Infinite limits and limits at infinity

Not every limit settles down to a finite real number as $x$ approaches a point — sometimes the function values grow without any bound at all. This section studies that behaviour, which is entirely different from "the limit does not exist because of disagreeing one-sided limits" (Section 9.2.1's third illustration): here, both sides can behave the same way, but that common behaviour is unboundedness rather than convergence to a number.

Motivating example. Consider $f(x)=\dfrac{1}{x^2}$, undefined at $x=0$. Tabulating $x$-values shrinking towards $0$ from either side shows $f(x)$ growing larger without limit — from $1$, to $100$, to $10{,}000$, to $10{,}00{,}000$, and beyond. We describe this by writing $\dfrac1{x^2}\to\infty$ as $x\to0^-$ and as $x\to0^+$ (Fig. 9.22 shows both branches of the curve rising without end on either side of the $y$-axis), and say $x=0$ is a vertical asymptote. A closely related but importantly different function is $f(x)=\dfrac1x$ (Fig. 9.23): here $\dfrac1x\to-\infty$ as $x\to0^-$ but $\dfrac1x\to+\infty$ as $x\to0^+$ — the two sides diverge to opposite infinities, unlike $1/x^2$ where both sides behave the same way.

Formally:

Definition 9.4. For $M>0$, an interval of the form $(M,\infty)$ is called a neighbourhood of $\infty$; for $K<0$, an interval of the form $(-\infty,K)$ is called a neighbourhood of $-\infty$.

Definition 9.5. $f(x)\to\infty$ as $x\to x_0$ if, for every $M>0$, there is a neighbourhood of $x_0$ within which $f(x)\in(M,\infty)$ whenever $x$ lies in that neighbourhood — and similarly $f(x)\to-\infty$ as $x\to x_0$ using a $K<0$ and the interval $(-\infty,K)$.

This gives rise to six possible one-sided and two-sided "infinite limit" statements — $f(x)\to\pm\infty$ as $x\to x_0$, or as $x\to x_0^-$, or as $x\to x_0^+$ — and whenever any one of them holds, the vertical line $x=x_0$ is a vertical asymptote of the graph of $f$.

The general odd/even-power pattern (Example 9.22, $f(x)=\dfrac{1}{(x-2)^3}$, Fig. 9.24). Near a point $x=a$ where the denominator is $(x-a)^n$:

• If $n$ is even, $\dfrac1{(x-a)^n}\to+\infty$ from both sides (exactly like $1/x^2$ above), so the two-sided infinite-limit statement holds.
• If $n$ is odd, $\dfrac1{(x-a)^n}\to-\infty$ from the left and $\to+\infty$ from the right (exactly like $1/x$ above): the two sides disagree in sign, so only the one-sided statements hold, and even the infinite-sense limit does not exist two-sidedly. …