Add content for functions/continuity-limits

Add actual content to the skeleton of the `functions/continuity-limits`
section.

Signed-off-by: Eggert Karl Hafsteinsson <[email protected]>
Signed-off-by: Teodor Dutu <[email protected]>
Signed-off-by: Razvan Deaconescu <[email protected]>

chapters/functions/continuity-limits/reading/read.md

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+# Continuity and Limits
+
+## The Concept of Continuity
+
+A function is continuous if it has no jumps.
+Thus, small changes in each $x_0$, the input, correspond to small changes in the output, $f(x_0)$.
+
+![Fig. 15](../media/11_1_The_concept_of_continuity.png)
+
+Figure: The above figure is an example of linear growth.
+
+Thomas Robert Malthus (1766-1834) warned about the dangers of uninhibited population growth.
+
+### Details
+
+A function is said to be discontinuous if it has jumps.
+The function is continuous if it has no jumps.
+Thus, for a continuous function, small changes in each $x_0$, the input, correspond to small changes in the output, $f(x_0)$.
+
+:::note Note
+
+Note that polynomials are continuous as are logarithms (for positive numbers).
+
+:::
+
+## Discrete Probabilities and Cumulative Distribution Functions
+
+The cumulative distribution function for a discrete random variable is discontinuous.
+
+![Fig. 16](../media/11_2_Discrete_probabilities_and_cumulative_distribution_functions.png)
+
+### Details
+
+:::note Definition
+
+If $X$ is a random variable with a discrete probability distribution and the probability mass function of
+
+$$p(x)=P[X=x]$$
+
+then the **cumulative distribution function**, defined by
+
+$$F(X)=P[X\leq x]$$
+
+is discontinuous, i.e. it jumps at points in which a positive probability occurs.
+
+:::
+
+:::note Note
+
+When drawing discontinuous functions, it is common practice to use a filled circle at $(x,f(x))$ to clarify what the function value is at a point $x$ of discontinuity.
+
+:::
+
+### Examples
+
+:::info Example
+
+If a coin is tossed three independent times and $X$ denotes the number of heads, then $X$ can only take on the values $0$, $1$, $2$ and $3$.
+The probability of landing exactly $x$ heads, $P(X=x)$, is $p(x) = \displaystyle\binom{n}{x} p^n (1-p)^{n-x}$.
+The probabilities are:
+
+$$
+\begin{array}{c c c}
+  \hline \\
+  x & p(x) & F(x) \\
+  \hline \\
+  0 & \displaystyle\frac{1}{8} & \displaystyle\frac{1}{8} \\
+  1 & \displaystyle\frac{3}{8} & \displaystyle\frac{4}{8} \\
+  2 & \displaystyle\frac{3}{8} & \displaystyle\frac{7}{8} \\
+  3 & \displaystyle\frac{1}{8} & 1 \\
+  \hline
+\end{array}
+$$
+
+The cumulative distribution function, $F(x)=P[X \leq x] = \displaystyle\sum_{t\leq x} p(t)$ has jumps and is therefore discontinuous.
+
+:::
+
+:::note Note
+
+Notice on the above figure how the circles are filled in, the solid circles indicate where the function value is.
+
+:::
+
+## Notes on Discontinuous Function
+
+A function is discontinuous for values or ranges of the variable that do not vary continuously as the variable increases.
+In other words, breaks or jumps.
+
+![Fig. 17](../media/11_3_Notes_on_discontinuous_function.png)
+
+Figure: $f(x) = \displaystyle\frac{1}{x}$, where $x\neq 0$
+
+### Details
+
+A function can be discontinuous in a number of different ways.
+Most commonly, it may jump at certain points or increase without bound in certain places.
+
+Consider the function $f$, defined by $f(x)= 1/x$ when $x\neq 0$.
+Naturally, $1/x$ is not defined for $x=0$.
+This function increases towards $+\infty$ as $x$ goes to zero from the right but decreases to $-\infty$ as $x$ goes to zero from the left.
+Since the function does not have the same limit from the right and the left, it cannot be made continuous at $x=0$ even if one tries to define $f(0)$ as some number.
+
+## Continuity of Polynomials
+
+All polynomials, $p(x)=a_0+a_1x+a_2x^2+\ldots +a_n x^n$ are continuous.
+
+![Fig. 18](../media/11_4_Continuity_of_polynomials.png)
+
+### Details
+
+It is easy to show that simple polynomials such as $p(x)=x$, $p(x)=a+bx$, $p(x)=ax^2+bx+c$ are continuous functions.
+
+It is generally true that a polynomial of the form
+
+$$p(x)=a_0+a_1x+a_2x^2+\ldots +a_n x^n$$
+
+is a continuous function.
+
+## Simple Limits
+
+A **limit** is used to describe the value that a function or sequence approaches as the input or index approaches some value.
+Limits are used to define continuity, derivatives and integrals.
+
+![Fig. 19](../media/11_5_Simple_limits.png)
+
+Figure: $f(x) = x^x$, for $x>0$
+
+### Details
+
+:::note Definition
+
+A **limit** describes the value that a function or sequence approaches as the input or index approaches some value.
+
+:::
+
+Limits are essential to calculus (and mathematical analysis in general) and are used to define continuity, derivatives and integrals.
+
+Consider a function $f(x)$ and a point ${x}_0$.
+If $f(x)$ gets steadily closer to some number $c$ as $x$ gets closer to a number $x_0$, then $c$ is called the limit of $f(x)$ as $x$ goes to $x_0$ and is written as:
+
+$$c = \lim_{x\to x_0}f(x)$$
+
+If $c = f(x_0)$ then $f$ is **continuous** at $x_0$.
+
+### Examples
+
+:::info Example
+
+Evaluate the limit of $f(x) = \displaystyle\frac{x^{2}-16}{x-4}$ when $x\rightarrow 4$, or
+
+$$\lim_{x\rightarrow 4} \displaystyle\frac{x^{2}-16}{x-4}$$
+
+Notice that in principle we cannot simply stick in the value $x=4$ since we would then get $0/0$ which is not defined.
+However we can look at the numerator and try to factor it.
+This gives us:
+
+$$\displaystyle\frac{x^{2}-16}{x-4} = \displaystyle\frac{(x-4)(x+4)}{x-4} = x+4$$
+
+and the result has the obvious limit of $4+4=8$ as $x\to 4$.
+
+:::
+
+:::info Example
+
+Consider the function:
+
+$$g(x) = \displaystyle\frac{1}{x}$$
+
+where $x$ is a positive real number.
+As $x$ increases, $g(x)$ decreases, approaching $0$ but never getting there since $\displaystyle\frac{1}{x}=0$ has no solution.
+One can therefore say: "The limit of $g(x)$, as $x$ approaches infinity, is $0$," and write:
+
+$$\lim_{x\to\infty} g(x)=0$$
+
+:::
+
+## More on Limits
+
+Limits impose a certain range of values that may be applied to the function.
+
+![Fig. 20](../media/11_6_More_on_limits.png)
+
+Figure: The function $f(x)= \displaystyle\frac{1}{1+e^{-x}}$
+
+### Examples
+
+:::info Example
+
+The Beverton-Holt stock recruitment curve is given by:
+
+$$R=\displaystyle\frac{\alpha S}{1+\displaystyle\frac{S}{K}}$$
+
+where $\alpha, K >0$ are constants and $S$ is biomass and $R$ is recruitment
+
+The behavior of this curve as $S$ increases $S\rightarrow\infty$ is
+
+$$\lim_{S\to\infty}\displaystyle\frac{\alpha S}{1+\displaystyle\frac{S}{K}} =\alpha K$$
+
+This is seen by rewriting the formula as follows:
+
+$$\lim_{S\to\infty}\displaystyle\frac{\alpha S}{1+\displaystyle\frac{S}{K}} = \lim_{S\to\infty}\displaystyle\frac{\alpha }{\displaystyle\frac{1}{S}+\displaystyle\frac{1}{K}} =\alpha K$$
+
+:::
+
+:::info Example
+
+A popular model for proportions is:
+
+$$f(x) = \displaystyle\frac{1}{1+e^{-x}}$$
+
+As x increases, $e^{-x}$ decreases which implies that the term $1+e^{-x}$ decreases and hence $\displaystyle\frac{1}{1+e^{-x}}$ increases, from which it follows that $f$ is an increasing function.
+
+Notice that $f(0)=\displaystyle\frac{1}{2}$ and further,
+
+$$\lim_{x\to\infty} f(x) = 1$$
+
+This is seen from considering the components.
+Since $e^{-x} = \displaystyle\frac{1}{e^{x}}$ and the exponential function goes to infinity as $x\to\infty$, $e^{-x}$ goes to $0$ and hence $f(x)$ goes to 1.
+
+Through a similar analysis one finds that:
+
+$$\lim_{x\to-\infty} f(x)=0$$
+
+since, as $x\rightarrow \infty$, first $-x\rightarrow \infty$ and second $e^{-x} \rightarrow \infty$.
+
+:::
+
+:::info Example
+
+Evaluate the limit of $f(x) = \displaystyle\frac{\sqrt{x + 4} - 2}{x}$ as $x \to 0$, i.e. solve:
+
+$$\lim_{x \to 0} \displaystyle\frac{\sqrt{x + 4} - 2}{x}$$
+
+Since there is an $x$ in the denominator we cannot just direct substitute the 0 as $x$ sunce that would give us $\displaystyle\frac{0}{0}$, which is an indeterminate form.
+We must do some algebra first.
+The square root makes this a little bit tricky.
+The way to get rid of the radical is to multiply the numerator by the conjugate.
+
+$$\displaystyle\frac{\sqrt{x + 4} - 2}{x} \cdot \displaystyle\frac{\sqrt{x + 4} + 2}{\sqrt{x + 4} + 2}$$
+
+This gives us:
+
+$$\displaystyle\frac{(\sqrt{x + 4})^2 + 2(\sqrt{x+4}) - 2(\sqrt{x+4}) -4}{x(\sqrt{x + 4} + 2)}$$
+
+The numerator reduces to $x$, and the $x$'s will cancel out, leaving us with:
+
+$$\displaystyle\frac{1}{\sqrt{x + 4} + 2}$$
+
+At this point we can direct substitute 0 for $x$, which will give us:
+
+$$\displaystyle\frac{1}{\sqrt{0 + 4} + 2}$$
+
+Therefore:
+
+$$\lim_{x \to 0} \displaystyle\frac{\sqrt{x + 4} - 2}{x} = \displaystyle\frac{1}{4}$$
+
+:::
+
+## One-sided Limits
+
+$f(x)$ may tend towards different numbers depending on whether $x$ approaches $x_0$ from left or right, usually written:
+
+$x \rightarrow x_{0+}$ (from the right)
+
+$x \rightarrow x_{0-}$ (from the left).
+
+![Fig. 21](../media/11_7_one-sided_limits.png)
+
+### Details
+
+Sometimes a function is such that $f(x)$ tends to different numbers depending on whether $x \rightarrow x_0$ from the right ( $x \rightarrow x_{0+}$ ) or from the left ($x \rightarrow x_{0-}$).
+
+If
+
+$$\lim_{x \to x_{0+}} f(x)=f(x_0)$$
+
+then we say that $f$ is continuous from the right at $x_0$.
+Same thing goes for the limit from the left.
+In order for the limit to exist at the point (that is the overall limit, regardless og direction) then it must hold true that
+
+$$\lim_{x \to x_{0-}} = \lim_{x to x_{0+}}$$
+
+i.e., the limit is the same from both directions.