Q.$\displaystyle\lim_{x\to\infty}\dfrac{\sqrt{x^2-1}}{2x+1}=$

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 $x \to \infty$: