cc}2x-2,\ &1\leq x<2\\3,\ &2\leq x\leq 4.\end{array}\right.$ $f(2)=3$ but $\displaystyle\lim_{x\to 2}f(x)$ does not exist because $\displaystyle\lim_{x\to 2-}f(x)=2$ while $\displaystyle\lim_{x\to 2+}f(x)=3$.

• $f(a)$ is defined and $\displaystyle\lim_{x\to a}f(x)$ exists, but $\displaystyle\lim_{x\to a}f(x)\ne f(a)$.

Example. Let $f(x)=\left\{\begin{array}{cc}\displaystyle\frac{x^2-4}{x-2},\ &x\ne 2\\3,\ &x=2.\end{array}\right.$ Then $f(2)=3$ and $\displaystyle\lim_{x\to 2}f(x)=4$.

From the properties of limits (Theorem 1, Lecture 4), we obtain the following properties of continuous functions.

Theorem 9. If functions $f(x)$ and $g(x)$ are continuous at $x=a$, then

1. $(f\pm g)(x)=f(x)\pm g(x)$ is continuous at $x=a$.
2. $f\cdot g(x)=f(x)\cdot g(x)$ is continuous at $x=a$.
3. $\displaystyle\frac{f}{g}(x)=\frac{f(x)}{g(x)}$ is continous at $x=a$ provided $g(a)\ne 0$.

There are some important classes of continous functions.

• Every polynomial function $p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0$ is continuous everywhere, because $\displaystyle\lim_{x\to a}p(x)=p(a)$ for any $-\infty<a<\infty$.
• If $p(x)$ and $q(x)$ are polynomials, then the rational function $\displaystyle\frac{p(x)}{q(x)}$ is continuous wherever it is defined ($q(a)\ne 0). • \(y=\sin x$ and $y=\cos x$ are continuous everywhere.
• $y=\tan x$ is continous where it is defined, i.e. everywhere except at the points $x=\pm\frac{\pi}{2},\pm\frac{3\pi}{2},\pm\frac{5\pi}{2},\cdots$.
• If $n$ is a positive integer, then $y=\root n\of{x}$ is continous where it is defined. That is, if $n$ is an odd integer, it is defined everywhere. If $n$ is an even integer,it is defined on $[0,\infty)$, the set of all non-negative real numbers.

Recall that the composite function $g\circ f(x)$ of two functions $f(x)$ and $g(x)$ (read $f$ followed by $g$) is defined by $g\circ f(x):=g(f(x)).$

Theorem 10. Suppose that $\displaystyle\lim_{x\to a}f(x)=L$ exists and $g(x)$is continuous function at $x=L$. Then$\lim_{x\to a}g\circ f(x)=g(\lim_{x\to a}f(x)).$

It follows from Theorem 10 that the composite function of two continuous functions is again a continuous function.

Corollary 11. If $f(x)$ is continuous at $x=a$ and $g(x)$ is continuous at $f(a)$, the the composite function $g\circ f(x)$ is continuous at $x=a$.

Example. The function$y=\sqrt{x^2-2x-5}$ is the composite function  $g\circ f(x)$ of two functions $f(x)=x^2-2x-5$ and $g(x)=\sqrt{x}$. The function $f(x)=x^2-2x-5$ is continuous everywhere while $g(x)=\sqrt{x}$ is continuous on $[0,\infty)$, so by Corollary 11, the composite function $g\circ f(x)=\sqrt{x^2-2x-5}$ is continuous on its repective domain which is $(-\infty,1-\sqrt{6}]$ or $[1+\sqrt{6},\infty)$. The following picture shows you the graph of $y=\sqrt{x^2-2x-5}$ on the intervals  $(-\infty,1-\sqrt{6}]$ and $[1+\sqrt{6},\infty)$.

Continuous functions exhibit many nice properties. I would like to introduced a couple of them here. The first is the so-called Max-Min Theorem.

Theorem 12. [Max-Min Theorem] If $f(x)$ is a continuous function on a closed interval $[a,b]$, $f(x)$ attains its maximum value and minimum value on $[a,b]$.

Another important property is the so-called Intermediate Value Theorem (IVT). The IVT has an important application in the study of equations.

Theorem 13. [The Intermediate Value Theorem] If $f(x)$ is continuous on a closed interval $[a,b]$ and $f(a)\ne f(b)$, then , then $f(x)$ takes on every value between $f(a)$ and $f(b)$. In other words, if $f(a)<k<f(b)$ (assuming that $f(a)<f(b)$), then $f(c)=k$ for some number $a<c<b$.

It follows from Theorem 13 that

Corollary 14. If $f(x)$ is continuous on a closed interval $[a,b]$ and $f(a)\cdot f(b)<0$, then $f(x)=0$ for some $a<x<b$.

Using  Corollary 14 we can tell if a root of the equation $f(x)=0$ can be found in some interval. For instance

Example. Show that the equation $x^3-x-1=0$ has a root in the interval $[-1,2]$.

Solution. Let $f(x)=x^3-x-1$. Then $f(x)$ is continuous on $[-1,2]$. Since $f(-1)=-1$ and $f(2)=5$ have different signs, by Corollary 14 there is a root of $x^3-x-1=0$ in the open interval $(-1,2)$.

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