Most familiar functions are continuous everywhere.
* For example, a monomial function $f(x) = ax^n$ for non-negative, integer $n$ will be continuous. This is because the limit exists everywhere, the domain of $f$ is all $x$ and there are no jumps.
* Similarly, the basic trigonometric functions $\sin(x)$, $\cos(x)$ are continuous everywhere.
* So are the exponential functions $f(x) = a^x, a > 0$.
* The hyperbolic sine ($(e^x - e^{-x})/2$) and cosine ($(e^x + e^{-x})/2$) are, as $e^x$ is.
* The hyperbolic tangent is, as $\cosh(x) > 0$ for all $x$.
Some familiar functions are *mostly* continuous but not everywhere.
* For example, $f(x) = \sqrt{x}$ is continuous on $(0,\infty)$ and right continuous at $0$, but it is not defined for negative $x$, so can't possibly be continuous there.
* Similarly, $f(x) = \log(x)$ is continuous on $(0,\infty)$, but it is not defined at $x=0$, so is not right continuous at $0$.
* The tangent function $\tan(x) = \sin(x)/\cos(x)$ is continuous everywhere *except* the points $x$ with $\cos(x) = 0$ ($\pi/2 + k\pi, k$ an integer).
* The hyperbolic co-tangent is not continuous at $x=0$ -- when $\sinh$ is $0$,
* The semicircle $f(x) = \sqrt{1 - x^2}$ is *continuous* on $(-1, 1)$. It is not continuous at $-1$ and $1$, though it is right continuous at $-1$ and left continuous at $1$.
##### Examples of discontinuity
There are various reasons why a function may not be continuous.
* The function $f(x) = \sin(x)/x$ has a limit at $0$ but is not defined at $0$, so is not continuous at $0$. The function can be redefined to make it continuous.
* The function $f(x) = 1/x$ is continuous everywhere *except* $x=0$ where *no* limit exists.
* A rational function $f(x) = p(x)/q(x)$ will be continuous everywhere except where $q(x)=0$. (The function ``f`` may still have a limit where ``q`` is ``0``, should factors cancel, but ``f`` won't be defined at such values.)
* The function
```math
f(x) = \begin{cases}
-1 & x < 0 \\
0 & x = 0 \\
1 & x > 0
\end{cases}
```
is implemented by `Julia`'s `sign` function. It has a value at $0$,
but no limit at $0$, so is not continuous at $0$. Furthermore, the
left and right limits exist at $0$ but are not equal to $f(0)$ so the
function is not left or right continuous at $0$. It is continuous everywhere except at $x=0$.
is not continuous at $x=0$. It has a limit of $0$ at $0$, a function
value $f(0) =1/2$, but the limit and the function value are not equal.
* The `floor` function, which rounds down to the nearest integer, is also not continuous at the integers, but is right continuous at the integers, as, for example, $\lim_{x \rightarrow 0+} f(x) = f(0)$. This graph emphasizes the right continuity by placing a point for the value of the function when there is a jump:
* The function $f(x) = 1/x^2$ is not continuous at $x=0$: $f(x)$ is not defined at $x=0$ and $f(x)$ has no limit at $x=0$ (in the usual sense).
* On the Wikipedia page for [continuity](https://en.wikipedia.org/wiki/Continuous_function) the example of Dirichlet's function is given:
```math
f(x) =
\begin{cases}
0 & \text{if } x \text{ is irrational,}\\
1 & \text{if } x \text{ is rational.}
\end{cases}
```
The limit for any $c$ is discontinuous, as any interval about $c$ will
contain *both* rational and irrational numbers so the function will
not take values in a small neighborhood around any potential $L$.
##### Example
Let a function be defined by cases:
```math
f(x) = \begin{cases}
3x^2 + c & x \geq 0,\\
2x-3 & x < 0.
\end{cases}
```
What value of $c$ will make $f(x)$ a continuous function?
We note that for $x < 0$ and for $x > 0$ the function is a simple polynomial, so is continuous. At $x=0$ to be continuous we need a limit to exists and be equal to $f(0)$, which is $c$. A limit exists if the left and right limits are equal. This means we need to solve for $c$ to make the left and right limits equal. We do this next with a bit of overkill in this case:
```julia;
@syms x c
ex1 = 3x^2 + c
ex2 = 2x-3
del = limit(ex1, x=>0, dir="+") - limit(ex2, x=>0, dir="-")
```
We need to solve for $c$ to make `del` zero:
```julia;
solve(del, c)
```
This gives the value of $c$.
## Rules for continuity
As we've seen, functions can be combined in several ways. How do these relate with continuity?
Suppose $f(x)$ and $g(x)$ are both continuous on $I$. Then
* The function $h(x) = a f(x) + b g(x)$ is continuous on $I$ for any real numbers $a$ and $b$;
* The function $h(x) = f(x) \cdot g(x)$ is continuous on $I$; and
* The function $h(x) = f(x) / g(x)$ is continuous at all points $c$ in $I$ **where** $g(c) \neq 0$.
* The function $h(x) = f(g(x))$ is continuous at $x=c$ *if* $g(x)$ is continuous at $c$ *and* $f(x)$ is continuous at $g(c)$.
So, continuity is preserved for all of the basic operations except when dividing by $0$.
##### Examples
* Since a monomial $f(x) = ax^n$ ($n$ a non-negative integer) is continuous, by the first rule, any polynomial will be continuous.
* Since both $f(x) = e^x$ and $g(x)=\sin(x)$ are continuous everywhere, so will be $h(x) = e^x \cdot \sin(x)$.
* Since $f(x) = e^x$ is continuous everywhere and $g(x) = -x$ is continuous everywhere, the composition $h(x) = e^{-x}$ will be continuous everywhere.
* Since $f(x) = x$ is continuous everywhere, the function $h(x) = 1/x$ - a ratio of continuous functions - will be continuous everywhere *except* possibly at $x=0$ (where it is not continuous).
* The function $h(x) = e^{x\log(x)}$ will be continuous on $(0,\infty)$, the same domain that $g(x) = x\log(x)$ is continuous. This function (also written as $x^x$) has a right limit at $0$ (of $1$), but is not right continuous, as $h(0)$ is not defined.
## Questions
###### Question
Let $f(x) = \sin(x)$ and $g(x) = \cos(x)$. Which of these is not continuous everywhere?
A value $x=c$ is a *removable singularity* for $f(x)$ if $f(x)$ is not
continuous at $c$ but will be if $f(c)$ is redefined to be $\lim_{x
\rightarrow c} f(x)$.
The function $f(x) = (x^2 - 4)/(x-2)$ has a removable singularity at
$x=2$. What value would we redefine $f(2)$ to be, to make $f$ a
continuous function?
```julia; hold=true; echo=false
f(x) = (x^2 -4)/(x-2);
numericq(f(2.00001), .001)
```
###### Question
The highly oscillatory function
```math
f(x) = x^2 (\cos(1/x) - 1)
```
has a removable singularity at $x=0$. What value would we redefine
$f(0)$ to be, to make $f$ a continuous function?
```julia; hold=true; echo=false
numericq(0, .001)
```
###### Question
Let $f(x)$ be defined by
```math
f(x) = \begin{cases}
c + \sin(2x - \pi/2) & x > 0\\
3x - 4 & x \leq 0.
\end{cases}
```
What value of $c$ will make $f(x)$ continuous?
```julia; hold=true; echo=false
val = (3*0 - 4) - (sin(2*0 - pi/2))
numericq(val)
```
###### Question
Suppose $f(x)$, $g(x)$, and $h(x)$ are continuous functions on $(a,b)$. If $a < c < b$, are you sure that $lim_{x \rightarrow c} f(g(x))$ is $f(g(c))$?
```julia; hold=true; echo=false
choices = [L"No, as $g(c)$ may not be in the interval $(a,b)$",
"Yes, composition of continuous functions results in a continuous function, so the limit is just the function value."
