Calculus 5: How to Calculate Limits II

In here, we studied how to calculate the limit of a rational function (corollary there). Let us state it here again:

Corollary. [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 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{align*}\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{align*}
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{align*}\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{align*}