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Welcome to the Wikibook of

Calculus

beginning with
Prerequisites and the Disciplines
of Differentiation and Integration,
and continuing with
Further Studies.

Mathematics is a universal study. It can be studied for its own sake as pure mathematics, or it can be applied to the natural world and to humanity as in the study of statistics. In the natural world, the nautilus shell presents one of the finest natural examples of a logarithmic spiral.

Mission

We at the Wikibook of Calculus aim to write a quality textbook to facilitate mastering the topics of differentiation, integration, infinite series, conics, and parametric and polar equations. Please contribute wherever you feel the need.

Printable Version

There is a printable version of this book.

Introduction

Precalculus

Contents

Limits

Contents

Differentiation

Basics of Differentiation

Applications of Derivatives

Important Theorems

Integration

Basics of Integration

Integration techniques

Applications of Integration

Parametric Equations

Contents

Polar Equations

Contents

Sequences and Series

Basics

Calculus and Series

Vector Calculations

Contents

Multivariable & Differential Calculus

Contents

Extensions

Further Analysis

Formal Theory of Calculus

References

Contents

Acknowledgements and Further Reading

Introduction

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Introduction

Calculus is a tool used almost everywhere in the modern world to describe change and motion. Its use is widespread in science, engineering, medicine, business, industry, and many other fields. Just as algebra introduces students to new ways of thinking about arithmetic problems (by way of variables, equations, functions and graphs), calculus introduces new ways of thinking about algebra problems (considering, for example, how the height of a point moving along a graph changes as its horizontal position changes).

For a different perspective, have a look at the Wikipedia entry for Calculus.

What is calculus?

Calculus is the branch of mathematics dealing with instantaneous rates of change of continuously varying quantities. For example, consider a moving car. It is possible to create a function describing the displacement of the car (where it is located in relation to a reference point) at any point in time as well as a function describing the velocity (speed and direction of movement) of the car at any point in time. If the car were traveling at a constant velocity, then algebra would be sufficient to determine the position of the car at any time; if the velocity is unknown but still constant, the position of the car could be used (along with the time) to find the velocity.

However, the velocity of a car cannot jump from zero to 35 miles per hour at the beginning of a trip, stay constant throughout, and then jump back to zero at the end. As the accelerator is pressed down, the velocity rises gradually, and usually not at a constant rate (i.e., the driver may push on the gas pedal harder at the beginning, in order to speed up). Describing such motion and finding velocities and distances at particular times cannot be done using methods taught in pre-calculus, but it is not only possible but straightforward with calculus.

Calculus has two basic applications: differential calculus and integral calculus. The simplest introduction to differential calculus involves an explicit series of numbers. Given the series (42, 43, 3, 18, 34), the differential of this series would be (1, -40, 15, 16). The new series is derived from the difference of successive numbers which gives rise to its name "differential". Rarely, if ever, are differentials used on an explicit series of numbers as done here. Instead, they are derived from a series of numbers defined by a continuous function which are described later.

Integral calculus, like differential calculus, can also be introduced via series of numbers. Notice that in the previous example, the original series can almost be derived solely from its differential. Instead of taking the difference, however, integration involves taking the sum. Given the first number of the original series, 42 in this case, the rest of the original series can be derived by adding each successive number in its differential (42, 42+1, 43+(-40), 3+15, 18+16). Note that knowledge of the first number in the original series is crucial in deriving the integral. As with differentials, integration is performed on continuous functions rather than explicit series of numbers, but the concept is still the same. Integral calculus allows us to calculate the area under a curve of almost any shape; in the car example, this enables you to find the displacement of the car based on the velocity curve. This is because the area under the curve is the total distance moved, as we will soon see.

Why learn calculus?

Calculus is essential for many areas of science and engineering. Both make heavy use of mathematical functions to describe and predict physical phenomena that are subject to continual change, and this requires the use of calculus. Take our car example: if you want to design cars, you need to know how to calculate forces, velocities, accelerations, and positions. All require calculus. Calculus is also necessary to study the motion of gases and particles, the interaction of forces, and the transfer of energy. It is also useful in business whenever rates are involved. For example, equations involving interest or supply and demand curves are grounded in the language of calculus.

Calculus also provided important tools in understanding functions and has led to the development of new areas of mathematics including real and complex analysis, topology, and non-euclidean geometry.

What is involved in learning calculus?

Learning calculus, like much of mathematics, involves two parts:

Understanding the concepts: You must be able to explain what it means when you take a derivative rather than merely apply the formulas for finding a derivative. Otherwise, you will have no idea whether or not your solution is correct. Drawing diagrams, for example, can help clarify abstract concepts.

Symbolic manipulation: Like other branches of mathematics, calculus is written in symbols that represent concepts. You will learn what these symbols mean and how to use them. A good working knowledge of trigonometry and algebra is a must, especially in integral calculus. Sometimes you will need to manipulate expressions into a usable form before it is possible to perform operations in calculus.

What you should know before using this text

There are some basic skills that you need before you can use this text. Continuing with our example of a moving car:

You will need to describe the motion of the car in symbols. This involves understanding functions.

You need to manipulate these functions. This involves algebra.

You need to translate symbols into graphs and vice verse. This involves understanding the graphing of functions.

It also helps (although it isn't necessarily essential) if you understand the functions used in trigonometry since these functions appear frequently in science.

Scope

The first three chapters of this textbook cover the topics taught in a typical high school or first year college course. The first chapter, Precalculus, reviews those aspects of functions most essential to the mastery of Calculus, introduces the concept of the limit process. It also discusses some applications of limits and proposes using limits to examine slope and area of functions. The next two chapters, differentiation and integration, apply limits to calculate derivatives and integrals. The Fundamental Theorem of Calculus is used, as are the essential formulae for computation of derivatives and integrals without resorting to the limit process. The second and third chapters include articles that apply the concepts previously learned to calculating volumes, and so on as well as other important formulae.

The remainder of the central Calculus chapters cover topics taught in higher level Calculus topics: multivariable calculus, vectors, and series (Taylor, convergence, divergence).

Finally, the other chapters cover the same material, using formal notation. They introduce the material at a much faster pace, and cover many more theorems than the other two sections. They assume knowledge of some set theory and set notation.

Navigation: Main Page · Precalculus · Limits · Differentiation · Integration · Infinite Series · Multivariable Calculus & Differential Equations · Extensions · References

Precalculus

Functions

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Functions

Classical understanding of functions

To provide the classical understanding of functions, a function can be thought of as a machine. Machines take in raw materials, change them in a predictable way, and give out a finished product. The kinds of functions we consider here, for the most part, take in a real number, change it in a formulaic way, and give out another real number. A function is usually named f or g or something similar except when it is a special, named function (such as the pi function, the Riemann Zeta function, or the random variable function). A function is always defined as "of a variable" which tells the reader what to replace in the function.

For instance, $f(x) = 3x+2 \$ tells the reader:

The function f is a function of x.
To evaluate the function at a certain number, replace the x with the desired number.
Replacing x with that number in the right side of the function will produce the function's output for that certain input.
In English, the definition of $f \$ is interpreted, "Given a number, $f$ will return two more than the triple of that number."

Thus, if we want to know the value (or output) of the function at 3:

$f(x) = 3x+2 \$

$f(3) = 3(3)+2 \$ We evaluate the function at x = 3.

$f(3) = 9+2 = 11 \$ The value of $f \$ at 3 is 11.

Note that $f(3) \$ means the value of the dependent variable when $x \$ takes on the value of 3. So we see that the number 11 is the output of the function when we give the number 3 as the input. We refer to the input as the argument of the function (or the independent variable), and to the output as the value of the function at the given argument (or the dependent variable). A good way to think of it is the dependent variable $f(x) \$ 'depends' on the value of the independent variable $x \$ . This is read as "the value of f of three is eleven", or simply "f of three equals eleven".

Notation

Functions are used so much that there is a special notation for them. The notation is somewhat ambiguous, so familiarity with it is important in order to understand the intention of an equation or formula.

Though there are no strict rules for naming a function, it is standard practice to use the letters $f$ , $g$ , and $h$ to denote functions, and the variable $x$ to denote an independent variable. $y$ is used for both dependent and independent variables.

When discussing or working with a function $f$ , it's important to know not only the function, but also its independent variable $x$ . Thus, when referring to a function $f$ , you usually do not write $f$ , but instead $f (x)$ . The function is now referred to as " $f$ of $x$ ". The dependent variable is adjacent to the independent variable (in parenthesis). This is useful for indicating the value of the function at a particular value of the independent variable. For instance, if

$f(x)=7x+1\,$ ,

and if we want to use the value of $f$ for $x$ equal to $2$ , then we would substitute 2 for $x$ on both sides of the definition above and write

$f(2)=7(2)+1=14+1=15\,$

This notation is more informative than leaving off the independent variable and writing simply ' $f$ ', but can be ambiguous since the parentheses can be interpreted as multiplication. Consistency in notation greatly improves the readability of mathematical text.

Modern understanding of functions

The formal definition of a function states that a function is actually a rule that associates elements of one set called the domain of the function, with the elements of another set called the range of the function. For each value we select from the domain of the function, there exists exactly one corresponding element in the range of the function. The definition of the function tells us which element in the range corresponds to the element we picked from the domain. Classically, the element picked from the domain is pictured as something that is fed into the function and the corresponding element in the range is pictured as the output. Since we "pick" the element in the domain for whose corresponding element in the range we want to find, we have control over what element we pick and hence this element is also known as the "independent variable". The element mapped in the range is beyond our control and is "mapped to" by the function. This element is hence also known as the "dependent variable", for it depends on which independent variable we pick. Since the elementary idea of functions is better understood from the classical viewpoint, we shall use it hereafter. However, it is still important to remember the correct definition of functions at all times.

To make it simple, for the function $f (x)$ , all of the $x$ values constitute the domain, and all of the values $f (x)$ ( $y$ on the x-y plane) constitute the range.

Remarks

The following arise as a direct consequence of the definition of functions:

By definition, functions can take as many "inputs" at a time as desired but return only one "output", corresponding to that set of input. While one set of inputs cannot correspond to more than one output, the same output may correspond to more than one set of inputs. This is interpreted graphically as the vertical line test: a line drawn parallel to the axis of the dependent variable (normally vertical) through the graph of a function will intersect that function only once. Equivalently, this has an algebraic (or formula-based) interpretation. We can always say if $a = b$ , then $f (a) = f (b)$ , but if we only know that $f (a) = f (b)$ then we can't be sure that $a = b$ .
Each function has a set of values, the function's domain, which it can accept as input. Perhaps this set is all positive real numbers; perhaps it is the set {pork, mutton, beef}. This set must be implicitly/explicitly defined in the definition of the function. You cannot feed the function an element that isn't in the domain, as the function is not defined for that input element.
Each function has a set of values, the function's range, which it can output. This may also be the set of real numbers, in which case the function is termed as a "real-valued" function. It may be the set of positive integers or even the set {0,1}. This set, too, must be implicitly/explicitly defined in the definition of the function.

This is an example of an expression which fails the vertical line test.

The vertical line test

The vertical line test is a systematic test to find out if an expression can serve as a function. Simply graph the expression and draw a vertical line along each point in the domain of the relation. If any vertical line ever touches the relation for more than one value, then the expression is not a function; if the line always touches only one value, then the expression is a function.

(There are a lot of useful curves, like circles, that aren't functions (see picture). Some people call these graphs with multiple intercepts, like our circle, "multi-valued functions"; they would refer to our "functions" as "single-valued functions".)

Important functions

$f(x)=a_n x^n + a_{n-1}x^{n-1} + \cdots + a_2 x^2 + a_1 x + a_0$ --Polynomial function

$f(x)=x\,$ --Identity function

Takes an input and gives it as output.

$g(x)=c\,$ --Constant function

Takes an input, ignores it, and always gives the constant c as an output. A polynomial of the zeroth degree

$g(x)=0\,$ --Zero function

A special case of the above. A polynomial of the undefined degree.

$f(x)=mx+b\,$ --Linear function (function of a line)--polynomial of the first degree.

Takes an input, multiplies by m and adds b

$\operatorname{sgn}(x) = \left\{ \begin{matrix} -1 & : & x < 0 \\ 0 & : & x = 0 \\ 1 & : & x > 0. \end{matrix} \right.$ -- The signum function

Determines the sign of the argument x.

$f(x)=ax^2+bx+c \,$ -- The quadratic function.

A particular case of the more general polynomial function.

Example functions

Some more simple examples of functions have been listed below.

$h(x)=\left\{\begin{matrix}1,&\mbox{if }x>0\\-1,&\mbox{if }x<0\end{matrix}\right.$

Gives 1 if input is positive, -1 if input is negative. Note that the function only accepts negative and positive numbers, not

0

. Mathematics describes this condition by saying

0

is not in the domain of the function.

$g(y)=y^2\,$

Takes an input and squares it.

$g(z)=z^2\,$

Exactly the same function, rewritten with a different independent variable. This is perfectly legal and sometimes done to prevent confusion (e.g. when there are already too many uses of

x

y

in the same paragraph.)

$f(y)=\left\{\begin{matrix}\int_{-y}^{y}e^{x^2}\,dx,&\mbox{if }y>0\\0,&\mbox{if }y\le0\end{matrix}\right.$

Takes an input and uses it as boundary values for an integration.

It is possible to replace the independent variable with any mathematical expression, not just a numeral. For instance, if the independent variable is itself a function of another variable, then it could be replaced with the value of that function. This is called composition, and is discussed next.

Manipulating functions

Functions can be manipulated in the same manner as any other variable; they can be added, multiplied, raised to powers, etc. For instance, let

$f(x)=3x+2\,$ and

$g(x)=x^2\,$ .

Then

$\begin{align} f+g &= (f+g)(x)\\ &= f(x)+g(x)\\ &= (3x+2)+(x^2)\\ &= x^2+3x+2\, \end{align}$ ,

$\begin{align} f-g &= (f-g)(x)\\ &= f(x)-g(x)\\ &= (3x+2)-(x^2)\\ &= -x^2+3x+2\, \end{align}$ ,

$\begin{align} f\times g &= (f\times g)(x)\\ &= f(x)\times g(x)\\ &= (3x+2)\times(x^2)\\ &= 3x^3+2x^2\, \end{align}$ ,

$\begin{align} \frac{f}{g} &= \left(\frac{f}{g}\right)(x)\\ &= \frac{f(x)}{g(x)}\\ &= \frac{3x+2}{x^2}\\ &= \frac{3}{x}+\frac{2}{x^2} \end{align}$ .

Composition of functions

However, there is one particular way to combine functions which cannot be done with other variables. The value of a function $f$ depends upon the value of another variable $x$ ; however, that variable could be equal to another function $g$ , so its value depends on the value of a third variable. If this is the case, then the first variable is a function $h$ of the third variable; this function ( $h$ ) is called the composition of the other two functions ( $f$ and $g$ ). Composition is denoted by

$f\circ g=(f\circ g)(x)=f(g(x))$ .

This can be read as either "f composed with g" or "f of g of x."

For instance, let

$f(x)=3x+2\,$ and

$g(x)=x^2\,$ .

Then

$\begin{align} h(x) &= f(g(x))\\ &= f(x^2)\\ &= 3(x^2)+2\\ &= 3x^2+2\, \end{align}$ .

Here, $h$ is the composition of $f$ and $g$ . Note that composition is not commutative:

$f(g(x))=3x^2+2\,$ , and

$\begin{align} g(f(x)) &= g(3x + 2)\\ &= (3x + 2)^2\\ &= 9x^2+12x+4\, \end{align}$

so $f(g(x))\ne g(f(x))\,$ .

Composition of functions is very common, mainly because functions themselves are common. For instance: addition, multiplication, etc., can be represented as functions of more than one independent variable:

$\operatorname{plus}(x,y)=x+y$ ,

$\operatorname{times}(x,y)=x\times y$ , etc.

Thus, the expression $2 \times 3 + 4$ is a composition of functions:

$2\times 3+4$	= $\operatorname{times}(2,3)+4$
	= $\operatorname{plus}( \operatorname{times}(2, 3), 4)$ .

Since the function times equals $6$ if $x = 2$ and $y = 3$ , then

$\operatorname{plus}(\operatorname{times}(2,3),4)= \operatorname{plus}(6,4)$ .

Since the function plus equals $10$ if $x = 6$ and $y = 4$ , then

$(2\times 3)+4=\operatorname{plus}(\operatorname{times}(2,3),4)=\operatorname{plus}(6,4)=10$ .

Transformations

Transformations are a type of function manipulation that are very common. They consist of multiplying, dividing, adding or subtracting constants to either the argument or the output. Multiplying by a constant is known as Dilation and adding a constant is called Translation. Here are a few examples:

$f(2\times x) \,$ Dilation

$f(x+2)\,$ Translation

$2\times f(x) \,$ Dilation

$2+f(x)\,$ Translation

Examples of horizontal and vertical translations

Examples of horizontal and vertical dilations

Translations and Dilations can be either horizontal or vertical. Examples of both vertical and horizontal translations can be seen at right. The red graphs represent functions in their 'original' state, the solid blue graphs have been translated (shifted) horizontally, and the dashed graphs have been translated vertically. Dilations are demonstrated in a similar fashion. The function

$f(2\times x) \,$

has had its input doubled. One way to think about this is that now any change in the input will be doubled. If I add one to x, I add two to the input of the function, so it will now change twice as quickly. However, this is a horizontal dilation by $\frac{1}{2}$ because the distance to the y-axis has been halved. A vertical dilation, such as

$2\times f(x) \,$

is slightly more straightforward. In this case, you double the output of the function. The output represents the distance from the x-axis, so in effect, you have made the graph of the function 'taller'. Here are a few basic examples where a is any postive constant:

Original Graph	$f(x)\,$	Reflection through origin	$-f(-x)\,$
Horizontal Translation by a units right	$f(x-a)\,$	Horizontal Translation by a units left	$f(x+a)\,$
Horizontal Dilation by a factor of a	$f(x\times \frac{1}{a}) \,$	Vertical Dilation by a factor of a	$a\times f(x) \,$
Vertical Translation by a units down	$f(x)-a\,$	Vertical Translation by a units up	$f(x)+a\,$
Reflection over x-axis	$-f(x)\,$	Reflection over y-axis	$f(-x)\,$

Domain and Range

Notation

The notation used to denote intervals is very simple, but sometimes ambiguous because of the similarity to ordered pair notation:

Meaning	Interval Notation	Set Notation
All values greater than or equal to $a$ and less than or equal to $b$	$\left[a,b\right]$	$\left\{x:a\le x\le b\right\}$
All values greater than $a$ and less than $b$	$\left(a,b\right)$	$\left\{x:a < x < b\right\}$
All values greater than or equal to $a$ and less than $b$	$\left[a,b\right)$	$\left\{x:a\le x < b\right\}$
All values greater than $a$ and less than or equal to $b$	$\left(a,b\right]$	$\left\{x:a < x\le b\right\}$
All values greater than or equal $a$ .	$\left[a,\infty\right)$	$\left\{x:x\ge a\right\}$
All values greater than $a$ .	$\left(a,\infty\right)$	$\left\{x:x > a\right\}$
All values less than or equal to $a$ .	$\left(-\infty,a\right]$	$\left\{x:x\le a\right\}$
All values less than $a$ .	$\left(-\infty,a\right)$	$\left\{x:x < a\right\}$
All values.	$\left(-\infty,\infty\right)$	$\left\{x: x\in\mathbb{R}\right\}$

Note that $\infty$ must always be unbounded (that is, have no inclusive bracket, but instead have an exclusive parentheses). This is because $\infty$ is not a number, and therefore cannot be in our set. $\infty$ is really just a symbol that makes things easier to write, like the intervals above.
Note: ( is also denoted by ], and ) by [, i.e., (a,b) is the same as ]a,b[, and [a,b) is [a,b[. This is source of funny misunderstandings.

Domain

The domain of the function is the interval from -1 to 1

The domain of a function is the set of all points over which it is defined. More simply, it represents the set of x-values which the function can accept as input. For instance, if

$f(x)=\sqrt{1-x^2}$

then $f (x)$ is only defined for values of $x$ between $- 1$ and $1$ , because the square root function is not defined (in real numbers) for negative values. Thus, the domain, in interval notation, is $\left[-1,1\right]$ . In other words,

$f(x) \mbox{is defined for } x\in [-1,1], \operatorname{ or } \{x:-1\le x\le 1\}$ .

The range of the function is the interval from 0 to 1

Range

The range of a function is the set of all values which it attains (i.e. the y-values). For instance, if:

$f(x)=\sqrt{1-x^2}$ ,

then $f (x)$ can only equal values in the interval from $0$ to $1$ . Thus, the range of $f$ is $\left[0,1\right]$ .

One-to-one Functions

A function $f (x)$ is one-to-one (or less commonly injective) if, for every value of $f$ , there is only one value of $x$ that corresponds to that value of $f$ . For instance, the function $f(x)=\sqrt{1-x^2}$ is not one-to-one, because both $x = 1$ and $x = - 1$ result in $f (x) = 0$ . However, the function $f (x) = x + 2$ is one-to-one because for every possible value of $f (x)$ , there is exactly one corresponding value of $x$ . Any function that looks like $f (x) = x 3 + a x$ , where $a\in \left[0,\infty\right)$ , is one-to-one. Note that if you have a one-to-one function and translates/dilates it, it remains one-to-one (Of course you can't multiply $x$ or $f$ by a zero factor).

If you know what the graph of a function looks like, it is easy to determine whether or not the function is one-to-one. If every horizontal line intersects the graph in at most one point, then the function is one-to-one. This is known as the Horizontal Line Test.

Inverse functions

Function $f (x)$ has an inverse function if and only if $f (x)$ is one-to-one. For $f (x)$ and $g (x)$ such that $g (x)$ is the inverse function of $f (x)$ :

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

For example, the inverse of $f (x) = x + 2$ is $g (x) = x - 2$ . The function $f(x)=\sqrt{1-x^2}$ has no inverse.

Notation

The inverse function of $f$ is denoted as $f - 1 (x)$ . The inverse of a function $f - 1 (x)$ is defined as the function that follows this rule

$f (f - 1 (x)) = f - 1 (f (x)) = x$ :

To determine $f - 1 (x)$ when given a function f, substitute $f - 1 (x)$ for $x$ and substitute $x$ for $f (x)$ . Then solve for $f - 1 (x)$ , provided that it is also a function.

Example: Given $f (x) = 2 x - 7$ , find $f - 1 (x)$ .

Substitute $f - 1 (x)$ for $x$ and substitute $x$ for $f (x)$ ., then solve for $f (x)$ :

$f(x) = 2x - 7\,$

$x = 2[f^{-1}(x)] - 7\,$

$x + 7 = 2[f^{-1}(x)]\,$

$\frac{x + 7}{2} = f^{-1}(x)\,$

To check your work, confirm that $f (f - 1 (x)) = x$ :

$f (f - 1 (x)) =$

$f \left (\frac{x + 7} {2}\right ) = {}$

$2\left (\frac{x + 7}{2}\right ) - 7 = (x+7) - 7 = x$

Graphing Functions

Graph of y=2x

It is sometimes difficult to understand the behavior of a function given only its definition; a visual representation or graph can be very helpful. A graph is a set of points in the Cartesian plane, where each point ( $x$ , $y$ ) indicates that $f (x) = y$ . In other words, a graph uses the position of a point in one direction (the vertical-axis or y-axis) to indicate the value of $f$ for a position of the point in the other direction (the horizontal-axis or x-axis).

Functions may be graphed by finding the value of $f$ for various $x$ and plotting the points ( $x$ , $f (x)$ ) in a Cartesian plane. Since functions that you will deal with are generally continuous (see below), the parts of the function between the points can be approximated by drawing a line or curve between the points. Extending the function beyond the set of points is also possible, but becomes increasingly inaccurate.

Plotting points like this is laborious. Fortunately, many functions' graphs fall into general patterns. For a simple case, consider functions of the form

$f(x)={a}x \,$

The graph of $f$ is a single line, passing through ( $0$ , $0$ ) and $(1, a)$ . Thus, after plotting the two points, a straightedge may be used to draw the graph as far as is needed.

Continuity

Most functions that you will deal with are not a random scattering of points. Rather, the graphs have one or more curves, which do not have sudden gaps in them, and may be drawn without lifting your pencil. Later, the principle of continuity will be defined formally using the concept of limits.

Algebraic manipulation

Purpose of review

This section is intended to review algebraic manipulation.

Rules of arithmetic and algebra

The following rules are always true (see: field theory).

Addition
- Commutative Law: $a+b=b+a \,$ .
- Associative Law: $(a+b)+c=a+(b+c)\,$ .
- Additive Identity: $a+0=a\,$ .
- Additive Inverse: $a+(-a)=0\,$ .
Subtraction
- Definition: $a-b = a+(-b)\,$ .
Multiplication
- Commutative law: $a\times b=b\times a\,$ .
- Associative law: $(a\times b)\times c=a\times (b\times c)\,$ .
- Multiplicative Identity: $a\times 1=a\,$ .
- Multiplicative Inverse: $a\times \frac{1}{a}=1,\qquad a \neq 0\,$
- Distributive law: $a\times (b+c)=a\times b+a\times c\,$ .
Division
- Definition: $\frac{a}{b}=a\times \frac{1}{b},\qquad b \neq 0\,$ .

The above laws are true for all $a$ , $b$ , and $c$ , whether $a$ , $b$ , and $c$ are numbers, variables, functions, or other expressions. For instance, although

$\frac{(x+2)(x+3)}{x+3}$	= $(((x+2)\times (x+3))\times (\frac{1}{x+3}))$
	= $((x+2)\times ((x+3)\times (\frac{1}{x+3})))$
	= $((x+2)\times (1)),\qquad x \neq -3 \,$
	= $x+2, \qquad x \neq -3\,$

is much longer than simply cancelling $x + 3$ out in both the numerator and denominator, it is important to understand the longer method. Occasionally people do the following, for instance, which is incorrect:

$\frac{2\times (x + 2)}{2}=\frac{2}{2}\times \frac{x+2}{2}=\frac{x+2}{2}$ .

The correct way is

$\frac{2\times (x + 2)}{2}=\frac{2}{2}\times \frac{x+2}{1}=1 \times \frac{x+2}{1}=x+2$ ,

where the number $2$ cancels out in both the numerator and the denominator.

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Limits

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Intuitive Look

A limit looks at what happens to a function when the input approaches a certain value. The general notation for a limit is as follows:

$\quad\lim_{x\to a} f(x) = L$

This is read as "The limit of f(x) as x approaches a is L". How we can determine whether a limit exists for f(x) at a and what the limit is is a technical point that we'll take up later. For now, we'll look at it from an intuitive standpoint.

Let's say that the function that we're interested in is $f (x) = x 2$ , and that we're interested in its limit as x approaches 2. We can write the limit that we're interested in using the above notation as follows:

$\quad\lim_{x\to 2} x^2$

One way to try to evaluate what this limit is would be to choose values near 2, compute f(x) for each, and see what happens as they get closer to 2. This is implemented as follows:

$x$	1.7	1.8	1.9	1.95	1.99	1.999
$f (x) = x 2$	2.89	3.24	3.61	3.8025	3.9601	3.996001

Here we chose numbers smaller than 2, and approached 2 from below. We can also choose numbers larger than 2, and approach 2 from above:

$x$	2.3	2.2	2.1	2.05	2.01	2.001
$f (x) = x 2$	5.29	4.84	4.41	4.2025	4.0401	4.004001

We can see from the tables that as x grows closer and closer to 2, f(x) seems to get closer and closer to four, regardless of whether we approach 2 with x from above or from below. For this reason, we feel reasonably confident that the limit of $x 2$ as x approaches 2 is 4, or written in limit notation,

$\quad\lim_{x\to 2} x^2=4$

Now let's look at another example. Suppose we're interested in the function $f(x)=\frac{1}{x-2}$ , and its behavior as x approaches 2. Here's the limit in limit notation:

$\quad\lim_{x\to 2} \frac{1}{x-2}$

Just as before, we can compute function values as x approaches 2 from below and from above. Here's a table, approaching from below:

$x$	1.7	1.8	1.9	1.95	1.99	1.999
$f(x)=\frac{1}{x-2}$	-3.333	-5	-10	-20	-100	-1000

And here from above:

$x$	2.3	2.2	2.1	2.05	2.01	2.001
$f(x)=\frac{1}{x-2}$	3.333	5	10	20	100	1000

In this case, the function doesn't seem to be approaching any value as x approaches 2. In this case we would say that the limit doesn't exist.

Both of these examples may seem trivial, but consider the following function:

$f(x) =\left\{\begin{matrix} x^2 & \mbox{if } x\neq 2 \\ \mbox{undefined} & \mbox{if } x=2\end{matrix}\right.$

$\begin{matrix} f(x) &=& \frac{x^2(x-2)}{x-2} \quad \end{matrix}$

Now, we see that these functions are completely identical; not just "almost the same," but actually, in terms of the definition of a function, completely the same. However, we are more comfortable with the second one algebraically, because it is easier to work with.

In algebra, we would simply say that we can cancel the term $(x - 2)$ , and then we have the function $f (x) = x 2$ . This, however, would be a bit dishonest; the function that we have now is not really the same as the one we started with, because it is defined at x=2, and our original function was, specifically, not defined at x=2. In algebra we were willing to ignore this difficulty because we had no better way of dealing with this type of function. Now, however, in calculus, we can introduce a better, more correct way of looking at this type of function. What we want is to be able to say that, even though at $x = 2$ the function doesn't exist, it works almost as though it does, and it's 4. It may not get there, but it gets really, really close. The only question that we have is: what do we mean by close?

Informal definition of a limit

As the precise definition of a limit is a bit technical, it is easier to start with an informal definition, and we'll explain the formal definition later.

We suppose that a function f is defined for x near c (but we do not require that it be defined when x=c).

Definition: (Informal definition of a limit)
We call L the limit of f(x) as x approaches c if f(x) becomes close to L when x is close (but not equal) to c.

When this holds we write

$\lim_{x \to c} f(x) = L$

$f(x) \to L \quad \mbox{as} \quad x \to c.$

Notice that the definition of a limit is not concerned with the value of f(x) when x=c (which may exist or may not). All we care about is the values of f(x) when x is close to c, on either the left or the right (i.e. less or greater).

Limit rules

Now that we have defined, informally, what a limit is, we will list some rules that are useful for manipulating a limit. These will all be proven, or left as exercises, once we formally define the fundamental concept of the limit of a function.

Suppose that $\lim_{x\to c} f(x) =L$ and $\lim_{x\to c} g(x) =M$ and that k is constant. Then

$\begin{align} \lim_{x\to c} k f(x) &=& k \cdot \lim_{x\to c} f(x) &=& k L \\ \lim_{x\to c} [f(x) + g(x)] &=& \lim_{x\to c} f(x) + \lim_{x\to c} g(x) &=& L + M \\ \lim_{x\to c} [f(x) - g(x)] &=& \lim_{x\to c} f(x) - \lim_{x\to c} g(x) &=& L - M \\ \lim_{x\to c} [f(x) \cdot g(x)] &=& \lim_{x\to c} f(x) \cdot \lim_{x\to c} g(x) &=& L\cdot M \\ \lim_{x\to c} {f(x) \over g(x)} &=& {\lim_{x\to c} f(x) \over \lim_{x\to c} g(x)} &=& {L \over M} & \hbox{as long as } M\neq 0 \end{align}$

Notice that in the last rule we need to require that M is not equal to zero (otherwise we would be dividing by zero which is an undefined operation).

These rules are known as identities; they are scalar multiplication, addition, subtraction, multiplication, and division of limits. (A scalar is a constant, and we say that when you multiply a function by a constant, you are performing scalar multiplication.)

Using these rules we can deduce some others. First, the constant rule states that if f(x) = b is constant for all x then the limit as x approaches c must be equal to b. In other words

$\lim_{x\to c} b = b$ .

Second, the identity rule states that if f(x) = x then the limit of f as x approaches c is equal to c. That is,

$\lim_{x\to c} x = c$ .

Third, using the rule for products many times we get that

$\lim_{x\to c} f(x)^n = \left(\lim_{x\to c} f(x) \right)^n$ for a positive integer n.

This is called the power rule.

Examples

Example 1 Find the limit $\lim_{x\to 2} {4x^3}.$

We need to simplify the problem, since we have no rules about this expression by itself. We know from the identity rule above that $\lim_{x\to 2} {x} = 2$ . By the power rule, $\lim_{x\to 2} {x^3} = 2^3 = 8$ . Lastly, by the scalar multiplication rule, we get $\lim_{x\to 2} {4x^3} = 4 \cdot 8=32$

Example 2

Find the limit $\lim_{x\to 2} 4x^3 + 5x +7.$

To do this informally, we split up the expression, once again, into its components. As above, $\lim_{x\to 2} 4x^3=32$ .

Also $\lim_{x\to 2} 5x = 5\cdot\lim_{x\to 2} x = 5\cdot2=10$

and $\lim_{x\to 2} 7 =7$ . Adding these together gives

$\lim_{x\to 2} 4x^3 + 5x +7 = 32 + 10 +7 =49.$

Example 3

Find the limit: $\lim_{x\to 2}$ $(4x^3 + 5x +7) \over (x-4)(x+10).$

From the previous example the limit of the numerator is $\lim_{x\to 2} 4x^3 + 5x +7 =49.$ The limit of the denominator is

$\lim_{x\to 2} (x-4)(x+10) = \lim_{x\to 2} (x-4) \cdot \lim_{x\to 2} (x+10) = (2-4)\cdot(2+10)=-24.$

As the limit of the denominator is not equal to zero we can divide which gives

$\lim_{x\to 2}$ $(4x^3 + 5x +7) \over (x-4)(x+10)$ $= - {49 \over 24}$ .

Example 4

Find the limit: $\lim_{x\to 4}$ $(x^4 - 16x + 7) \over (4x-5).$

We apply the same process here as we did in the previous set of examples;

$\lim_{x\to 4}$ $(x^4 - 16x + 7) \over (4x-5)$

=

$\lim_{x\to 4} (x^4 - 16x + 7) \over \lim_{x\to 4} (4x-5)$

=

$( \lim_{x\to 4} (x^4) - \lim_{x\to 4} (16x) + \lim_{x\to 4} (7) ) \over \lim_{x\to 4} (4x) - \lim_{x\to 4} (5)$

We can evaluate each of these; $\lim_{x\to 4} (x^4) = 256.$ $\lim_{x\to 4} (16x) = 64.$ $\lim_{x\to 4} (7) = 7.$ $\lim_{x\to 4} (4x) = 16.$ $\lim_{x\to 4} (5) = 5.$ $= {199 \over 11}$

Example 5

Evaluate the limit $\lim_{x\to 0}$ $1-\cos(x) \over x$ .

To evaluate this seemingly complex limit, be aware of your sine and cosine identities. We will also have to use two new facts. First, if f(x) is a trigonometric function (that is, one of sine, cosine, tangent, cotangent, secant or cosecant) and is defined at a, then $\lim_{x\to a} f(x) = f(a)$ . Second, $\lim_{x\to 0}$ $\sin(x) \over (x)$ $= 1$ .

To evaluate the limit, recognize that $1 - cos(x)$ can be multiplied by $1 + c o s (x)$ to obtain $(1 - c o s 2 (x))$ which, by our trig identities, is $s i n 2 (x)$ . So, multiply the top and bottom by $1 + c o s (x)$ (This is allowed because it is identical to multiplying by one). This is a standard trick for evaluating limits of fractions; multiply the numerator and the denominator by a carefully chosen expression which will make the expression simplify somehow. In this case, we should end up with:

$\lim_{x\to 0}$ $1-\cos(x) \over x$ $=$ $1-\cos(x) \over x$ $\cdot$ $1 \over 1$ $=$ $1-\cos(x) \over x$ $\cdot$ $1 + \cos(x) \over 1+ \cos(x)$ $=$ $[(1 - \cos(x)) \cdot 1] + [(1 - \cos(x)) \cdot \cos(x)] \over x \cdot (1+ \cos(x))$ $=$ $1 - \cos(x) + \cos(x) - \cos^2(x) \over x \cdot (1+ \cos(x))$ $=$ $1 - \cos^2(x) \over x \cdot (1+ \cos(x))$ $=$ $\sin^2(x) \over x \cdot (1+ \cos(x))$ $=$ $\sin(x) \cdot \sin(x) \over x \cdot (1+ \cos(x))$ $=$ $\sin(x) \over x$ $\cdot$ $\sin(x) \over (1+ \cos(x))$

Your next step shall be to break this up into $\lim_{x\to 0}$ $\sin(x) \over (x)$ $\cdot$ $\lim_{x\to 0}$ $\sin(x) \over (1+\cos(x))$ by the limit rule of multiplication. By the fact mentioned above, $\lim_{x\to 0}$ $\sin(x) \over (x)$ $= 1$ .

Next, $\lim_{x\to 0}$ $\sin(x) \over (1+\cos(x))$ $=$ $\lim_{x\to 0}\sin x \over \lim_{x\to 0} (1+\cos x)$ $=$ $0 \over 1 + \cos 0$ $= 0$ .

Thus, by multiplying these two results, we obtain 0.

We will now present an amazingly useful result, even though we cannot prove this yet. We can find the limit of any polynomial or rational function, as above, as long as that rational function is defined at c (so we are not dividing by zero). More precisely, c must be in the domain of the function.

Limits of Polynomials and Rational functions

If f is a polynomial or rational function that is defined at c then

$\lim_{x \rightarrow c} f(x) = f(c)$

We already learned this for trigonometric functions, so we see that it is easy to find limits of polynomial, rational or trigonometric functions wherever they are defined. In fact, this is true even for combinations of these functions; thus, for example, $\lim_{x\to 1} (\sin (x^2) + 4\cos^3(3x-1)) = \sin 1^2 + 4\cos^3 (3(1)-1)$ .

The Squeeze Theorem

Graph showing f being squeezed between g and h

The squeeze theorem is very important in calculus proofs, where it is typically used to confirm the limit of a function via comparison with two other functions whose limits are known.

It is called the Squeeze Theorem because it refers to a function f whose values are squeezed between the values of two other function g and h. g and h both have the same limit L at trapped between the values of the two functions that approach L, the values of f must also approach L.

A more mathematical definition is:

Suppose that $g(x) \le f(x) \le h(x)$ hold for all x in some open interval containing a, except possibly at $x = a$ itself. Suppose also that $\lim_{x\rightarrow a}g(x)=\lim_{x\rightarrow a}h(x)=L$ . Then $\lim_{x\rightarrow a}f(x)=L$ also. Similar statements hold for left and right limit.

Finding limits

Now, we will discuss how, in practice, to find limits. First, if the function can be built out of rational, trigonometric, logarithmic and exponential functions, then if a number c is in the domain of the function, then the limit is simply the value of the function at c.

If c is not in the domain of the function, then in many cases (as with rational functions) the domain of the function includes all the points near c, but not c. An example would be if we wanted to find $\lim_{x\to 0} \frac{x}{x}$ , where the domain includes all numbers besides 0. In that case, we want to find a similar function, except with the hole filled in. The limit of this function at c will be the same, as can be seen from the definition of a limit. The function is the same as the previous except at a point c. The limit definition depends on f(x) only at the points where x is close to c but not equal to it. When $x \ne c$ , this condition doesn't hold, and so the limit at c does not depend on the value of the function at c. Therefore, the limit of the new function is the same as of the previous function. And since the domain of our new function includes c, we can now (assuming it's still built out of rational, trigonometric, logarithmic and exponential functions) just evaluate the function at c as before.

In our example, this is easy; canceling the xs gives 1, which equals x/x at all points except 0. Thus, we have $\lim_{x\to 0}\frac{x}{x} = \lim_{x\to 0} 1 = 1$ . In general, when computing limits of rational functions, it's a good idea to look for common factors in the numerator and denominator.

Lastly, note that the limit might not exist at all. There are a number of ways in which this can occur:

"Gap": There is a gap (more than a point wide) in the function where the function is not defined. As an example, in

$f(x) = \sqrt{x^2 - 16}$

f(x) does not have any limit when -4 ≤ x ≤ 4. There is no way to "approach" the middle of the graph. Note also that the function also has no limit at the endpoints of the two curves generated (at x = -4 and x = 4). For the limit to exist, the point must be approachable from both the left and the right. Note also that there is no limit at a totally isolated point on the graph.

"Jump": If the graph suddenly jumps to a different level, there is no limit. This is illustrated in the floor function (in which the output value is the greatest integer not greater than the input value).

Asymptote: In

$f(x) = {1 \over x^2}$

the graph gets arbitrarily high as it approaches 0, so there is no finite limit. In this case we say the limit is infinite.

Infinite oscillation: These next two can be tricky to visualize. In this one, we mean that a graph continually rises above and below a horizontal line. In fact, it does this infinitely often as you approach a certain x-value. This often means that there is no limit, as the graph never approaches a particular value. However, if the height (and depth) of each oscillation diminishes as the graph approaches the x-value, so that the oscillations get arbitrarily smaller, then there might actually be a limit.

The use of oscillation naturally calls to mind the trigonometric functions. An example of a trigonometric function that does not have a limit as x approaches 0 is

$f(x) = \sin {1 \over x}.$

As x gets closer to 0 the function keeps oscillating between -1 and 1. It is also true that sin(1/x) oscillates an infinite number of times on the interval between 0 and any positive value of x. The sine function, sin(x), is equal to zero whenever x=kπ, where k is a positive integer. Between each value of k, sin(x) oscillates between 0 and -1 or 0 and 1. Hence, sin(1/x)=0 for every x=1/(kπ). In between consecutive pairs of these values, 1/(kπ} and 1/[(k+1)π], sin(1/x) oscillates from 0, to -1, to 1 and back to 0 and so on. We may also observe that there are an infinite number of such pairs, and they are all between 0 and 1/π. There are a finite number of such pairs between any positive value of x and 1/π which implies that there must be infinitely many between x and 0. From our reasoning we may conclude that as x approaches 0 from the right, the function sin(1/x) does not approach any value. We say that the limit as x approaches 0 from the right does not exist. (In contrast, in the case of a "jump", the limits from each side did exist. Since they weren't equal, though, the regular limit didn't exist.)

Incomplete graph: Let us consider two examples.

First, let f be the constant function f(q)=2 where we specify that q is only allowed to be a rational number. Let $q 0$ be some rational number. Then $\lim_{q\to q_0} f(q) = 2$ , since, for q close to $q 0$ , f(q) is close to (in fact equals) 2.

Now let g be the similar-looking function defined on the entire real line, but we change the value of the function based on whether x is rational or not.

$g(x)=\left\{\begin{matrix} 2, & \mbox{if }x \mbox{ is rational} \\ 0, & \mbox{if }x \mbox{ is irrational} \end{matrix}\right.$

Now g has a limit nowhere! For let x be a real number; we show that g can't have a limit at x. No matter how close we get to x, there will be rational numbers (where g will be 2) and irrational numbers (where g will be 0). Thus g has no limit at any real number!

Using limit notation to describe asymptotes

Now consider the function

$g(x) = \frac {1}{x^2}.$

What is the limit as x approaches zero? The value of g(0) does not exist; the value of g(0) is not defined.

$\qquad g(0) = \frac {1}{0^2}$

Notice that we can make the function g(x) as large as we like, by choosing a