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\documentclass[12pt,a4paper]{article}
\usepackage[latin1]{inputenc} 
\usepackage[T1]{fontenc}
\usepackage{amsfonts,amsmath,amssymb,graphicx,amsthm,mathtools,systeme}
\usepackage[tikz]{bclogo}
\usepackage{pifont} %bouni
\usepackage{fancybox} %pour faire l'encadrement
\newcommand{\isEquivTo}[1]{\underset{#1}{\sim}}
\begin{document}
\begin{center}
\section*{{\shadowbox{\bctakecare \;Développements limités et équivalences usuels à apprendre}}}
\end{center}
Les formules suivantes sont vrai {\bf \underline{uniquement} au voisinage de $0$}.
\begin{center}
\begin{tabular}[t]{|c|c@{\vrule depth 1.2ex height 3ex width 0mm \ }|}
\hline
\textbf{Les développements limités usuels en $0$}         & \textbf{Les équivalences usuels en $0$} \\
\hline
   (1) $e^u= 1 + u+\frac{u^2}{2}+\frac{u^3}{6}+ o(u^3)$         & (1) $e^u-1\isEquivTo{0} u$\\
     (2)   $\ln(1+u)= u - \frac{u^2}{2}+\frac{u^3}{3}+ o(u^3)$        &  (2) $\ln(1+u)\isEquivTo{0} u$\\
    (3)    $\sin  u = u - \frac{u^3}{6}+ o(u^3)$         & (3) $\sin  u\isEquivTo{0} u$\\
   (4)  $\cos u = 1 - \frac{u^2}{2}+o(u^3)$         & (4) $1-\cos u \isEquivTo{0}\frac{u^2}{2}$\\
  (5)   $\frac{1}{1-u}=1+u+u^2+u^3+ o(u^3)$      & (5) $(1+u)^{\alpha}-1\isEquivTo{0} \alpha u,\quad\quad\forall\,\alpha\in \mathbb{R}$\\
(6) $f(x)=f(0)+f'(0)x+\frac{f''(0)}{2}x^2+\frac{f^{3}(0)}{6}x^3+o(x^3)$ & En particulier on a: $\sqrt{1+u}-1\isEquivTo{0} \frac{u}{2}$\\
 (vrai si $f$ est $3$-fois dérivable en $0$)       &  (6) $\tan  u\isEquivTo{0} u$
   \\ \hline
\end{tabular}
\end{center}
\end{document}

How to align the equations in the table as follows?

enter image description here

1
  • 1
    unrelated to the question but \usepackage[latin1]{inputenc} are you sure your file is latin1 (ISO-8859-1) not UTF-8 ? certainly the example as posted to this site is (like all code here) UTF-8 so that line would need to be deleted. Nov 28 '21 at 15:30
3

A simple solution would be to put the preceding numbers in an own left-aligned column.

As shown in the modified version of your code below, the tabular declaration changes to \begin{tabular}[t]{|lc|lc@{\vrule depth 1.2ex height 3ex width 0mm \ }|}, including an extra left-aligned column before each of your two center-aligned columns.

In order to keep the headings in the first line of your table, you can use multicolumn to glue together the left- and center-aligned column: \multicolumn{2}{|c|}{\textbf{Les développements limités usuels en $0$}}.

Separate attention has to be paid for the lines of your table where you do not have a preceding number. I guess, the content of these lines shall not propagate to the column preserved for the numbering, so I just added an empty cell, as in the second-last line: & & En particulier on a: $\sqrt{1+u}-1\isEquivTo{0} \frac{u}{2}$\\.

\documentclass[12pt,a4paper]{article}
\usepackage[latin1]{inputenc} 
\usepackage[T1]{fontenc}
\usepackage{amsfonts,amsmath,amssymb,graphicx,amsthm,mathtools,systeme}
\usepackage[tikz]{bclogo}
\usepackage{pifont} %bouni
\usepackage{fancybox} %pour faire l'encadrement
\newcommand{\isEquivTo}[1]{\underset{#1}{\sim}}
\begin{document}
\begin{center}
\section*{{\shadowbox{\bctakecare \;Développements limités et équivalences usuels à apprendre}}}
\end{center}
Les formules suivantes sont vrai {\bf \underline{uniquement} au voisinage de $0$}.
\begin{center}
\begin{tabular}[t]{|lc|lc@{\vrule depth 1.2ex height 3ex width 0mm \ }|}
\hline
\multicolumn{2}{|c|}{\textbf{Les développements limités usuels en $0$}}         & \multicolumn{2}{c|}{\textbf{Les équivalences usuels en $0$}} \\
\hline
   (1) & $e^u= 1 + u+\frac{u^2}{2}+\frac{u^3}{6}+ o(u^3)$         & (1) & $e^u-1\isEquivTo{0} u$\\
     (2) &  $\ln(1+u)= u - \frac{u^2}{2}+\frac{u^3}{3}+ o(u^3)$        &  (2) & $\ln(1+u)\isEquivTo{0} u$\\
    (3) &   $\sin  u = u - \frac{u^3}{6}+ o(u^3)$         & (3) & $\sin  u\isEquivTo{0} u$\\
   (4) & $\cos u = 1 - \frac{u^2}{2}+o(u^3)$         & (4) & $1-\cos u \isEquivTo{0}\frac{u^2}{2}$\\
  (5) &  $\frac{1}{1-u}=1+u+u^2+u^3+ o(u^3)$      & (5) & $(1+u)^{\alpha}-1\isEquivTo{0} \alpha u,\quad\quad\forall\,\alpha\in \mathbb{R}$\\
(6) & $f(x)=f(0)+f'(0)x+\frac{f''(0)}{2}x^2+\frac{f^{3}(0)}{6}x^3+o(x^3)$ & & En particulier on a: $\sqrt{1+u}-1\isEquivTo{0} \frac{u}{2}$\\
 & (vrai si $f$ est $3$-fois dérivable en $0$)       &  (6) & $\tan  u\isEquivTo{0} u$
   \\ \hline
\end{tabular}
\end{center}
\end{document}

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