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.

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*}

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+a_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 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?

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.

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.