For example we have suggested that if $f(x) = x^2 +x + 1$ then as

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.

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 $x$

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 $x \in (a-\delta,a) \cup (a, a+\delta)$.

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

So it appears that for these $x$, $P(x)$ is relatively close to $5$, within $1$ of $5$.$2\cdot(2.5) \lt 2\cdot x\lt 2\cdot(3.5)$ 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 $| P(x) -5| \lt 1$.

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.

We say that

Another way to express this defining statement:

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

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.

- 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 $(L-\epsilon, L + \epsilon)$.

- 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 $(a-\delta^*,a+\delta^*)$.

- 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.

- 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 $|P(x)-L| \lt \epsilon$.

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

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 $P$.

We say that

$P$ is continuous at $a$ if

Another way to express this defining statement:

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.

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

i. $L = L^*$. [Limits are unique.]ii. $ \lim\limits_{x \rightarrow a} \ P(x) \pm Q(x)= L \pm M $.iii. $ \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$ |

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 $P(x) \ne L$

Suppose further that

If $Q(P(x)) = x$, then $M = a$.

Suppose $a$ is a real number in the domain of functions $P$, $Q$ and $R$.

Suppose further that

i. $S(x) = P(x) \pm Q(x) $ is continuous at $a$.ii. $S(x) = c \cdot P(x)$ 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 $a$ |

Suppose $r (\gt 0)$, $a$,and $b=P(a)$ are real numbers and $P$ is defined on an interval containing $a$,

Suppose further that