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, $$\begin{eqnarray*}\lim_{x\to 2}\frac{4-x^2}{3-\sqrt{x^2+5}}&=&\lim_{x\to 2}\frac{4-x^2}{3-\sqrt{x^2+5}}\cdot\frac{3+\sqrt{x^2+5}}{3+\sqrt{x^2+5}}\\&=&\lim_{x\to 2}\frac{(4-x^2)(3+\sqrt{x^2+5})}{4-x^2}\\&=&\lim_{x\to 2}(3+\sqrt{x^2+5})\\&=&6.\end{eqnarray*}$$