MA005: Calculus I

If the limit, as $x$ approaches $c$, exists and equals $L$, then we can guarantee that the values of $f(x)$ are as close to $L$ as we want by restricting the values of $x$ to be very, very close to $\mathrm{c}$. To show that a limit, as $x$ approaches c, does not exist, we need to show that no matter how closely we restrict the values of $x$ to $\mathrm{c}$, the values of $\mathrm{f}(\mathrm{x})$ are not all close to a single, finite value $\mathrm{L}$. One way to demonstrate that $\lim\limits_{x \rightarrow c} \mathrm{f}(\mathrm{x})$ does not exist is to show that the left and right limits exist but are not equal.

Another method of showing that $\lim\limits_{x \rightarrow c} \mathrm{f}(\mathrm{x})$ does not exist is to find two infinite lists of numbers, $\left\{\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \mathrm{a}_{4}, \ldots\right\}$ and $\left\{\mathrm{b}_{1}, \mathrm{~b}_{2}, \mathrm{~b}_{3}, \mathrm{~b}_{4}, \ldots\right\}$, which approach arbitrarily close to the value $\mathrm{c}$ as the subscripts get larger, but so that the lists of function values, $\left\{\mathrm{f}\left(\mathrm{a}_{1}\right), \mathrm{f}\left(\mathrm{a}_{2}\right), \mathrm{f}\left(\mathrm{a}_{3}\right), \mathrm{f}\left(\mathrm{a}_{4}\right), \ldots\right\}$ and $\left\{\mathrm{f}\left(\mathrm{b}_{1}\right), \mathrm{f}\left(\mathrm{b}_{2}\right), \mathrm{f}\left(\mathrm{b}_{3}\right), \mathrm{f}\left(\mathrm{b}_{4}\right), \ldots\right\}$, approach two different numbers as the subscripts get larger.

Example 6: For $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{ll}1 & \text { if } \mathrm{x} < 1 \\ \mathrm{x} & \text { if } 1 < \mathrm{x} < 3 \quad \text { if } 3 < \mathrm{x}\end{array} \quad\right.$, show that $\lim\limits_{x \rightarrow 3} \mathrm{f}(\mathrm{x})$ does not exist.

Solution: We can use one-sided limits to show that this limit does not exist, or we can use the list method by selecting values for one list to approach 3 from the right and values for the other list to approach 3 from the left.

One way to define values of $\left\{a_{1}, a_{2}, a_{3}, a_{4}, \ldots\right\}$ which approach 3 from the right is to define $\mathrm{a}_{1}=3+1, \mathrm{a}_{2}=3+\frac{1}{2}, \mathrm{a}_{3}=3+\frac{1}{3}, \mathrm{a}_{4}=3+\frac{1}{4}$ and, in general, $\mathrm{a}_{\mathrm{n}}=3+\frac{1}{\mathrm{n}}$. Then $\mathrm{a}_{\mathrm{n}} > 3$ so $\mathrm{f}\left(\mathrm{a}_{\mathrm{n}}\right)=2$ for all subscripts $\mathrm{n}$, and the values in the list $\left\{\mathrm{f}\left(\mathrm{a}_{1}\right), \mathrm{f}\left(\mathrm{a}_{2}\right), \mathrm{f}\left(\mathrm{a}_{3}\right), \mathrm{f}\left(\mathrm{a}_{4}\right), \ldots\right\}$ are approaching 2. In fact, all of the $\mathrm{f}\left(\mathrm{a}_{\mathrm{n}}\right)=2$.

We can define values of $\left\{b_{1}, b_{2}, b_{3}, b_{4}, \ldots\right\}$ which approach 3 from the left by $b_{1}=3-1$, $\mathrm{b}_{2}=3-\frac{1}{2}, \mathrm{~b}_{3}=3-\frac{1}{3}, \mathrm{~b}_{4}=3-\frac{1}{4}$ and, in general, $\mathrm{b}_{\mathrm{n}}=3-\frac{1}{\mathrm{n}}$. Then $\mathrm{b}_{\mathrm{n}} < 3$ so $\mathrm{f}\left(\mathrm{b}_{\mathrm{n}}\right)=\mathrm{b}_{\mathrm{n}}=3-\frac{1}{\mathrm{n}}$ for each subscript $\mathrm{n}$, and the values in the list $\left\{\mathrm{f}\left(\mathrm{b}_{1}\right), \mathrm{f}\left(\mathrm{b}_{2}\right), \mathrm{f}\left(\mathrm{b}_{3}\right), \mathrm{f}\left(\mathrm{b}_{4}\right), \ldots\right\}=\left\{2,2.5,2.67,2.75,2.8, \ldots, 3-\frac{1}{\mathrm{n}}, \ldots\right\}$ approach 3.

Since the values in the lists $\left\{f\left(a_{1}\right), f\left(a_{2}\right), f\left(a_{3}\right), f\left(a_{4}\right), \ldots\right\}$ and $\left\{f\left(b_{1}\right), f\left(b_{2}\right), f\left(b_{3}\right), f\left(b_{4}\right), \ldots\right\}$ approach two different numbers, we can conclude that $\lim\limits_{x \rightarrow 3} \mathrm{f}(\mathrm{x})$ does not exist.

Example 7: Let $h(x)=\left\{\begin{array}{ll}2 & \text { if } x \text { is a rational number } \\ 1 & \text { if } x \text { is an irrational number }\end{array} \quad\right.$ be the "holey" function. Use the list method to show that $\lim\limits_{x \rightarrow 3} \mathrm{~h}(\mathrm{x})$ does not exist.

Solution: Let $\left\{\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \mathrm{a}_{4}, \ldots\right\}$ be a list of rational numbers which approach 3, for example, $\mathrm{a}_{1}=3+1, \mathrm{a}_{2}=3+$ $1 / 2, \ldots, \mathrm{a}_{\mathrm{n}}=3+1 / \mathrm{n}.$ Then $\mathrm{f}\left(\mathrm{a}_{\mathrm{n}}\right)$ always equals 2 so $\left\{\mathrm{f}\left(\mathrm{a}_{1}\right), \mathrm{f}\left(\mathrm{a}_{2}\right), \mathrm{f}\left(\mathrm{a}_{3}\right), \mathrm{f}\left(\mathrm{a}_{4}\right), \ldots\right\}=\{2,2,2, \ldots\}$ and the $f\left(a_{n}\right)$ values "approach" 2. If $\left\{b_{1}, b_{2}, b_{3}, b_{4}, \ldots\right\}$ is a list of irrational numbers which approach 3, for example, $b_{1}=3+\pi, b_{2}=3+\pi / 2, \ldots, b_{n}=3+\pi / \mathrm{n}$. then $\left\{f\left(b_{1}\right), f\left(b_{2}\right), f\left(b_{3}\right), f\left(b_{4}\right), \ldots\right\}=\{1,1,1, \ldots\}$ and the $f\left(b_{n}\right)$ "approach" 1. Since the $f\left(a_{n}\right)$ and $f\left(b_{n}\right)$ values approach different numbers, the limit as $x \rightarrow 3$ does not exist.

A similar argument will work as $x$ approaches any number $c$, so for every $c$ we have that $\lim\limits_{x \rightarrow c} h(x)$ does not exist. The "holey" function does not have a limit as $x$ approaches any value $c$.