Limits at infinity

Section 9.2.4 let $x$ approach a finite point and asked what happens to $y$ when $y$ becomes unbounded. This section reverses the roles: now $x$ itself is allowed to grow without bound (positively or negatively), and we ask what value, if any, $y=f(x)$ settles towards.

Motivating example. For $f(x)=\dfrac{x^2-1}{x^2+1}$, tabulating $f$ at $x=0,\pm1,\pm2,\dots,\pm1000$ shows the values climbing steadily towards (but never quite reaching) $1$ as $|x|$ grows (Fig. 9.25 shows the curve flattening out and hugging the horizontal line $y=1$ on both ends). We write $\lim_{x\to\pm\infty}\dfrac{x^2-1}{x^2+1}=1$.

Definition 9.6 (horizontal asymptote). The line $y=l$ is a horizontal asymptote of $y=f(x)$ if $\lim_{x\to-\infty}f(x)=l$ or $\lim_{x\to+\infty}f(x)=l$ — the two limits, at $-\infty$ and at $+\infty$, need not agree, so a single curve can genuinely have two different horizontal asymptotes, one on each end.

Illustration 9.4 applies this to the inverse-tangent function $f(x)=\tan^{-1}x$ on $\left(-\tfrac\pi2,\tfrac\pi2\right)$ (Fig. 9.26): reading the graph gives $\lim_{x\to-\infty}\tan^{-1}x=-\dfrac\pi2$ and $\lim_{x\to+\infty}\tan^{-1}x=\dfrac\pi2$ — two different horizontal asymptotes, one at each end, since the arctangent curve never actually reaches $\pm\dfrac\pi2$ but approaches each of them on its own side.

The indeterminate-form trap and its fix (Illustration 9.5). For $\dfrac{2x^2-2x+3}{x^2+4x+3}$ as $x\to\infty$, simply noting "numerator $\to\infty$" and "denominator $\to\infty$" and writing $\dfrac\infty\infty$ tells us nothing — $\dfrac\infty\infty$ is called an indeterminate form, precisely because different expressions sharing that same superficial pattern can tend to entirely different limits. A table of values does suggest the answer is $2$, but the reliable route is algebraic: divide every term, top and bottom, by the highest power of $x$ appearing in the denominator — here $x^2$:

$\frac{2x^2-2x+3}{x^2+4x+3}=\frac{2-\frac2x+\frac3{x^2}}{1+\frac4x+\frac3{x^2}}\;\xrightarrow[x\to\infty]{}\;\frac{2-0+0}{1+0+0}=2,$ …