# What do we mean by “limit undefined”?

There may be a confusion regarding the meaning of “limit undefined”. It is actually a matter of opinion. My notion of “limit undefined” is different from that of your textbook. In your textbook, the limit is said to be undefined if it fails to exist as a number. So for instance the limit $\lim_{x\to 0}\frac{1}{x^2}=\infty$ is undefined according to textbook. In my case however I would still say that the limit exists as $\infty$ since the left-hand limit and the right-hand limit coincide as $\infty$. It wouldn’t matter whichever definition you follow as long as you are clear about it.

# How to Calculate Limits II

In the previous posting, we studied how to calculate limit of a rational function (Corollary 3). Let us state it here again:

Corollary 3. [Limit of a Rational Function] Let $p(x)$ and $q(x)$ be two polynomials. Then for any real number $b$, $\lim_{x\to b}\frac{p(x)}{q(x)}=\frac{p(b)}{q(b)}$ provided $q(b)\ne 0$.

But what if $q(b)=0$? To answer this question let us take a look at the following example.

Example. Find the limit $\displaystyle\lim_{x\to -1}\frac{x^2+3x+2}{x^2-x-2}$.

Solution. Let $p(x)=x^2+3x+2$ and $q(x)=x^2-x-2$. Then $p(-1)=0$ and $q(-1)=0$. Since $q(-1)=0$, we cannot use Corollary 3 to calculate the limit. So what do we do? Note that $p(-1)=0$ and $q(-1)=0$ means that both $p(x)$ and $q(x)$ contains a power of $(x+1)$ in them. Let us factor out the maximum common power of $(x+1)$ from $p(x)$ and $q(x)$. Since $x\to -1$, $x\ne -1$ i.e. $x+1\ne 0$. So we can cancel the maximum common power of $(x+1)$ and then calculate limit of the resulting function as $x\to -1$: \begin{eqnarray*}\lim_{x\to -1}\frac{x^2+3x+2}{x^2-x-2}&=&\lim_{x\to -1}\frac{(x+1)(x+2)}{(x-2)(x+1)}\\&=&\lim_{x\to -1}\frac{x+2}{x-2}\ \mbox{since $x\ne -1$}\\&=&-\frac{1}{3}.\end{eqnarray*}

Remark. [Indeterminate Form] In the above example, $\frac{\displaystyle\lim_{x\to -1}(x^2+3x+2)}{\displaystyle\lim_{x\to -1}(x^2-x-2)}=\frac{0}{0}.$ What is this? and how do we understand it? It turns out that the quantity $\frac{0}{0}$ is not undefined but something else. Remember that here $0$ is not a number but an infinitesimal, a state that is extremely close to the number $0$. The quantity $\frac{0}{0}$ is called an indeterminate form. There are other types of indeterminate forms, to name a few, $\frac{\infty}{\infty}$, $0\cdot\infty$, $0^0$, etc. We will study them later. There are four possibilities for the value of an indeterminate form: $0$, $\pm\infty$, or a non-zero real number. Although we denote infinitesimals by the same symbol $0$, some infinitesimals dominate others. For instance, consider the limit of a rational function $\displaystyle\lim_{x\to a}\frac{p(x)}{q(x)}$. Suppose that $\displaystyle\lim_{x\to a}p(x)=\lim_{x\to a}q(x)=0$. There can be three possible scenarios then:

1. If $p(x)$ approaches $0$ way faster than $q(x)$ does, then $\displaystyle\lim_{x\to a}\frac{p(x)}{q(x)}=0$.
2. If $q(x)$ approaches $0$ way faster than $p(x)$ does, then $\displaystyle\lim_{x\to a}\frac{p(x)}{q(x)}=\pm\infty$.
3. If $p(x)$ and $q(x)$ approaches $0$ at about the same rate (speed), then $\displaystyle\lim_{x\to a}\frac{p(x)}{q(x)}$ may be a non-zero real number.

Example. Find the limit $\lim_{x\to 2}\frac{4-x^2}{3-\sqrt{x^2+5}}.$

Solution. $\displaystyle\lim_{x\to 2}(4-x^2)=\lim_{x\to 2}(3-\sqrt{x^2+5})=0$. This means that both the numerator and the denominator have a power of $x-2$ as a common factor. As we did in the previous example, we would attempt to factor both the numerator and the denominator. Only problem is that the denominator is not a polynomial and we don’t know how to factor it. Well, we learned about rationalizing the denominator in algebra. We multiply the numerator and the denominator by the conjugate $3+\sqrt{x^2+5}$ of the denominator. More specifically,\begin{eqnarray*}\lim_{x\to 2}\frac{4-x^2}{3-\sqrt{x^2+5}}&=&\lim_{x\to 2}\frac{4-x^2}{3-\sqrt{x^2+5}}\cdot\frac{3+\sqrt{x^2+5}}{3+\sqrt{x^2+5}}\\&=&\lim_{x\to 2}\frac{(4-x^2)(3+\sqrt{x^2+5})}{4-x^2}\\&=&\lim_{x\to 2}(3+\sqrt{x^2+5})\\&=&6.\end{eqnarray*}

# How to Calculate Limits I

When you calculate limits, the following theorem plays a crucial role.

Theorem 1. Suppose that $c$ is a constant and the limits $\lim_{x\to a}f(x)\ {\rm and}\ \lim_{x\to a}g(x)$ exist. Then the following properties hold:

1. $\displaystyle\lim_{x\to a}\{f(x)+g(x)\}=\lim_{x\to a}f(x)+\lim_{x\to a}g(x)$
2. $\displaystyle\lim_{x\to a}cf(x)=c\lim_{x\to a}f(x)$
3. $\displaystyle\lim_{x\to a}f(x)g(x)=\lim_{x\to a}f(x)\cdot\lim_{x\to a}g(x)$
4. $\displaystyle\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{\displaystyle\lim_{x\to a}f(x)}{\displaystyle\lim_{x\to a}g(x)}$ provided $\displaystyle\lim_{x\to a}g(x)\ne 0$

Before we discuss how we can calculate limits, there are very basic limits we need to know. They are: $\lim_{x\to a}x=a\ {\rm and}\ \lim_{x\to a}c=c,$ where $c$ is a constant. The limits are trivial from our intuition and also from their graphs. However, those who have analytical mind may want prove them. They can be proved using the Cauchy’s definition of a limit we discussed in the previous posting. Let us first prove $\displaystyle\lim_{x\to a}x=a$.

Proof. Let $\epsilon>0$ be given. Choose $\delta=\epsilon$. Then for all $x$ that satisfies the inequality $|x-a|<\delta,$ it is true that $|x-a|<\epsilon.$

Now we prove $\displaystyle\lim_{x\to a}c=c,$ where $c$ is a constant.

Proof. Let $\epsilon>0$ be given. Choose $\delta>0$ to be any positive real number. Then for all $x$ that satisfies the inequality $|x-a|<\delta,$ we have $|c-c|=0<\epsilon.$

Using these two limits, we can now show the following useful theorem for calculating limits of polynomial functions.

Theorem 2. Let $p(x)$ be a polynomial. Then for any real number $b$, $\lim_{x\to b}p(x)=p(b).$

Proof. Let $p(x)$ be a polynomial of degree $n$. Then $p(x)$ can be written as $p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_2x^2+a_1x+a_0,$ where $a_n,a_{n-1},\cdots,a_2,a_1,a_0$ are constant real coefficients. Now \begin{eqnarray}\lim_{x\to b}p(x)&=&\lim_{x\to b}(a_nx^n+a_{n-1}x^{n-1}+\cdots+a_2x^2+a_1x+a_0)\\&=&\lim_{x\to b}a_nx^n+\lim_{x\to a}a_{n-1}x^{n-1}+\cdots+\lim_{x\to a}a_2x^2+\lim_{x\to a}a_1x+\lim_{x\to a}a_0,\ \mbox{by property 1}\\&=&a_n\lim_{x\to b}x^n+a_{n-1}\lim_{x\to b}x^{n-1}+\cdots+a_2\lim_{x\to b}x^2+a_1\lim_{x\to b}x+a_0,\ \mbox{by property 2 and $\lim_{x\to b}a_0=a_0$}\\&=&a_nb^n+a_{n-1}b^{n-1}+\cdots+a_2b^2+a_1b+b_0,\ \mbox{by $\lim_{x\to b}x=b$ and property 3}\\&=&p(b).\end{eqnarray}

Theorem 2 also implies that any polynomial function is continuous everywhere. We will discuss the notion of continuity later.

Due to theorem 2 and property 4 of theorem 1, the following corollary about the limit of a rational function holds.

Corollary 3. [Limit of a Rational Function] Let $p(x)$ and $q(x)$ be two polynomials. Then for any real number $b$, $\lim_{x\to b}\frac{p(x)}{q(x)}=\frac{p(b)}{q(b)}$ provided $q(b)\ne 0$.

Theorem 4. [Other Important Limits] Suppose that $\displaystyle\lim_{x\to a}f(x)$ exists. Then the following properties hold:

1. $\displaystyle\lim_{x\to a}f(x)^n=[\displaystyle\lim_{x\to a}f(x)]^n$
2. $\displaystyle\lim_{x\to a}\root n\of{f(x)}=\root n\of{\displaystyle\lim_{x\to a}f(x)}$
3. $\displaystyle\lim_{x\to a}\ln f(x)=\ln[\displaystyle\lim_{x\to a}f(x)]$
4. $\displaystyle\lim_{x\to a}\sin f(x)=\sin[\displaystyle\lim_{x\to a}f(x)]$
5. $\displaystyle\lim_{x\to a}\cos f(x)=\cos[\displaystyle\lim_{x\to a}f(x)]$

It is assumed that $\displaystyle\lim_{x\to a} f(x)$ belongs to the domain of each function in each property. For example, in property 2 if $n$ is an even integer, it is required that $\displaystyle\lim_{x\to a} f(x)\geq 0$. The first property is a direct consequence of property 3 of theorem 1. The rest of the properties are related to the continuity of a function. This will be discussed later.

# The Precise Definition of a Limit

The definition of a limit we previously discussed here is intuitive and qualitative rather than quantitative. It may be helpful for us to conceptually understand the notion of a limit, but it is useless when you try to prove some fundamental properties of limits, for instance the properties described in the following theorem.

Theorem. Suppose that $c$ is a constant and the limits $\lim_{x\to a}f(x)\ {\rm and}\ \lim_{x\to a}g(x)$ exist. Then the following properties hold:

1. $\displaystyle\lim_{x\to a}\{f(x)+g(x)\}=\lim_{x\to a}f(x)+\lim_{x\to a}g(x)$
2. $\displaystyle\lim_{x\to a}cf(x)=c\lim_{x\to a}f(x)$
3. $\displaystyle\lim_{x\to a}f(x)g(x)=\lim_{x\to a}f(x)\cdot\lim_{x\to a}g(x)$
4. $\displaystyle\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{\displaystyle\lim_{x\to a}f(x)}{\displaystyle\lim_{x\to a}g(x)}$ provided $\displaystyle\lim_{x\to a}g(x)\ne 0$

For a long time, mathematicians have believed that the above properties were true and have used them without having been able to prove them. A limit is a mathematical quantity and in order to deal with a mathematical quantity, one needs to have a quantitative definition. And a quantitative definition must consist of quantities that are finite. Finally a French mathematician Augustin-Louis Cauchy came up with such a definition.

Definition. [$\epsilon-\delta$ Argument] A function $f(x)$ is said to approach a value $A$ as $x$ approaches $a$, if for any given positive number $\epsilon>0$ (no matter how small it is) there exists a positive number $\delta>0$ such that $|f(x)-A|<\epsilon$ is true for every $x$ that satisfies the inequality $0<|x-a|<\delta.$

The following picture also shows you why this definition makes sense.

The limit of f(x) is A when x approaches a.

Example. Prove that $\displaystyle\lim_{x\to 2}(5x-2)=8$.

Solution: Let $\epsilon>0$ be given. Then we want to show that there exists $\delta>0$ such that $|(5x-2)-8|=|5x-10|<\epsilon$ is satisfied, whenever $x$ is in the domain $0<|x-2|<\delta.$ Now divide the inequality $|5x-10|<\epsilon$ by $5$. Then we obtain $|x-2|<\frac{\epsilon}{5}.$ Hence, $\delta=\frac{\epsilon}{5}$is an adequate choice for $\delta$ and the proof is complete.

If $\epsilon=0.005$, the proceeding result tells us that the function $5x-2$ will lie in the range $7.995<5x-2<8.005$ whenever the domain of $x$ is $1.999<x<2.001$.

Example. In order to see why the above definition does not work for non-existing limits, let us recall the second example from my previous posting: $f(x)=\left\{\begin{array}{ccc}x-1 & {\rm if} & x<2\$x-2)^2+3 & {\rm if} & x\geq 2.\end{array}\right.$ Let us show that the left-hand limit \(1$ of $f(x)$ at $x=2$ cannot be a limit of $f(x)$ when $x$ approaches $2$. Let $\epsilon=1$. Then no matter how small $\delta>0$ one chooses there is a number $2<x_0<2+\delta$ and $f(x_0)>3$. That is, $f(x_0)$ does not satisfy the inequality $0<f(x)<2$. Hence, $2$ cannot be a limit. Similarly, one can show that the right-hand limit $3$ of $f(x)$ at $x=1$ cannot be a limit either.

In order to see how the Cauchy’s definition of a limit can be used to prove some fundamental properties of limits, we prove the first property of the above theorem: If $\lim_{x\to a}f(x)=A\ {\rm and}\ \lim_{x\to a}g(x)=B$ then $\lim_{x\to a}\{f(x)+g(x)\}=A+B.$

Proof. Let $\epsilon>0$ be given. Since $\displaystyle\lim_{x\to a}f(x)=A$ and $\displaystyle\lim_{x\to a}g(x)=B$, there exist $\delta_1>0$ and $\delta_2>0$ such that $|f(x)-A|<\frac{\epsilon}{2},$ whenever $0<|x-a|<\delta_1$ and such that $|g(x)-B|<\frac{\epsilon}{2},$ whenever $0<|x-a|<\delta_2$. Choose $\delta=\min(\delta_1,\delta_2)$, i.e. we choose $\delta$ to be the smaller of $\delta_1$ and $\delta_2$. Now if $x$ satisfies the inequality $0<|x-a|<\delta$ then $|f(x)+g(x)-(A+B)|=|f(x)-A+g(x)-B|\leq |f(x)-A|+|g(x)-B|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.$ This completes the proof.

The rest of the properties in the above theorem can be proved in a similar manner. I stumbled upon the Cauchy’s definition of a limit back when I was a high school senior. Once I understood what it means I was so amazed by its beauty. After I proved all the properties listed in the above theorem, I nearly experienced Nirvana and Enlightenment or more precisely speaking I tasted very slightly what they would be like. It was a long time ago, but I still remember the joy and the excitement from that little experience. It was an awakening moment for me that opened my eyes to the beauty of the Nature and the Universe. I can’t say I am spiritual but that experience surely led me to mathematics. Won’t you try to discover and experience some of it too?

# Examples of Non-Existing Limits

Limit of a function does not necessarily exists. Possible cases of non-existing limits would be when

1. at least one of the one-sided limits does not exist;
2. both one-sided limits exist but they are not the same.

Here are a couple of examples of non-existing limits.

Example. Let $f(x)$ be the function defined by $f(x)=\sin\frac{1}{x}$ for $x\ne 0$. The graph of this function is given by

.

As $x$ approaches to $0$, $\sin\frac{1}{x}$ keeps oscillating near the $y$-axis but it does not approach to anywhere. This is the case when neither $\lim_{x\to 0-}\sin\frac{1}{x}$ nor $\lim_{x\to 0+}\sin\frac{1}{x}$ exists. The following picture shows you a closer look at the graph near the $y$-axis.

The function $f(x)=\sin\frac{1}{x}$ is called topologist’s sine curve.

Example. Let $f(x)$ be the function defined by $f(x)=\left\{\begin{array}{ccc}x-1 & {\rm if} & x<2\$x-2)^2+3 & {\rm if} & x\geq 2.\end{array}\right.$ The graph of \(f(x)$ is

Let us calculate the left-hand and the right-hand limit of $f(x)$ at $x=2$: \begin{eqnarray*}\lim_{x\to 2-}f(x)&=&\lim_{x\to 2-}(x-1)\\&=&1,\\\lim_{x\to 2+}f(x)&=&\lim_{x\to 2+}(x-2)^2+3\\&=&3.\end{eqnarray*} Both the left-hand and the right-hand limits of $f(x)$ exist, however they do not coincide. Hence the limit $\lim_{x\to 2}f(x)$ does not exist.

# Limits of Functions

It is very important for students to get familiar with notion of function, some important classes of functions (polynomial functions, rational functions, and trigonometric functions, etc.) and their properties, and trigonometry before they begin to study calculus. If you are not comfortable with any of those I mentioned above, you should brush up as soon as possible.

If you have not studied calculus before, you will notice that it is very different from any subject of mathematics you have encountered before. Calculus deals with a notion of closeness or nearness. Imagine that the variable $x$ does not just represent a fixed number as a solution of an equation but it keeps changing and approaching to a number, say $2$, closer and closer. This describes a state in which $x$ is very near the number $2$. Such a state will be denoted by $x\to 2$. To get some better picture, you may imagine a flying arrow that keeps getting closer and closer to the target but it never hits the target. Such notion of nearness plays a very important role in mathematics and it is developed into topology, one of the most important and fundamental subjects of mathematics. In particular, the state $x\to 0$ is called infinitesimal and is also denoted by $0$. Infinitesimal means “extremely small.” What would be confusing to students who learn calculus for the first time is that infinitesimal $0$ and the number $0$ are denoted by the same symbol. There is a branch of advanced mathematics, so-called non-standard analysis. In there, infinitesimal is treated as a number and is denoted by $0^\ast$ in order to distinguish it from the number $0$. But we are not going to use this notation. So how do we distinguish them? It turns out that it is not difficult to distinguish them from the context as we will see later. So there is absolutely no reason for you to panic. For the same reason, we often write $x\to 2$ by $2$, however it is very important for you to see it, from the context, as a state in which $x$ is very near 2, not as the fixed number 2. Using the notations from non-standard analysis, it can be written $2^\ast=2+0^\ast.$ Before we move on, I would like to point out that there are clear distinctions between an infinitesimal $0$ and the number $0$. First, an infinitesimal $0$ can be positive or negative while the number $0$ is neutral. A positive infinitesimal and a negative infinitesimal may be represented by the notations $x\to 0+$ and $x\to 0-$, respectively. The notation $x\to 0+$ means that $x$ is approaching to $0$ from the right (or from above $0$). Similarly, $x\to 0-$ means that $x$ is approaching to $0$ from the left (or from below $0$). Second, one can divide a number by an infinitesimal while the division of a number by the number $0$ is not defined. For instance, $\frac{1}{0}$ is not defined when $0$ is a number but $\frac{1}{0}=\infty$ when $0$ is a positive infinitesimal. Similarly, $\frac{1}{0}=-\infty$ if $0$ is a negative infinitesimal.

In calculus, we are interested in the behavior of a function near a point. For instance, we want to study how the function $f(x)=\frac{x^2-1}{x-1}$ behaves near $x=1$. Note that the function is not defined at $x=1$. The most direct way to study this is to use sample values of $x$ that get closer to 1 and see how the values of $f(x)$ changes.

 Values of $x$ below and above $1$ $f(x)=\frac{x^2-1}{x-1}$ 0.9 1.9 0.99 1.99 0.999 1.999 0.999999 1.999999 1.1 2.1 1.01 2.01 1.001 2.001 1.0000001 2.0000001

One can guess from the table that the values of $f(x)=\frac{x^2-1}{x-1}$ approach to $2$ as both $x\to 1-$ and $x\to 1+$. In this case, we say that the limit of $f(x)=\frac{x^2-1}{x-1}$ is $2$ as $x$ approaches to $1$ and write $\lim_{x\to 1}f(x)=2$ or $\lim_{x\to 1}\frac{x^2-1}{x-1}=2.$ The following graph of $f(x)=\frac{x^2-1}{x-1}$ also clearly shows this.

In our example, $\lim_{x\to 1-}\frac{x^2-1}{x-1}=\lim_{x\to 1+}\frac{x^2-1}{x-1}=2.$ The one-sided limits $\lim_{x\to a-}f(x)$ and $\lim_{x\to a+}f(x)$ are called, respectively, the left-hand limit and the right-hand limit of $f(x)$ at $x=a$. Clearly we have the following definition.

Definition. We say that the limit $\lim_{x\to a}f(x)$ exists if both $\lim_{x\to a-}f(x)$ and $\lim_{x\to a+}f(x)$ exist and they are the same. The limit $\lim_{x\to a}f(x)$ is defined to be the value of $\lim_{x\to a-}f(x)$ (or that of $\lim_{x\to a+}f(x)$).

Now we have got some picture of what the limit of a function is. The next important question would be “how do we find the limit?” if it exists. In practice, we do not try to find the limit of a function as we did in the table for reasons. We could only guess the limit by the method and often it can be very difficult to make a guess if the given function is a complicated one. As a result, in many cases you cannot be certain that your guess is correct even if you come up with one. Such method sometimes can easily lead to a wrong answer. If limits exist, there are ways to calculate them exactly. In the above example, again $x\to 1$ means that $x$ is very near $1$ but not exactly equal to $1$. Since $x\ne 1$,

\begin{eqnarray*}\lim_{x\to 1}\frac{x^2-1}{x-1}&=&\lim_{x\to 1}\frac{(x+1)(x-1)}{x-1}\\&=&\lim_{x\to 1}(x+1)\\&=&2.\end{eqnarray*}

The method of finding exact values of limits can vary depending on the types of functions. We will discuss more about it later.

# MAT 167 Calculus 1 Class Orientation

Welcome to my MAT 167 class.

I need to mention a few important things for start-up:

1. The course syllabus is on-line and it is a part of our class home page. The main page contains general information that you need to know about the course. In there, you will find a link to course outline page. The course outline page contains weekly schedule, homework assignments, weekly quizzes and their solutions, and important notices regarding quizzes and exams.