Q.An important problem in fishery science is to estimate the number of fish presently spawning in streams and use this information to predict the number of mature fish or "recruits" that will return to the rivers during the reproductive period. If $S$ is the number of spawners and $R$ the number of recruits, the "Beverton-Holt spawner recruit function" is

[!FORMULA] $R(S)=\dfrac{S}{\alpha S+\beta}$

where $\alpha$ and $\beta$ are positive constants. Show that this function predicts approximately constant recruitment when the number of spawners is sufficiently large.

Two Different Kinds of "Infinity" in a Limit

This concept covers two distinct situations that both involve the symbol $\infty$, and it's important to keep them apart:

1. Infinite limits — $x$ approaches a finite point, but $f(x)$ itself grows without bound.
2. Limits at infinity — $x$ itself grows without bound (positively or negatively), and we ask what $f(x)$ settles toward.

Infinite limits and vertical asymptotes

Consider $f(x) = \dfrac{1}{x^2}$ near $x=0$. As $x \to 0$ from either side, $f(x)$ grows without bound. We write

meaning the limit does not exist (there is no finite $L$) — but this particular flavour of non-existence is worth naming, because it tells us $x=0$ is a vertical asymptote.

Definitions 9.4 & 9.5 (informal). A neighbourhood of $+\infty$ is any interval $(M,\infty)$ for large $M>0$; a neighbourhood of $-\infty$ is any $(-\infty,K)$ for very negative $K$. We say $f(x) \to \infty$ as $x \to x_0$ if $f(x)$ eventually lands in every such neighbourhood of $+\infty$ as $x$ gets close enough to $x_0$ — and similarly for $f(x)\to -\infty$, and for one-sided versions ($x\to x_0^-$, $x \to x_0^+$).

General pattern for $\dfrac{1}{(x-a)^n}$:

• If $n$ is even, $\dfrac{1}{(x-a)^n} \to +\infty$ as $x \to a$ from either side (both one-sided "limits" blow up the same way).
• If $n$ is odd, $\dfrac{1}{(x-a)^n} \to -\infty$ as $x\to a^-$ but $\to +\infty$ as $x \to a^+$ (the two sides disagree in sign — the two-sided limit fails to exist even in this loose infinite sense).

In every such case, the line $x=a$ is a vertical asymptote of the graph.

Limits at infinity and horizontal asymptotes

Now let $x$ itself run away to $\pm\infty$, and ask what $f(x)$ approaches.

Definition 9.6. The line $y = l$ is a horizontal asymptote of $y=f(x)$ if $\displaystyle\lim_{x\to -\infty} f(x) = l$ or $\displaystyle\lim_{x\to +\infty} f(x) = l$.

Illustration: $\tan^{-1}x$ has two different horizontal asymptotes — $\lim_{x\to-\infty}\tan^{-1}x = -\frac{\pi}{2}$ and $\lim_{x\to+\infty}\tan^{-1}x = \frac{\pi}{2}$ — a reminder that a function can have (at most) two horizontal asymptotes, one per direction, and they need not agree.

The core technique: divide by the highest power of $x$

Trying to apply the ordinary limit laws to something like $\dfrac{2x^2+2x+3}{x^2+4x+3}$ as $x\to\infty$ produces $\frac{\infty}{\infty}$ — an indeterminate form: not a valid computation, just a signal that you must rewrite before proceeding.

The fix: divide numerator and denominator by the highest power of $x$ appearing in the denominator. For the example above, dividing through by $x^2$ gives

since every term of the form $\dfrac{c}{x^k}\to 0$ as $x\to\infty$.

Degree comparison for rational functions (§9.2.6)

For $R(x) = \dfrac{p(x)}{q(x)}$ as $x \to \infty$: