The Timpson Lyceum · Calculus 1 (2024)

Limits Involving Infinity or Zero, and Asymptotes - Lecture 7 (Calculus 1)

Overview

This lecture extends the idea of a limit to the two ways infinity enters calculus: limits taken as the input runs off to $+\infty$ or $-\infty$ , and limits in which the output shoots up to $+\infty$ or down to $-\infty$ . These two behaviors correspond exactly to the two kinds of straight-line "guide rails" a graph can have — horizontal asymptotes (what a function approaches far to the left or right) and vertical asymptotes (where a function blows up). There are no new theorems here; the same limit laws (limit of a sum, product, quotient) still apply. The payoff is a small set of practical rules for rational functions plus techniques — dividing by the highest-power term, multiplying by a conjugate, the squeeze theorem, and the substitution $t = 1/x$ — for evaluating these limits.

Key Concepts

Limits at infinity. $\lim_{x \to \infty} f(x) = L$ means: for every $\varepsilon > 0$ there is a large number $M$ such that $x > M \implies |f(x) - L| < \varepsilon$ . In words, once $x$ is far enough to the right, every output sits inside any chosen band around $L$ . The limit as $x \to -\infty$ is identical except the requirement becomes $x < -M$ .

Infinite limits. $\lim_{x \to c} f(x) = \infty$ means: for every large $B$ there is a $\delta > 0$ such that $0 < |x - c| < \delta \implies f(x) > B$ . The function can be forced above any target by squeezing the inputs near $c$ . For $-\infty$ , the conclusion becomes $f(x) < -B$ . Note that $\infty$ is not a number — the notation records unbounded growth, not equality with a value.

Horizontal asymptote. The line $y = b$ is a horizontal asymptote of $f$ if $\lim_{x \to \infty} f(x) = b$ or $\lim_{x \to -\infty} f(x) = b$ . Because each one-sided-at-infinity limit gives at most one value, a function has 0, 1, or 2 horizontal asymptotes (two when the left- and right-hand limits differ). A graph may cross a horizontal asymptote — e.g. $\tfrac{1}{x}\sin x$ crosses $y = 0$ infinitely often while still approaching it.

Vertical asymptote. The line $x = a$ is a vertical asymptote if $\lim_{x \to a^+} f(x) = \pm\infty$ or $\lim_{x \to a^-} f(x) = \pm\infty$ . A graph never crosses a vertical asymptote, and these occur only where the function is undefined. A function may have infinitely many (e.g. $\tan x$ ).

Hole vs. vertical asymptote (rational functions). Where the denominator is zero you get either a hole or a vertical asymptote. If the offending factor cancels with one in the numerator, it is a removable hole; if it does not cancel, it is a vertical asymptote. For example $\dfrac{x^2-1}{x-1} = x+1$ has a hole at $x = 1$ , but $\dfrac{1}{x}$ has a vertical asymptote at $x = 0$ .

End-behavior rules for a rational function $f(x) = \dfrac{p(x)}{q(x)}$ , with numerator degree $n$ and denominator degree $m$ :

$n < m$ : horizontal asymptote $y = 0$ (denominator outgrows numerator).
$n = m$ : horizontal asymptote $y = \dfrac{a_n}{b_m}$ , the ratio of the leading coefficients.
$n = m + 1$ : an oblique (slant) asymptote — divide $p$ by $q$ via polynomial long division; the quotient line is the asymptote and the remainder term vanishes at infinity.

Standard technique. To evaluate a rational limit at infinity, divide every term top and bottom by the highest power of $x$ present, then apply the limit laws using $\lim_{x \to \pm\infty} \tfrac{1}{x^k} = 0$ .

Other tools.

Conjugate trick for differences like $x - \sqrt{x^2 + 16}$ (an $\infty - \infty$ form): multiply by the conjugate over itself.
Substitution $t = 1/x$ turns an $x \to \infty$ problem into a $t \to 0^+$ problem (useful for $\infty \cdot 0$ forms such as $x\sin(1/x)$ ).
Sign of $|x|$ matters: for $x \to -\infty$ , $\sqrt{x^2} = |x| = -x$ and $|x^3| = -x^3$ , which can flip a sign and produce a different horizontal asymptote on the left.

Worked Example

Equal degrees ( $n = m$ ): divide by the highest power. Evaluate $\displaystyle \lim_{x \to \infty} \frac{5x^2 + 8x - 3}{3x^2 + 2}$ . Divide every term by $x^2$ (the highest power), then send each $1/x^k$ term to $0$ :

\begin{aligned} \lim_{x \to \infty} \frac{5x^2 + 8x - 3}{3x^2 + 2} &= \lim_{x \to \infty} \frac{5 + \dfrac{8}{x} - \dfrac{3}{x^2}}{3 + \dfrac{2}{x^2}} \\ &= \frac{5 + 0 - 0}{3 + 0} = \frac{5}{3}. \end{aligned}

This matches the equal-degree rule (leading coefficients $5$ and $3$ ), so $y = \tfrac{5}{3}$ is the horizontal asymptote.

Conjugate trick ( $\infty - \infty$ ). Evaluate $\displaystyle \lim_{x \to \infty}\bigl(x - \sqrt{x^2 + 16}\bigr)$ by multiplying by the conjugate, then dividing by $x$ :

\begin{aligned} x - \sqrt{x^2 + 16} &= \frac{\bigl(x - \sqrt{x^2+16}\bigr)\bigl(x + \sqrt{x^2+16}\bigr)}{x + \sqrt{x^2+16}} = \frac{-16}{x + \sqrt{x^2 + 16}} \\ \lim_{x \to \infty}\frac{-16}{x + \sqrt{x^2+16}} &= \lim_{x \to \infty}\frac{-16/x}{1 + \sqrt{1 + 16/x^2}} = \frac{0}{1 + 1} = 0. \end{aligned}

Summary

Two ideas: limits at infinity (input $\to \pm\infty$ ) describe horizontal asymptotes; infinite limits (output $\to \pm\infty$ ) describe vertical asymptotes. $\infty$ is a description of unbounded behavior, not a number.
A function has at most two horizontal asymptotes and may cross them; it can have infinitely many vertical asymptotes and never crosses one.
For a rational function: a zero of the denominator that cancels is a hole; one that does not cancel is a vertical asymptote.
End behavior by degree: $n<m \Rightarrow y=0$ ; $n=m \Rightarrow y=$ ratio of leading coefficients; $n=m+1 \Rightarrow$ oblique asymptote found by polynomial long division.
Core techniques: divide by the highest power of $x$ ; multiply by a conjugate for $\infty - \infty$ ; substitute $t = 1/x$ for $\infty \cdot 0$ ; use the squeeze theorem for bounded-times-decaying products.
Watch signs for $x \to -\infty$ : $\sqrt{x^2} = -x$ and $|x^3| = -x^3$ , which can give a different left-hand asymptote than the right-hand one.