Infinite Limits and Limits at Infinity

Infinite limits and limits at infinity are the backbone of asymptotic analysis: determining the long-run behavior of algorithms (Big-$O$ notation in computer science is literally a statement about $\lim_{n\to\infty}$ ratios and bounds), the decay of probability tails in statistics (e.g. showing a density $\to 0$ as $x\to\pm\infty$ so that it can integrate to $1$), and the classification of singularities in complex analysis, where poles are precisely points at which $|f(z)|\to\infty$. In physics, limits at infinity describe boundary conditions at spatial infinity (e.g. potentials vanishing far from a charge), while infinite limits capture divergences that signal a breakdown of a model (e.g. an electric field blowing up at a point charge's location). In cryptography and complexity theory, statements like "the running time grows as $\Theta(n\log n)$" are limits-at-infinity statements dressed in different notation. Finally, this material directly feeds the Classification of Discontinuities: a function has an infinite discontinuity exactly when one of its one-sided limits at a point is $+\infty$ or $-\infty$, distinguishing this case sharply from removable and jump discontinuities.

Ask the tutor: Can you walk through how to determine, using only the $M$-$\delta$ definition, whether $\lim_{x\to 3} \frac{2}{(x-3)^3}$ is $+\infty$, $-\infty$, or does not exist as an infinite limit at all?

Look again at the classical definition: for every tolerance $\varepsilon>0$ around the target $L$, there is a window of radius $\delta$ around the input $a$ that lands inside the tolerance. Each of the new definitions is obtained by replacing one or both "closeness" requirements with an "eventually beyond a bound" requirement. If the target is infinite, "$|f(x)-L|<\varepsilon$" is replaced by "$f(x)>M$" — there is no finite target to be close to, only a bound to exceed, and $M$ can be taken arbitrarily large. If the input tends to infinity, "$0<|x-a|<\delta$" is replaced by "$x>N$" — there is no finite point to approach, only a threshold to exceed. This uniformity is the entire content of the topic: once you see that $M$ plays the role $\varepsilon$ used to play (a challenge to be met) and $N$ plays the role $\delta$ used to play (a response that meets it), all four definitions are the same skeleton with different flesh. The direction of the inequality with $M$ (whether $f(x)>M$ or $f(x)<M$) encodes $+\infty$ versus $-\infty$; whether the domain condition is $0<|x-a|<\delta$ or $x>N$ (or $x<N$) encodes whether the input approaches a finite point or $\pm\infty$. Geometrically, an infinite limit at $a$ produces a vertical asymptote at $x=a$: the graph shoots up or down without bound as $x\to a$. A finite limit at infinity produces a horizontal asymptote at height $L$: the graph flattens out approaching the line $y=L$. An infinite limit at infinity means the graph keeps climbing (or falling) forever, without leveling off — no horizontal asymptote, but a definite long-run trend.

Example 1 (an infinite limit at a finite point, verified directly from the definition). Let $f(x) = \dfrac{3}{(x-1)^2}$. Claim: $\displaystyle\lim_{x\to 1} f(x) = \infty$. Proof from the definition. Let $M\in\mathbb R$ be given. We must find $\delta>0$ such that $0<|x-1|<\delta \implies \dfrac{3}{(x-1)^2} > M$. If $M\le 0$, any $\delta>0$ works since $\dfrac{3}{(x-1)^2}>0\ge M$ automatically for $x\ne 1$. So assume $M>0$. We want

So choose $\delta = \sqrt{3/M}$. Then for $0<|x-1|<\delta$, reversing the (valid, since both sides are positive) chain of equivalences above gives $f(x) > M$. Since $M$ was arbitrary, $\lim_{x\to1} f(x)=\infty$, confirmed directly, not just by "plugging in." Numerically: to force $f(x) > 10{,}000$, we need $\delta = \sqrt{3/10000} = \sqrt{0.0003}\approx 0.01732$; indeed at $x=1.01$, $f(1.01) = 3/(0.01)^2 = 30{,}000 > 10{,}000$. ✓.

Example 2 (a finite limit at infinity and an infinite limit at infinity, combined via the algebra theorem). Let

We compute $\lim_{x\to\infty} g(x)$ and $\lim_{x\to\infty} h(x)$, then combine them via part (a) of the Theorem: Algebra of infinite limits and limits at infinity. Step 1: $\lim_{x\to\infty} g(x)$. Divide numerator and denominator by $x^2$ (valid for $x\ne 0$):

We use (without reproof, as it is the elementary case $L=0$ analog of part (c) applied to $f(x)=x$, or direct verification): $\lim_{x\to\infty} 1/x = 0$ and $\lim_{x\to\infty} 1/x^2=0$. Both follow directly from the $N$–$\varepsilon$ definition: given $\varepsilon>0$, take $N=1/\varepsilon$; then $x>N \implies 0<1/x<\varepsilon$, and similarly $x>N=\max(1,1/\sqrt\varepsilon)\implies 1/x^2<\varepsilon$. By the ordinary (finite) algebra of limits — sums and quotients of finite limits, applicable here since all four numerator/denominator pieces have finite limits and the denominator's limit $2\neq 0$ — we get

So $g$ has a horizontal asymptote $y=7/2$ as $x\to\infty$. Let's double check numerically: $g(1000) = \frac{7,000,000 - 2000+5}{2,000,000+3} = \frac{6{,}998{,}005}{2{,}000{,}003} \approx 3.49900$, close to $3.5$. ✓ Step 2: $\lim_{x\to\infty} h(x) = \infty$. Write $h(x) = x^3(1 - 100/x^2)$. Given $M\in\mathbb R$, first note $1-100/x^2 \to 1$ as $x\to\infty$ (again by the $1/x^2\to0$ fact above and finite-limit algebra), so there is $N_1$ with $x>N_1 \implies 1-100/x^2 > 1/2$. Also $x^3 \to \infty$ (direct: given any bound $M'$, take $N=\max(1,M'^{1/3})$; then $x>N\implies x^3>M'$). By part (b) of the theorem (product of an infinite limit with a finite positive limit $1/2$... more precisely we apply it with the roles $f(x)=x^3\to\infty$, $g(x) = 1-100/x^2 \to 1 > 0$), $\lim_{x\to\infty} h(x) = \infty$. Numerically: $h(20) = 8000 - 2000 = 6000$; $h(100)=1{,}000{,}000-10{,}000 = 990{,}000$, confirming rapid, unbounded growth despite the "$-100x$" correction term. ✓ Step 3: combine via part (a). Consider $k(x) = h(x) + g(x) = x^3-100x + \dfrac{7x^2-2x+5}{2x^2+3}$. Since $\lim h=\infty$ and $\lim g = 7/2$ (finite), part (a) of the theorem gives immediately, with no further computation,

This illustrates the payoff of the algebra theorem: once the two pieces' limits are known, the combination's limit requires no new $\varepsilon$-$M$-$N$ argument.