Math Analysis 1

Limits in depth: the number e, the hierarchy of infinities, little-o and the tables to remember

Existence of the number e proved, from sequences to functions, hierarchy of infinities, little-o rules and substitution of infinitesimals, tables of standard limits, equivalences, Taylor expansions and indeterminate forms.

ielettronica 25 October 2026 8 min read 0 comments

This page goes deeper into the lessons on limits and standard limits. It proves step by step the results that are usually used without proof: the existence of the number $e$ , the passage from sequences to functions, the hierarchy of infinities, the rules of little-o notation and the principle of substituting infinitesimals. At the end are the tables to remember: standard limits, equivalences, the most used Taylor expansions, indeterminate forms with the right technique for each. The geometric proof of $\sin x/x \to 1$ is in a separate in-depth page, linked from the lesson.

The number e exists

Theorem. The sequence $a_n = \left(1 + \frac{1}{n}\right)^n$ is increasing and bounded, $2 \le a_n \lt 3$ ; so it has a limit, called $e$ , with $2 \lt e \le 3$ .

Proof. By the binomial formula:

$a_n = \sum_{k=0}^{n}\binom{n}{k}\frac{1}{n^k} = \sum_{k=0}^{n}\frac{1}{k!}\cdot\frac{n(n - 1)\cdots(n - k + 1)}{n^k} = \sum_{k=0}^{n}\frac{1}{k!}\left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right)\cdots\left(1 - \frac{k - 1}{n}\right)$

Increasing. Going from $n$ to $n + 1$ , each factor $1 - j/n$ becomes $1 - j/(n + 1)$ , which is larger; so each term of the sum increases. In addition a positive term is added (the one with $k = n + 1$ ). Hence $a_{n+1} \gt a_n$ .

Bounded. Every product of factors less than 1 is less than 1, so $a_n \lt \sum_{k=0}^{n}\frac{1}{k!}$ . For $k \ge 1$ , $k! = 1\cdot 2\cdot 3\cdots k \ge 2^{k-1}$ , so

$a_n \lt 1 + \sum_{k=1}^{n}\frac{1}{2^{k-1}} \lt 1 + \frac{1}{1 - 1/2} = 3$

An increasing sequence bounded above has a limit (monotone sequence theorem, lesson on sequences). Since $a_1 = 2$ and the sequence increases, the limit is greater than 2. ∎

The value is $e = 2.718\,281\,8\ldots$ ; the same proof also shows that $e = \sum_{k \ge 0} 1/k!$ , a series that converges very fast.

From sequences to functions

Theorem. $\displaystyle\lim_{x\to\pm\infty}\left(1 + \frac{1}{x}\right)^x = e$ and $\displaystyle\lim_{x\to 0}(1 + x)^{1/x} = e$ .

Proof. *Towards $+\infty$ .* For $x \ge 1$ let $n = \lfloor x\rfloor$ be the integer part, so $n \le x \lt n + 1$ . Then $\frac{1}{n + 1} \lt \frac{1}{x} \le \frac{1}{n}$ , and since base and exponent both increase:

$\left(1 + \frac{1}{n + 1}\right)^{n} \le \left(1 + \frac{1}{x}\right)^{x} \le \left(1 + \frac{1}{n}\right)^{n + 1}$

On the left there is $a_{n+1}/(1 + \frac{1}{n + 1}) \to e/1$ ; on the right $a_n(1 + \frac{1}{n}) \to e\cdot 1$ . By the squeeze theorem the middle term also tends to $e$ .

*Towards $-\infty$ .* Set $x = -(t + 1)$ , with $t \to +\infty$ :

$\left(1 - \frac{1}{t + 1}\right)^{-(t+1)} = \left(\frac{t + 1}{t}\right)^{t + 1} = \left(1 + \frac{1}{t}\right)^{t}\left(1 + \frac{1}{t}\right) \to e\cdot 1$

*Towards 0.* With $x = 1/t$ : if $x \to 0^+$ , $t \to +\infty$ ; if $x \to 0^-$ , $t \to -\infty$ . In both cases $(1 + x)^{1/x} = (1 + 1/t)^t \to e$ . ∎

Graph of (1 + 1/x) to the power x — The function tends to e both as x goes to plus infinity, from below, and as x goes to minus infinity, from above. Between minus 1 and 0 it is not defined. The dots are the values of the sequence for integer n.

From here follow, as the lesson shows, $\ln(1 + x)/x \to 1$ , $(e^x - 1)/x \to 1$ , $(a^x - 1)/x \to \ln a$ and $((1 + x)^\alpha - 1)/x \to \alpha$ . A useful corollary: $\left(1 + \frac{a}{x}\right)^{bx} \to e^{ab}$ , obtained by setting $x = a\,t$ .

The hierarchy of infinities, proved

Theorem. For every $\alpha \gt 0$ and $a \gt 1$ :

$\text{(i)}\ \lim_{x\to+\infty}\frac{\ln x}{x^\alpha} = 0,\qquad \text{(ii)}\ \lim_{x\to+\infty}\frac{x^\alpha}{a^x} = 0,\qquad \text{(iii)}\ \lim_{n\to\infty}\frac{a^n}{n!} = 0$

Proof. *(i)* First the case $\alpha = 1$ . For $x \ge 1$ , since $1/t \le 1/\sqrt t$ for $t \ge 1$ :

$\ln x = \int_1^x\frac{dt}{t} \le \int_1^x\frac{dt}{\sqrt t} = 2(\sqrt x - 1) \lt 2\sqrt x$

so $0 \le \frac{\ln x}{x} \lt \frac{2}{\sqrt x} \to 0$ . For any $\alpha$ write $\frac{\ln x}{x^\alpha} = \frac{1}{\alpha}\cdot\frac{\ln(x^\alpha)}{x^\alpha}$ and set $y = x^\alpha \to +\infty$ : it is $\frac{1}{\alpha}\cdot\frac{\ln y}{y} \to 0$ .

*(ii)* Set $x = \log_a y$ , i.e. $y = a^x \to +\infty$ : $\frac{x^\alpha}{a^x} = \frac{(\ln y/\ln a)^\alpha}{y} = \frac{1}{(\ln a)^\alpha}\left(\frac{\ln y}{y^{1/\alpha}}\right)^\alpha$ , which tends to 0 by (i) with exponent $1/\alpha$ .

*(iii)* Let $N$ be an integer greater than $2a$ . For $n \gt N$ , every factor $a/k$ with $k \gt N$ is less than $1/2$ :

$\frac{a^n}{n!} = \frac{a^N}{N!}\cdot\frac{a}{N + 1}\cdots\frac{a}{n} \lt \frac{a^N}{N!}\left(\frac{1}{2}\right)^{n - N} \to 0$

∎

The first one is what is used for $x\ln x \to 0$ as $x \to 0^+$ (set $x = 1/t$ ), and hence for $x^x = e^{x\ln x} \to 1$ .

Ratios between infinities of different order — The blue curve is x cubed over e to the x: it rises to about 1.34 at x equal to 3 and then collapses to zero, because the exponential wins. The orange curve is log x over the square root of x: it peaks at x equal to e squared and then slowly tends to zero, because the power wins over the logarithm.

Little-o notation and the substitution principle

Definition. As $x \to x_0$ one writes $f = o(g)$ (“f is little-o of g”) if $f(x)/g(x) \to 0$ : $f$ is negligible compared with $g$ . For example $x^3 = o(x^2)$ as $x \to 0$ , but $x^2 = o(x^3)$ as $x \to \infty$ . The link with equivalences: $f \sim g$ if and only if $f = g + o(g)$ .

Calculation rules (as $x \to 0$ , with $m, n$ positive), all immediate consequences of the definition:

$o(x^n) \pm o(x^n) = o(x^n)$ : the difference of two little-o terms is not zero;
$c\cdot o(x^n) = o(x^n)$ for every constant $c \ne 0$ ;
$x^m\cdot o(x^n) = o(x^{m+n})$ and $o(x^m)\cdot o(x^n) = o(x^{m+n})$ ;
$o(x^n)/x^m = o(x^{n-m})$ for $m \le n$ ;
if $m \gt n$ , $o(x^m) = o(x^n)$ and $x^m = o(x^n)$ : in a sum the lowest power “wins”.

Theorem (substitution of infinitesimals). If $f \sim f_1$ and $g \sim g_1$ , then $\lim\frac{f}{g} = \lim\frac{f_1}{g_1}$ , in the sense that one of the two limits exists if and only if the other does, and they are equal. The same holds for products.

Proof. $\frac{f}{g} = \frac{f}{f_1}\cdot\frac{f_1}{g_1}\cdot\frac{g_1}{g}$ , and the first and last factors tend to 1. ∎

In sums, on the other hand, substitution is allowed only if the leading terms do not cancel: $f + g \sim f_1 + g_1$ if $f_1 + g_1$ is not “smaller” than the individual terms. When they cancel, as in $\tan x - \sin x$ , the next-order terms must be kept, that is, Taylor expansions must be used with the remainder $o(\cdot)$ written explicitly. This rule avoids 90 % of the mistakes in written exams.

Infinitesimals of different order near zero — The curves x, x squared and x cubed: near the origin each power is negligible compared with the previous one. This is the graphical meaning of x cubed being little-o of x squared as x tends to zero.

Tables to remember

Standard limits as $x \to 0$ and equivalences

Limit	Value	Equivalence as $x \to 0$
$\dfrac{\sin x}{x}$ , $\dfrac{\tan x}{x}$ , $\dfrac{\arcsin x}{x}$ , $\dfrac{\arctan x}{x}$	1	$\sin x \sim \tan x \sim \arcsin x \sim \arctan x \sim x$
$\dfrac{1 - \cos x}{x^2}$	$\frac{1}{2}$	$1 - \cos x \sim \frac{x^2}{2}$
$\dfrac{\ln(1 + x)}{x}$	1	$\ln(1 + x) \sim x$
$\dfrac{\log_a(1 + x)}{x}$	$\frac{1}{\ln a}$	$\log_a(1 + x) \sim \frac{x}{\ln a}$
$\dfrac{e^x - 1}{x}$	1	$e^x - 1 \sim x$
$\dfrac{a^x - 1}{x}$	$\ln a$	$a^x - 1 \sim x\ln a$
$\dfrac{(1 + x)^\alpha - 1}{x}$	$\alpha$	$(1 + x)^\alpha - 1 \sim \alpha x$
$\dfrac{\sinh x}{x}$	1	$\sinh x \sim x$
$\dfrac{\cosh x - 1}{x^2}$	$\frac{1}{2}$	$\cosh x - 1 \sim \frac{x^2}{2}$
$(1 + x)^{1/x}$	$e$	$\ln f \sim f - 1$ if $f \to 1$

Each row also holds with $x$ replaced by a function $h(x) \to 0$ : for example $\sin(3x^2) \sim 3x^2$ and $e^{\sin x} - 1 \sim \sin x \sim x$ .

At infinity ( $\alpha \gt 0$ , $a \gt 1$ , $b$ real)

Limit	Value
$\left(1 + \dfrac{b}{x}\right)^x$ as $x \to \pm\infty$	$e^b$
$\dfrac{\ln x}{x^\alpha}$ , $\dfrac{x^\alpha}{a^x}$ , $\dfrac{a^n}{n!}$ , $\dfrac{n!}{n^n}$	0
$x^\alpha\ln x$ and $x^x$ as $x \to 0^+$	0 and 1
$\sqrt[n]{n}$ , $\sqrt[n]{a}$ as $n \to \infty$	1
$n\left(\sqrt[n]{a} - 1\right)$	$\ln a$
$\dfrac{\sin x}{x}$ as $x \to \infty$	0

Order of growth, from slowest to fastest: $\ln x \ll x^\alpha \ll a^x \ll x! \ll x^x$ (the symbol $\ll$ means “negligible compared with”).

Maclaurin expansions most used in limit exercises

Function	Expansion as $x \to 0$
$e^x$	$1 + x + \frac{x^2}{2} + \frac{x^3}{6} + o(x^3)$
$\sin x$	$x - \frac{x^3}{6} + o(x^4)$
$\cos x$	$1 - \frac{x^2}{2} + \frac{x^4}{24} + o(x^5)$
$\tan x$	$x + \frac{x^3}{3} + o(x^4)$
$\ln(1 + x)$	$x - \frac{x^2}{2} + \frac{x^3}{3} + o(x^3)$
$(1 + x)^\alpha$	$1 + \alpha x + \frac{\alpha(\alpha - 1)}{2}x^2 + o(x^2)$
$\sqrt{1 + x}$	$1 + \frac{x}{2} - \frac{x^2}{8} + o(x^2)$
$\dfrac{1}{1 + x}$	$1 - x + x^2 - x^3 + o(x^3)$
$\arctan x$	$x - \frac{x^3}{3} + o(x^4)$
$\arcsin x$	$x + \frac{x^3}{6} + o(x^4)$

Indeterminate forms and technique

Form	Typical example	Technique
$\frac{0}{0}$	$\frac{\sin 3x}{\ln(1 + 2x)}$	equivalences; if terms cancel, Taylor
$\frac{\infty}{\infty}$	$\frac{3x^2 - x}{2x^2 + 5}$	factor out the dominant power or infinity
$\infty - \infty$	$\sqrt{x^2 + x} - x$	rationalise or factor out
$0\cdot\infty$	$x\ln x$	rewrite as a quotient, hierarchy of infinities
$1^\infty$	$\left(\frac{x - 1}{x + 2}\right)^x$	$f^g = e^{g\ln f}$ and $\ln f \sim f - 1$
$0^0$ , $\infty^0$	$x^x$ , $x^{1/x}$	$f^g = e^{g\ln f}$ , then the limit of the exponent

The forms $\frac{1}{0}$ are not indeterminate: they give $\pm\infty$ , and the sign of the denominator matters (right and left limits). Neither are $0^\infty$ and $\infty^\infty$ : they are 0 and $\infty$ .

Exercises on all these techniques, from basic to exam level, are on the exercise page linked at the top.

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The number e exists

From sequences to functions

The hierarchy of infinities, proved

Little-o notation and the substitution principle

Tables to remember

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