MA005: Calculus I

Fortunately, the situations which we encounter most often in applications and the functions which model those situations are either continuous everywhere or continuous everywhere except at a few places, so any result which is true of all continuous functions will be true of most of the functions we commonly use.

Theorem: The following functions are continuous everywhere, at every value of $\mathrm{x}$:

(a) polynomials, (b) $\sin (x)$ and $\cos (x)$ and (c) $|\times|$.

Proof: (a) This follows from the

Substitution Theorem for Polynomials and the definition of continuity.

(b) The graph of $\mathrm{y}=\sin (\mathrm{x})$ (Fig. 6) clearly indicates that $\sin (\mathrm{x})$ does not have any holes or breaks so $\sin (\mathrm{x})$ is continuous everywhere. Or we could justify that result analytically:

for every real number $\mathrm{a}$,

$\begin{aligned} \lim\limits_{h \rightarrow 0} \sin (a+h) &=\lim\limits_{h \rightarrow 0} \sin (a) \cos (\mathrm{h})+\cos (a) \sin (h) \\ &=\lim\limits_{h \rightarrow 0} \sin (a) \cdot \lim\limits_{h \rightarrow 0} \cos (\mathrm{h})+\lim\limits_{h \rightarrow 0} \cos (\mathrm{a}) \cdot \lim\limits_{h \rightarrow 0} \sin (\mathrm{h}) \end{aligned}$

(recall from section $1.2$ that $\quad \lim\limits_{h \rightarrow 0} \cos (\mathrm{h})=1$ and $\lim\limits_{h \rightarrow 0} \sin (\mathrm{h})=0$)

$=\lim\limits_{h \rightarrow 0} \sin (a) \cdot \mathbf{1}+\lim\limits_{h \rightarrow 0} \cos (\mathrm{a}) \cdot \mathbf{0}=\sin (\mathrm{a})$

so $f(x)=\sin (x)$ is continuous at every point. The justification of $f(x)=\cos (x)$ is similar.

(c) $\mathrm{f}(\mathrm{x})=|\mathrm{x}|$. When $\mathrm{x}>0$, then $|\mathrm{x}|=\mathrm{x}$ and its graph (Fig. 7) is a straight line and is continuous since $x$ is a polynomial function. When $x$, then $|x|=-x$ and it is also continuous. The only questionable point is the "corner" on the graph when $\mathrm{x}=0$, but the graph there is only bent, not broken:

$\lim\limits_{h \rightarrow 0^{+}}|x|=\lim\limits_{h \rightarrow 0^{+}} x=0$

and $\lim\limits_{h \rightarrow 0^{-}}|x|=\lim\limits_{h \rightarrow 0^{-}}-x=0$ so $\lim\limits_{h \rightarrow 0}|x|=0=10$,

and $\mathrm{f}(\mathrm{x})=|\mathrm{x}|$ is also continuous at $\mathrm{0}$.

A continuous function can have corners but not holes or breaks (jumps). 

Several results about limits of functions can be written in terms of continuity of those functions. Even

functions which fail to be continuous at some points are often continuous most places.

Theorem: (a) A rational function is continuous except where the denominator is $\mathrm{0}$.

(b) Tangent, cotangent, secant and cosecant are continuous except where they are undefined.

(c) The greatest integer function $\mathrm{[ x ] = INT(x)}$ is continuous except at each integer.

(d) But the "holey" function $\mathrm{h}(\mathrm{x})= \begin{cases}2 & \text { if } \mathrm{x} \text { is a rational number } \\ 1 & \text { if } \mathrm{x} \text { is an irrational number }\end{cases}$ is discontinuous everywhere.