Infinite Limits and Asymptotes

The function $f(x)=\frac{1}{x^{2}}$ is undefined at $x=0$, but we can still ask about the behavior of $\mathrm{f}(\mathrm{x})$ for values of $\mathrm{x}$ "close to" $0$. Fig. 6 indicates that if $\mathrm{x}$ is very small, close to $0$, then $f(x)$ is very large. As the values of $x$ get closer to $0$, the values of $\mathrm{f}(\mathrm{x})$ grow larger and can be made as large as we want by picking $\mathrm{x}$ to be close enough to $0$. Even though the values of $f$ are not approaching any number, we use the "infinity" notation to indicate that the values of $\mathrm{f}$ are growing without bound, and write $\lim \limits_{x \rightarrow 0} \frac{1}{\mathrm{x}^{2}}=\infty$.

Fig. 6

The values of $\frac{1}{\mathrm{x}^{2}}$ do not equal "infinity:" $\lim \limits_{x \rightarrow 0} \frac{1}{\mathrm{x}^{2}}=\infty$ means that the values of $\frac{1}{\mathrm{x}^{2}}$ can be made arbitrarily large by picking values of $x$ very close to $0$.

The limit, as $x \rightarrow 0$, of $\frac{1}{x}$ is slightly more complicated. If $x$ is close to $0$, then the value of $f(x)=1 / x$ can be a large positive number or a large negative number, depending on the sign of $\mathrm{x}$.

The function $\mathrm{f}(\mathrm{x})=1 / \mathrm{x}$ does not have a (two-sided) limit as $\mathrm{x}$ approaches $0$, but we can still ask about one-sided limits:$\lim \limits_{x \rightarrow 0^{+}} \frac{1}{x}=\infty \text { and } \lim \limits_{x \rightarrow 0^­} \frac{1}{x}=-\infty$

Example 5: Determine $\lim \limits_{x \rightarrow 3^{+}} \frac{x-5}{x-3}$ and $\lim \limits_{x \rightarrow 3^­} \frac{x-5}{x-3}$.

Solution: (a) As $\mathrm{x} \rightarrow 3^{+}$, then $\mathrm{x}-5 \rightarrow-2$ and $\mathrm{x}-3 \rightarrow 0$. Since the denominator is approaching $0$ we cannot use the Main Limit Theorem, and we need to examine the functions more carefully. If $\mathrm{x} \rightarrow 3^{+}$, then $x > 3$ so $x-3 > 0$. If $x$ is close to $3$ and slightly larger than $3$, then the ratio of $x-5$ to $x-3$ is the ratio $\frac{\text { a number close to }-2}{\text { small positive number }} = \text{large negative number}$. As $x > 3$ gets closer to $3, \frac{x-5}{x-3}$ is $\frac{\text { a number closer to }-2}{\text { positive and closer to } 0} = \text{large negative number}$. By taking $\mathrm{x} > 3$ closer to $3$, the denominator gets closer to $0$ but is always positive, so the ratio gets arbitrarily large and negative: $\lim \limits_{x \rightarrow 3^{+}} \frac{x-5}{x-3}=-\infty$.
(b) As $x \rightarrow 3^­$, then $x-5 \rightarrow-2$ and $x-3$ gets arbitrarily close to $0$, and $x-3$ is negative. The value of the ratio $\frac{x-5}{x-3}$ is $\frac{\text { a number close to }-2}{\text { arbitrarily small negative number }} = \text{arbitrarily large positive number}$: $\lim \limits_{x \rightarrow 3^­} \frac{x-5}{x-3}=+\infty$.

Practice 5: Determine $\lim \limits_{x \rightarrow 2^{+}} \frac{7}{2-x}, \lim \limits_{x \rightarrow 2^{+}} \frac{3 x}{2 x-4}, \lim \limits_{x \rightarrow 2^{+}} \frac{3 x^{2}-6 x}{x-2}$.