I. H. Limits in More Detail. (Draft-Work in Progress.)
© 2014 M. Flashman.

Preface: Central to all the work in this chapter related to the derivative is the concept of limit, which we have been using at this early stage without a formal definition. In this work many of the results we have obtained have been presented relying on the sensibility of the estimations that have been presented for motivation.
For example we have suggested that if $f(x) = x^2 +x + 1$ then as $x \rightarrow 3, x^2 \rightarrow 9$,  and thus $f(x) \rightarrow 13.$ We have expressed this with the notation $ \lim\limits_{x \rightarrow 3} \ x^2 +x + 1 = 13$.

In this section we will explore in greater detail the meaning and definition of the limit concept and introduce the important related concepts of continuity and continuous functions.

Limits: A short sensible trip from the informal to the formal.
In our first explorations of estimates of the derivative we considered a function, $f$, a number $a$ and the numerical expressions $\frac {f(x)-f(a)}{x-a}$ for numbers $x$ close to but not equal to $a$. the number that these numbers approached as $x$ was taken closer to$a$ was described as the limit of $\frac {f(x)-f(a)}{x-a}$ as $x$ approached $a$. The expression $\frac {f(x)-f(a)}{x-a}$ was often simplified with algebra so the analysis of the estimation made sense without much further discussion. In delving more deeply into the meaning of limits we will simplify the issues by recognizing that for $x \ne a$, $\frac {f(x)-f(a)}{x-a}$ is a function of $x$, and so we replace the quotient with a consideration of a function $P$ which is defined for all numbers $x \ne a$ in an open interval that contains the number $a$.

Informally we say that $L$ is the limit of $P(x)$ as $x$ approaches $a$ if the values $P(x)$ are close to $L$ when $x$ is close to $a$, or dynamically,  as $x$ is taken closer and closer to $a$, the value $P(x)$ is closer and closer to $L$. Our task is to make the use of the terms for closeness more rigorous in developing a definition for the limit concept in its usage in this context.

"Close": To measure the closeness of numbers $x$ and $a$ we use the absolute value of the difference $|x-a|$ to give a non-negative measure of the distance between the numbers. [$|x-a| = 0$ if and only if $x=a$.] We say numbers are close when $|x-a|$ is small. Of course this translates the question of "what is close?" to"what is small". More important is what is means to be "closer" and the related comparison, "smaller".  So we will say that $x_2$ is closer to $a$ than $x_1$ if $0\le |x_2-a| < |x_1-a|$.

We can also describe special intervals that contain $a$ for a positive number $\delta \gt 0 $, $(a-\delta, a+\delta)$ as $\{x : |x-a|\le \delta\}$. If we remove the number $a$ from this interval we have $\{x : |x-a|\lt \delta\}$.

Our interest in the function $P$ is concerned with its values $P(x)$ when $x$ is close to but not equal to $a$. We use the positive number, $\delta (\gt 0)$ to control the closeness of $x$ to $a$ by requiring that $0 \lt |x-a| \lt \delta$ so that $x \in (a-\delta,a) \cup (a, a+\delta)$.

Example I.H.1. From an informal point of view, it should make sense  that $ \lim\limits_{x \rightarrow 3} \ 2x-1 = 5$.
We explore the meaning of this more carefully.
Suppose $P(x) = 2x-1$, $a = 3$, and $\delta ^*= 0.5$.
Then $0 \lt |x-a| \lt \delta^*$ means $x \ne 3$ and $x \in (2.5,3.5)$ or $2.5 \lt x\lt 3.5$.
Using simple properties of inequalities we have that 
$2\cdot(2.5) \lt 2\cdot x\lt 2\cdot(3.5)$ and
$2\cdot(2.5)-1 \lt 2\cdot x-1\lt 2\cdot(3.5)-1$ so
$4 \lt P(x) \lt 6$ or 
$-1 \lt P(x)-5 \lt 1$.
Thus $| P(x) -5| \lt 1$.
So it appears that for these $x$, $P(x)$ is relatively close to $5$, within $1$ of $5$.
We can generalize this analysis by replacing $0.5$ in our work with an unknown $\delta^* (\gt 0)$.
Now $0 \lt |x-a| \lt \delta^*$ means $x \ne 3$ and $x \in (3-\delta^*,3+\delta^*)$ or $3-\delta^* \lt x\lt 3+\delta^*$.
Thus again using simple properties of inequalities we have  that
$2\cdot(3-\delta^*) \   \lt 2\cdot x\lt 2\ \cdot(3+\delta^*)$ and
$2\cdot(3-\delta^*)-1 \lt 2\cdot x-1\lt 2\cdot(3+\delta^*)-1$ so
$5-2\delta^* \ \ \lt P(x) \ \ \lt 5 + 2\delta^*$ and
$\ \ -2\delta^* \ \lt P(x)-5  \   \lt  2\delta^*$.
Thus $|P(x)-5| \lt 2\delta^*$

So for these $x$, $P(x)$ is within $2\delta^*$ of $5$.

So for smaller values of $\delta^*$, $x$ is closer to $a$ and $P(x)$ is closer to 5.
So it should make sense that $ \lim\limits_{x \rightarrow 3} \ 2x-1 = 5$.

Example I.H.1 -continued:
Find $\delta^*$ so that if (or when), $0 \lt |x-3| \lt \delta^*$,  $|P(x)-5| \lt 0.01$.
Solution: from the previous work, for an arbitrary choice of $\delta^*$, $|P(x)-5| \lt 2\delta^*$, so we can use any value for $\delta^* $ with $2\delta^* \le 0.01$. For convenience, we choose $\delta^* = 0.005$ and the previous analysis shows that when  $0 \lt |x-3| \lt \delta^*$,  $|P(x)-5| \lt 0.01$ as required.

Reflecting  on the work in Example I.H.1, here is the commonly used definition of the limit concept for functions.

Definition (Limit): Suppose $r (\gt 0)$, $a$ and $L$ are real numbers and $P$ is a function defined on the deleted interval, ${x: 0 \lt |x-a| \lt r}.
We say that $L$ is the limit of $P$ as $x$ approaches $a$ , denoted $\lim \limits_{x \rightarrow a}f(x) = L$ if the following statement is true:
For any $\epsilon > 0$, there is a number $\delta^* >0$ so that if $x$ is in the domain of $P$ and$0 \lt |x-a| \lt \delta^*$, then $|P(x) -L| \lt \epsilon $.
Another way to express this defining statement:
For any $\epsilon > 0$, there is a number $\delta^* >0$ so that $|P(x) -L| \lt \epsilon $ whenever $x$ is in the domain of $P$ and $0 \lt |x-a| \lt \delta^*$.

Example I.H.1 revisited.
Suppose $P(x) = 2x-1$, $a = 3$ and $L = 5$. Start with assuming we have been given a number $\epsilon > 0$.
We want to find a value for $\delta^*$ so that when $0 \lt |x-a| \lt \delta^*$ we can show that $|f(x) -L| \lt \epsilon $.
From the previous work  with $0 \lt |x-a| \lt \delta$ we  can choose $\delta^* = \frac {\epsilon}2$.
SO: if $0 \lt |x-a| \lt \delta^* = \frac {\epsilon}2$ then $|P(x)-5| \lt 2\delta^* = \epsilon$.
Thus $ \lim\limits_{x \rightarrow 3} \ 2x-1 = 5$.
Comments on the structure of verifying the definition of limit:
The logical structure for the defining statement of the limit definition is subtle and sometimes confusing at first use.
  1. The statement starts with "for any $\epsilon >0$ ". In developing a demonstration that the statement is true for a particular function $P$ and numbers $a$ and $L$, an argument usually begins by selecting an arbitrary number (thinking of it as a small number) designated by the Greek letter $\epsilon.$ This number, $\epsilon$, is considered as a control on the difference allowed ("error") in the value of the function  from the candidate for the limit value, $L$. The numbers $L$ and $L \pm \epsilon$ can be visualized in the target for the function all related to the interval $(L-\epsilon, L + \epsilon)$.

  2. After establishing the number $\epsilon$, the statement asserts "there is a number $\delta^*>0$" i.e., the existence of a second positive number, denoted $\delta^*$, with a particular property related to the number $\epsilon$. Finding a satisfactory $\delta^*$ for the chosen $\epsilon$ can be a serious challenge requiring insight into the nature of the function $P$. In a sense, finding a value for $\delta^*$ that depends systematically on the choice of $\epsilon$ can be thought of as finding a way to determine $\delta^*$ as a function of $\epsilon$. The numbers $a$ and $a\pm \delta^*$ can be visualized in the domain of the function, all related to the interval $(a-\delta^*,a+\delta^*)$.

  3. The actual search for $\delta^*$ that works for the chosen $\epsilon$ can involve investigations solving the inequality $a-\delta^* \lt P(x) \lt a+\delta^*$ for $x$ or investigating what happens to $x$ in the inequality $a-\delta^* < x< a+\delta^*$ when the function $P$ is applied.

  4. The key property of the number $\delta^*$ is something that must be demonstrated to complete the argument that $L$ is the "limit". To demonstrate that $\delta^*$ satisfies the required property, the argument now must consider an number $x$ which is within $\delta^*$ of $a$ but not equal to $a$. So the argument proceeds to suppose a number $x$ where $x \ne a$ and $|x-a| \lt \delta^*$, i.e., $a-\delta^* < x< a+\delta^*$. From this assumption the argument must conclude that the function's value for $x$, namely $P(x)$, is within $\epsilon$ of  $L$, i.e., $L-\epsilon \lt P(x) \lt L + \epsilon$ or  $|P(x)-L| \lt \epsilon$.

Example I.H.2. We can also use the definition to show that a number is not the limit. Is it possible that $ \lim\limits_{x \rightarrow 3} \ 2x-1 = 6$? To show this statement is false,
we use the definition with again with $P(x) = 2x-1$, $a = 3$, but now with $L = 6$. If the statement is true then it should be verifiable using any number for $\epsilon$. So in particular choose $\epsilon = 0.5$.
If we choose any $\delta ^* \gt 0$ then consider $x^*=3-\frac {\delta^*}2$ so that $0 \lt |x^*-3| \lt \delta^*$. Furthermore we have that $f(x^*) = 5 -\delta^* \lt 6- 0.5 = 5.5$, so $|f(x*) - 6| > 0.5$. Thus for  $\epsilon = 0.5$ there is no $\delta^*$ that will satisfy the required condition that $|P(x) -6| \lt \epsilon $ whenever $0 \lt |x-a| \lt \delta^*$f or $6$ to be the limit.

Once the definition of limit has been understood, we can look a little further and define the concept of continuity for a function  $P$ at a number $a$ in the domain of $P$.
Definition (Continuous) : Suppose $P$ is a function and $a$ is a number in the domain of $P$.
We say that $P$ is continuous at $a$ if $ \lim\limits_{x \rightarrow a} \ P(x) = P(a)$. This can be translated into a more complex statement using the definition of limit:
$P$ is continuous at  $a$ if
for any $\epsilon > 0$, there is a number $\delta^* >0$ so that if $x$ is in the domain of $P$ and  $0 \lt |x-a| \lt \delta^*$, then $|P(x) -P(a)| \lt \epsilon $.
Another way to express this defining statement:
For any $\epsilon > 0$, there is a number $\delta^* >0$ so that  $|P(x) -P(a)| \lt \epsilon $ whenever $x$ is in the domain of $P$ and $0 \lt |x-a| \lt \delta^*

Once these two definitions are established and understood, there are several related results about limits and continuity that can be proven rigorously using the definitions and basic properties of the algebra of real numbers and inequalities.
Some of these will be proven in detail in Appendix I.A. For now these properties will be stated and used in subsequent work as being sensible tools of thought.

Theorem I.H.1: Basic Limits for Core Functions:
$ \lim\limits_{x \rightarrow a} \ c = c$ where $c$ is a constant.
$ \lim\limits_{x \rightarrow a} \ x = a$.
$ \lim\limits_{x \rightarrow a} \sin(x) = \sin(a)$.
$ \lim\limits_{x \rightarrow a} \cos(x) = \cos(a)$.
$ \lim\limits_{x \rightarrow a} \ b^x = b^a$ where $b >0 , b\ne 1$.
$ \lim\limits_{x \rightarrow a} \ log_b(x) = log_b(a)$ where $b >0 , b\ne 1$.

Theorem I.H.2: Basic Algebraic Properties of Limits. 
Suppose $r (\gt 0)$, $a, c, L, L^*$ and $M$ are real numbers and $P$, $Q$ and $R$ are function defined on the deleted interval, $\{x: 0 \lt |x-a| \lt r\}$.
Suppose further that  
$ \lim\limits_{x \rightarrow a} \ P(x) = L$, $ \lim\limits_{x \rightarrow a} \ P(x) = L^*$, and $ \lim\limits_{x \rightarrow a} \ Q(x) = M$. Then:
i. $L = L^*$.  [Limits are unique.]
ii. $ \lim\limits_{x \rightarrow a} \ P(x) \pm Q(x)= L \pm M $.
$ \lim\limits_{x \rightarrow a} \ c \cdot P(x) = c\cdot L$.
iv. $ \lim\limits_{x \rightarrow a} \ P(x)\cdot Q(x) = L\cdot M$.
v. If $M \ne 0$ then
$ \lim\limits_{x \rightarrow a} \ \frac {P(x)}{Q(x)} = \frac LM$.
vi. If $R(x)$ is between $P(x)$ and $Q(x)$ for the deleted interval
and $M = L$ then 
$ \lim\limits_{x \rightarrow a} \ R(x) = L$

Theorem I.H.3: Limits for Composed and Inverse Functions:

Suppose $r (\gt 0)$, $a, L$,and $M$ are real numbers and $P$  is defined on the deleted interval, $\{x: 0 \lt |x-a| \lt r\}$ with $P(x) \ne L$
and $Q$ is defined on the deleted interval $\{u: 0 \lt |u-L| \lt r\}$ .
Suppose further that 
$ \lim\limits_{x \rightarrow a} \ P(x) = L$, and $ \lim\limits_{u \rightarrow L} \ Q(u) = M$. Then $ \lim\limits_{x \rightarrow a} \ Q(P(x)) = M$.
If  $Q(P(x)) = x$, then $M = a$.

Theorem I.H.4: Continuity for Core Functions: the following core functions are continuous for every $x$ in their domain:
$f(x) = c$ where $c$ is a constant.
$f(x) = x$.
$f(x) = \sin(x) $.
$f(x) = \cos(x) $.
$f(x) = b^x $ where $b >0 , b\ne 1$.
$f(x) = \log_b(x) $ where $b >0$, $b\ne 1$.

Theorem I.H.5: Basic Algebra of Continuous Functions. 
Suppose $a$ is a real number in the domain of functions $P$, $Q$ and $R$.
Suppose further that  
$P$ and $Q$ are continuous at $a$ Then:
i. $S(x) = P(x) \pm Q(x) $ is continuous at $a$.
$S(x) = c  \cdot P(x)$ is continuous at $a$.
iii $S(x) =P(x)\cdot Q(x)$ is continuous at a$a$ .
v. If $Q(a) \ne 0$ then $ S(x) =\frac {P(x)}{Q(x)}$ is continuous at $a$.
vi. If $R(x)$ is between $P(x)$ and $Q(x)$ for the interval
and $P(a)= Q(a)$ then 
$R$ is continuous at $a$

Theorem I.H.6: Continuity for Composed Functions:

Suppose $r (\gt 0)$, $a$,and $b=P(a)$ are real numbers and $P$  is defined on an interval containing $a$,
and $Q$ is defined on an interval containing $b=f(A)$ .
Suppose further that 
$P$ is continuous at $a$ , and $Q$ is continuous at $b=P(a)$.  Then $S(x) = Q(P(x))$ is continuous at $a$.