## MA101.12 The exponential functions

We define

$\exp{x} = 1 +x+\frac{x^2}{2!}+\frac{x^3}{3!}+...+\frac{x^n}{n!}+...$ which in fact converges $\forall x \in \mathbb{R}$.

\begin{aligned} \exp{x} \cdot \exp{y} &= (1 +x+\frac{x^2}{2!}+...)(1 +y+\frac{y^2}{2!}+...) \\&= 1 +x+y+\frac{x^2}{2!}+xy+\frac{y^2}{2!}+...+\frac{x^n}{n!}+\frac{x^{n-1}y}{n-1!}+\frac{x^{n-2}y^2}{(n-2)!2!}+...+\frac{y^2}{2!}+... \\&= 1 +(x+y)+\frac{(x+y)^2}{2!}+...+\frac{(x+y)^n}{n!}+... \\&= \exp{(x+y)} \end{aligned}

We have assumed that we can multiply infinite series as if they were finite algebraic expressions. This is not always true, but it is for convergent power series.

The relationship $\exp{x}\cdot\exp{y} = \exp{(x+y)}$ leads us to write $\exp{x}=e^x$ so that $e^x\cdot e^y = e^{x+y}$.

Note, $\exp{1} = e^1 = e = 1+1+\frac{1}{2!}+\frac{1}{3!}+...$

On the assumption that a power series can be differentiated term by term $\frac{d e^x}{dx} = 0 + 1 +\frac{x^2}{2!}+...=e^x$.

##### Properties of exp

The graph of $y=e^x$ is as shown, (with series definition this is not easy to see), but there is some evidence from the property $e^{x_1}\cdot e^{x_2}=e^{x_1+x_2}$

Observe from the graph the properties:

1. $\displaystyle \lim_{x\to\infty}e^x=\infty$.
2. $\displaystyle \lim_{x\to-\infty}e^x=0$.
3. $\forall x, e^x >0$.
4. $e^x$ is continuous everywhere.
5. $D(e^x)=\mathbb{R}$.
6. $e^x$ is an injection (or one to one).
ie. $x_1 \neq x_2 \Rightarrow f(x_1) \neq f(x_2)$.

This means there will be an inverse function. We call it the log function. We have $\log{y}$ equal to the unique x such that $y=e^x$.

$y=e^x$ and $x=\ln{y}$ mean of course the same thing. The graph of $y=\ln{x}$ will be a reflection of $y=e^x$ in the line $y=x$.

Note that the domain of ln = range of exp = $\mathbb{R}^+$.

##### Properties of ln
1. If $y=\ln{x}$ then $x=e^y$, so $\frac{dx}{dy} = e^y$.
Therefore, $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{1}{e^y} = \frac{1}{x}$.
2. $\ln{x}$ is an injection.
Proof:
Let $\ln{x}=\ln{y}$.
Put $\ln{x}=p$ and $\ln{y}=p$.
So, $e^p=x$ and $e^p=y$.
Whence $x=y$.
3. $\forall x, x=e^{\ln{x}}$.
Proof:
Let $y=e^{ln{x}}$.
Then $\ln{y} = \ln{x}$.
Whence $x=y$ by 2.
So, $x=e^{\ln{x}}$.
4. $\ln{xy}=\ln{x}+\ln{y}$.
Proof:
\begin{aligned} \ln{xy}&=\ln{e^{\ln{x}}\cdot e^{\ln{y}}} \\&= \ln{e^{\ln{x}+\ln{y}}} \\ xy &= e^{ln{x}+\ln{y}} \\ \mbox{Hence}\; \ln{x}+\ln{y} &= \ln{xy} \end{aligned}.
5. $\forall a \in \mathbb{R}^+$
(i) if  m in an integer then $a^m = m\ln{a}$.
(ii) if $\frac{m}{n}$ is in $\mathbb{Q}$, then equally we have $\ln{a\frac{m}{n}}=\frac{m}{n}\ln{a}$. ($a^{\frac{m}{n}} \mbox{ means } \sqrt[n]{a^m}$.
##### Definition

For $a \in \mathbb{R}^+$ and $x \in \mathbb{R}$ we define $a^x = e^{x\ln{a}}$.

For example $3^{\sqrt{2}}=e^{\sqrt{2}\ln{3}}$.

Note that this definition fits in with the usual one when x is rational.

For fixed $a \in \mathbb{R}^+, a^x$ is a function.

Note: $\frac{d(a^x)}{dx} = \frac{d(e^{\ln{a^x}}}{dx} = \frac{d(e^{x\ln{a}})}{dx} = \ln{a(e^{x\ln{a}})} = a^x\ln{a}$.

## MA101.11 Finding Limits – Examples

A number of examples of how to find limits.

\begin{aligned} \displaystyle \lim_{x\to 1} \{ x^2 + \frac{x^3+1}{x^2 +1} + 2x\sqrt{1+x^2} \} &= 1 + \frac{1+1}{1+1}+2\cdot 1\cdot\sqrt{1+1} \\&= 2 + 2\sqrt{2} \end{aligned}

\begin{aligned} \displaystyle \lim_{x\to \infty} \frac{x^2-3x+2}{2x^2+4x-7} &= \lim_{x\to \infty} \frac{1 - \frac{3}{x} + \frac{2}{x^2}}{2 + \frac{4}{x} - \frac{7}{x^2}} \\&= \frac{1}{2} \end{aligned}

\begin{aligned} \displaystyle \lim_{x\to 2} \frac{x^2 - 4}{x^2-5x-6} &= \lim_{x\to 2} \frac{(x-2)(x+2)}{(x-2)(x-3)} \\&= \lim_{x\to 2} \frac{(x+2)}{(x-3)} \\&= -4 \end{aligned}

\begin{aligned} \displaystyle \lim_{x\to \infty} (\sqrt{x^2+6x+4}-x) &= \lim_{x\to \infty} \frac{(\sqrt{x^2+6x+4}-x)(\sqrt{x^2+6x+4}+x)}{(\sqrt{x^2+6x+4}+x)} \\&= \lim_{x\to \infty} \frac{x^2 + 6x + 4 - x^2}{(\sqrt{x^2+6x+4}+x)} \\&= \lim_{x\to \infty} \frac{6+\frac{4}{x}}{\sqrt{1+\frac{6}{x}+\frac{4}{x^2}}+1} \\&= \frac{6}{\sqrt{1}+1} = 3 \end{aligned}

\begin{aligned} \displaystyle \lim_{x\to 1^+} \frac{\sqrt{x-1}}{x^2-1} &= \lim_{x\to 1^+} \frac{1}{\sqrt{x-1}(x+1)} \\&= \infty \end{aligned}

\begin{aligned} \displaystyle \lim_{x\to 0} \frac{x}{\sqrt{1-x}-\sqrt{1+x}} &= \lim_{x\to 0} \frac{x(\sqrt{1-x}+\sqrt{1+x})}{(1-x)-(1+x)} \\&= \lim_{x\to 0} \frac{x(\sqrt{1-x}+\sqrt{1+x}}{-2x} \\&= -1 \end{aligned}

\begin{aligned} \displaystyle \lim_{x\to 1} \frac{ln x}{x^2-1} &= \lim_{x\to 1} \frac{\frac{1}{x}}{2x} \:\mbox{ - (Using L'Hopital's Rule)} \\&= \frac{1}{2} \end{aligned}

\begin{aligned} \displaystyle \lim_{x\to 0} \frac{sin(2x)}{x} &= \lim_{x\to 0} \frac{2cos(2x)}{1} \\&= 1 \end{aligned}

\begin{aligned} \displaystyle \lim_{x\to 0} \frac{1-cos(x)}{x^2} &= \lim_{x\to 0} \frac{sin(x)}{2x} \\&= \frac{1}{2} \end{aligned}

\begin{aligned} \displaystyle \lim_{x\to 0} (cos(x))^\frac{1}{x} &= \lim_{x\to 0} e^{ln(cos(x)^\frac{1}{x}} \\&= \lim_{x\to 0} e^{\frac{1}{x}ln(cos(x))} \\&= e^{\displaystyle\lim_{x\to 0} \frac{ln(cos(x))}{x}} \end{aligned}

Now,

\begin{aligned} \displaystyle \lim_{x\to 0} \frac{ln(cos(x))}{x} &= \lim_{x\to 0} \frac{\frac{-sin(x)}{cos(x)}}{1} \\&= \lim_{x\to 0} -tan(x) = 0 \end{aligned}

So, $\displaystyle \lim_{x\to 0} (cos(x))^\frac{1}{x} = 1$

\begin{aligned} \displaystyle \lim_{x\to 0^+} x lnx &= \lim_{x\to 0^+} \frac{lnx}{\frac{1}{x}} \\&= \lim_{x\to 0^+} \frac{\frac{1}{x}}{-\frac{1}{x^2}} \\&= \lim_{x\to 0^+} -x = 0 \end{aligned}

\begin{aligned} \displaystyle \lim_{x\to \infty} x^3 e^{-x^2} &= \lim_{x\to \infty} \frac{x^3}{e^{x^2}} \\&= \lim_{x\to \infty} \frac{3x^2}{2x e^{x^2}} \\&= \lim_{x\to \infty} \frac{3}{4x e^{x^2}} = 0 \end{aligned}

\begin{aligned} \displaystyle \lim_{x\to \infty} (1+\frac{1}{x})^x &= \lim_{x\to \infty} e^{ln(1+\frac{1}{x})^x} \\&= \lim_{x\to \infty} e^{xln(1+\frac{1}{x})} \\&= e^{\displaystyle \lim_{x\to \infty} xln(1+\frac{1}{x})} \end{aligned}

Now,

\begin{aligned} \displaystyle \lim_{x\to \infty} x ln(1 + \frac{1}{x}) &= \lim_{x\to \infty} \frac{ln(1+\frac{1}{x})}{\frac{1}{x}} \\&= \lim_{x\to \infty} \frac{(-\frac{1}{x^2}\cdot\frac{1}{1+\frac{1}{x}})}{-\frac{1}{x^2}} \\&= \lim_{x\to \infty} \frac{1}{1+\frac{1}{x}} = 1 \end{aligned}

So, $\displaystyle \lim_{x\to \infty} (1+\frac{1}{x})^x = e^1 = e$

## MA101.10 How to find limits

If a function is know to be  continuous at $x=a$, then by the definition of continuity we have, $\displaystyle\lim_{t\to a} f(x) = f(a)$.

We will take is as known that the elementary functions – polynomial, rational (polynomial/polynomial), trigonometrical, exponential, logarithmic) – are continuous everywhere.

Therefore, although not very exciting, we are able to say such things as:

$\displaystyle\lim_{x\to \frac{\pi}{2}} sin x = sin \frac{\pi}{2}=1$

$\displaystyle\lim_{x\to 3} e^x = e^3$

$\displaystyle\lim_{x\to 2} \frac{x(x-1)}{x-3} = \frac{2(2-1)}{2-3} = -2$

A very useful rule (it’s a theorem really) for finding limits is L’Hôpitals rule (LH), which is really a collection of rules. The most comprehensive form of the rule is as follows:

Suppose you wish to find something like $\displaystyle\lim_{x\to[\;]} \frac{f(x)}{g(x)}$ where [ ] may be $a, a^+, a^-, \infty, -\infty$.

Suppose further that $\displaystyle\lim_{x\to[\;]} f(x) = \lim_{x\to [\;]} g(x) = \{\}$ where either

$\{\}=0$  – ($\frac{0}{0}$ type), or
$\{\}=\infty$ – ($\frac{\infty}{\infty}$ type).

Then the LH rule says look at $\displaystyle\lim_{x\to [\;]} \frac{f'(x)}{g'(x)}$ and if you can do it, say find it to be ( ) then the original limit is the same.

Danger: Do not use LH when it is not applicable.

For example $\displaystyle\lim_{x\to 0} \frac{sin(x) + 1}{sin(x)+2} = \frac{1}{2}$ is not the same as $\displaystyle\lim_{x\to 0} \frac{cos(x)}{cos(x)} = 1$.

The LH rule gives us with no effort some useful results. For example:

$\displaystyle\lim_{x\to 0} \frac{sin(x)}{x} = \lim_{x\to 0} \frac{cos(x)}{1} = 1$

## MA101.9 Continuity

Let a be a point in the domain of a function. The function is said to be continuous at $x=a$ if $\displaystyle\lim_{x\to a^+} f(x)= \lim_{x\to a^-} f(x) = f(a)$.

Otherwise f is said to be discontinuous at $x=a$. The function may be discontinuous at a for various reasons:

i) $\displaystyle\lim_{x\to a^-} f(x)$ may not exist, $\displaystyle\lim_{x\to a^+} f(x)$ may not exist, or both may not exist.
ii) both may exist, but may not be equal.
iii) both may exist, and be equal, but not be equal to f(a).

If f is continuous at all points of its domain it is simply said to be continuous. (Sometimes for emphasis mathematicians say f is continuous everywhere).

Note that it is a silly question to ask whether or not a function is continuous at a point which is not in the domain. For example is $\frac{x(x-1)}{x-1}$ continuous at x=1?

## MA101.8 Limits

Consider the function $f(x) = \begin{cases} \frac{-x}{2} + 4&\mbox{} x < 2 \\ -3 &\mbox{} x = 2 \\ x-1 &\mbox{} x > 2\end{cases}$

As x gets closer to 2 from above we notice that f(x) gets as close as we like to 1. We say that f has limit 1 as it tends to 2 from above (some people say from the right).

We write $\displaystyle\lim_{x\to 2^+} f(x) = 1$ or, $f(x) \rightarrow 1$ as $x \rightarrow a^+$.

Note that the value of f at x=2 is not relevant, for the concept of limits we do not even need x=2 to be in the domain for f.

In the same spirit,

$\displaystyle\lim_{x\to 2^-} f(x) = 3$.

Consider the function $f(x) = \begin{cases} \frac{1}{x} &\mbox{} x \neq 0 \\ 2 &\mbox{} x = 2\end{cases}$

As we get closer to 0 from above f(x) gets as large as you like, we say f(x) tends towards infinity.

We write, $\displaystyle\lim_{x\to 0^+} f(x) = \infty$, or
$f(x) \rightarrow \infty$ as $x \rightarrow 0^+$.

It does not mean that f(x) gets close to infinity. The value of f at x=0 is again irrelevant.

In the same spirit we have, $\displaystyle\lim_{x\to0^-} f(x) = -\infty$.

If for a function f(x) and a point x=a we have

$\displaystyle\lim_{x\to a^+} f(x) = \lim_{x\to a^-} f(x) = h \in \mathbb{R}$

then we say that f has a limit h as x tends to a, and we write $\displaystyle\lim_{x\to a} f(x) = h$ or $f(x) \rightarrow h$ as $x \rightarrow a$.

Likewise if $\displaystyle\lim_{x\to a^+} f(x) = \lim_{x\to a^-} f(x) = \infty$ we write simply $\displaystyle\lim_{x\to a} f(x) = \infty$.

For example, $\displaystyle\lim_{x\to 0} \frac{1}{x^2} = \infty$.

Consider the function $f(x) = \frac{2x+3}{x-1},x\neq 1$.

Then $f(x) = \frac{2+\frac{3}{x}}{1- \frac{1}{x}}, x \neq 0, x \neq 1$

As x gets large and positive we see that f(x) gets as close to 2 as we like.

We write $\displaystyle\lim_{x\to \infty} f(x) = 2$ or $f(x) \rightarrow 2$ as $x \rightarrow \infty$.

Likewise $\displaystyle\lim_{x\to -\infty} f(x) =2$ or $f(x) \rightarrow 2$ as $x \rightarrow -\infty$.

## MA101.7 Composition of functions

Also know as function of a function.

Let $f:A \rightarrow B$ and $g:U \rightarrow V$ be two functions such that $R(f) \subseteq D(g) = U$.

Then the function $h:A \rightarrow V$ defined by $h(x) = g(f(x))$ is called the composite of f and g. The function h is often denoted by $g \cdot f$ (read  g and f).

Thus $(g \cdot f)(x) = g(f(x))$.

## MA101.6 Monotonic Functions

A function f is said to be monotonic increasing in (a,b) if $\forall x_1,x_2$ in this interval, $x_1 < x_2 \Rightarrow f(x_1) \leq f(x_2)$.

The function is said to be strictly monotonic increasing in (a,b) if $\forall x_1,x_2$ in this interval, $x_1 < x_2 \Rightarrow f(x_1) < f(x_2)$.

A function f is said to be monotonic decreasing in (a,b) if $\forall x_1,x_2$ in this interval, $x_1 < x_2 \Rightarrow f(x_1) \geq f(x_2)$.

The function is said to be strictly monotonic decreasing in (a,b) if $\forall x_1,x_2$ in this interval, $x_1 < x_2 \Rightarrow f(x_1) > f(x_2)$.

Example

Let f be a strictly monotonic increasing function. Prove that f is an injection.

Solution

Let f be a strictly monotonic increasing function.
Let $x_1,x_2 \in D(f)$ be such that $f(x_1) = f(x_2)$.
Now $x_1 < x_2 \Rightarrow f(x_1) < f(x_2)$ which is a contradiction, and $x_2 < x_1 \Rightarrow f(x_2) < f(x_1)$ which also is a contradiction.
So $x_1 = x_2$ and thus f is an injection.

## MA101.5 Odd and even functions

A function is said to be an even function if,

$\forall x \in D(f), f(x) = f(-x)$

and to be an odd function if,

$\forall x \in D(f), f(x) = -f(x)$.

Note that these definitions require D(f) to lie symmetrically about x=0 (ie $x \in D(f) \Rightarrow -x \in D(f)$ ). Clearly even functions have graphs that are symmetrical about the y-axis, whilst for odd functions the origin is a centre of rotational symmetry.

Examples are $x^2, \; |x|$ for even, and $x^3, \; sin(x)$ for odd. Most functions are neither odd nor even. However every function (subject to having a symmetrical domain) may be expressed as the sum of an odd and even function since $f(x) = \frac{1}{2}\{f(x) +f(-x)\} + \frac{1}{2}\{f(x) - f(-x)\}$, which is even + odd.

Example

\begin{aligned} f(x) &= e^x sin(x) \\ &= \frac{1}{2}\{e^x sin(x) + e^{-x} sin(-x)\} + \frac{1}{2}\{e^x sin(x) - e^{-x} sin(-x)\} \\&= sin(x) \frac{e^x-e^{-x}}{2} + sin(x) \frac {e^x + e^{-x}}{2} \\&= sin(x)sinh(x) + sin(x)cosh(x) \end{aligned}

ie. even + odd.

## MA101.4 Inverse Functions

Let f be a function and let $y=f(x), x \in D(f)$. We often call x the independent variable and y the dependent variable. Now consider specifying a number $y \in R(f)$ and working out what x’s are in the domain of the f, for which $f(x)=y$. There may of course be many such x’s corresponding to a particular y, but functions for which there is only ever one such x have a special name, they are called injections (their old name was one-to-one).

##### Definition

A function is an injection if for all x,y $f(x)=f(y) \Rightarrow x=y$

Or if you prefer $\forall x,y:\; x \neq y \Rightarrow f(x) \neq f(y)$

($\forall$ – for all and $\Rightarrow$ – implies.)

Injections are easily recognised from their graphs in that any line parallel to the x axis will meet the graph in at most one point.

Let f be an injection, then we use $f^{-1}(y), y \in R(y)$ to denote that single element $x \in D(f)$ which is such that $f(x)=y$.

The function $f^{-1} : R(f) \rightarrow D(f)$ is called the inverse function of the f.

By the definition of $f^{-1}$ we have, $D(f^{-1}) = R(f)$.

It is also clear that $R(f^{-1}) = D(f)$.

Note that functions that are not injections do not have inverse functions.

Example

Find the inverse function of the function f given by:

$f(x) = \begin{cases} (1 - x)^2 &\mbox{}0 \leq x \leq 1 \\ x+1 &\mbox{} 1 < x < 2 \end{cases}$

Solution: To have a sketch is a great help

$D(f)=[0,2)$ and $R(f)=[0,1] \cup (2,3)$

So $D(f^{-1}) = R(f) = [0,1] \cup (2,3)$.

Since the rule for f is given in two pieces we find the rule for $f^{-1}$ in two pieces.

$Let\; y \in [0,1] \\ So,\\ y=(1-x)^2 \\ y = 1 - 2x +x^2\\ x^2 - 2x + 1 - y = 0 \\ x = \frac{2-\sqrt{4-4(1-y)}}{2} \\ x = 1 - \sqrt{y}$

$Let\; y \in (2,3) \\So,\\ y = x+1 \; thus \; x = y-1$.

Whence $f^{-1}$ is given by

$f^{-1}(y) = \begin{cases} 1-\sqrt{y} &\mbox{}0 \leq y \leq 1 \\ y-1 &\mbox{} 2 < y < 3 \end{cases}$

Note that f is an injection and so $f^{-1}$ exists then the graphs f and $f^{-1}$ will be reflections of one another in the line y = x.