缩放包含合并单元格的表格以适合整个页面

Question 1

无需使用adjustbox环境。最好使用\scriptsize指令并降低的值\setcellgapes。

在下面的代码中，我还清理并简化了代码的各个部分。最大的变化应该在最后一列中显而易见，因为用替换了\rm R。\mathrm{R}我还为用于表示开区间或半开区间的左侧边缘的\mathopen情况提供了前缀。（默认情况下，的数学代码是；这种区别会影响符号的解释方式：它可以是二元运算符或一元运算符。）]]mathclose-

\documentclass[12pt,a4paper]{report}
 %This code was written by Mohcine
\usepackage[margin=1cm]{geometry}
\usepackage[french]{babel}
\usepackage{fontspec}
\usepackage{makecell}
\usepackage{rotating}
\setcellgapes{4pt}
\usepackage{amsthm,amssymb,amsfonts,mathtools}
\DeclareMathOperator{\arctanh}{arctanh}
\DeclareMathOperator{\ch}{ch}
\DeclareMathOperator{\sh}{sh}
\newcommand{\eqx}{\mathrel{\underset{\scriptscriptstyle x\to0}{=}}}

\begin{document}
\begin{sidewaystable}
\centering
\makegapedcells
\scriptsize
\begin{tabular}{ | c | *{4}{>{$\displaystyle}c<{$}|}}
\hline
Mohcine    
&   \multicolumn{2}{c|}{Développement limité en $0$ de fonctions usuelles }    & 
\text{Equivalents} & 
\text{Développements en série entière usuels} \\ 
\hline
Théorème &   
\multicolumn{2}{c|}{$\displaystyle f(x)\underset{\scriptscriptstyle x\to a}{=} \sum_{k=0}^{n}a_k(x-1)^{k}+o\bigl((x-a)^{n}\bigr)$}
&  &   
f(x)=\sum_{n=0}^{+\infty} a_n(x)^{n} \\ 
\hline
Formulaire        &   
\multicolumn{2}{c|}{$\displaystyle f(x)\underset{\scriptscriptstyle x\to a}{=}\sum_{k=0}^{n}a_k(x-1)^{k}+o\bigl((x-a)^{n}\bigr)$}
& &  
f(x)=\sum_{n\geq 0} a_k(x-1)^{k} \\ 
\hline
Ordre   &   DL_{n}(0)   &   DL_{5}(0)   &  & \\ 
\hline
$\dfrac{1}{1-x}$   &  
\frac{1}{1-x}\eqx \sum_{k=0}^{n} {x^k} +o\bigl(x^{n}\bigr) &  
\frac{1}{1-x}\eqx 1+x+x^2+x^3+x^4+x^5+o\bigl(x^{5}\bigr) &  & 
\mathrm{R}=1,\ \forall x\in \mathopen]-1;1[ \quad 
\frac{1}{1-x}=\sum_{n=0}^{+\infty} {x^n} \\   
\hline
$\dfrac{1}{1+x}$   &  
\frac{1}{1+x}\eqx \sum_{k=0}^{n} (-1)^{k}{x^k} +o\bigl(x^{n}\bigr) &  
\frac{1}{1+x}\eqx  1-x+x^2-x^3+x^4-x^5+o\bigl(x^{5}\bigr) &  & 
\mathrm{R}=1,\ \forall x\in \mathopen]-1;1[\quad 
\frac{1}{1+x}=\sum_{n=0}^{+\infty} (-1)^{n}{x^n} \\   
\hline
$-\ln(1-x)$   &   
-\ln(1-x) \eqx \sum_{k=0}^{n} \frac{x^{k+1}}{k+1} +o\bigl(x^{n}\bigr)  
&  
-\ln(1-x)\eqx  x+{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}+{\frac {x^{4}}{4}}+{\frac {x^{5}}{5}}+o\bigl(x^{5}\bigr) & & 
\mathrm{R}=1,\ \forall x\in [-1;1[\quad 
-\ln(1-x)=\sum_{n=0}^{+\infty} \frac{x^{n+1}}{n+1} \\   
\hline
$\ln(1+x)$   &   
\ln(1+x)\eqx \sum_{k=0}^{n} (-1)^{k}\frac{x^{k+1}}{k+1} +o\bigl(x^{n}\bigr)   &  
\ln(1+x)\eqx  x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-{\frac {x^{4}}{4}}+{\frac {x^{5}}{5}}+o\bigl(x^{5}\bigr)  &  & 
\mathrm{R}=1,\ \forall x\in \mathopen]-1;1]\quad 
\ln(1+x)=\sum_{n=0}^{+\infty}(-1)^{n}\frac{x^{n+1}}{n+1}\\   
\hline
& \multicolumn{4}{c|}{On en déduit la somme de la série harmonique alternée $\displaystyle\sum_{n=1}^{+\infty} \frac{(-1)^{n-1}}{n}=\ln(2)$} \\  
\hline
$\exp(x)$   & 
\mathrm{e}^{x}\eqx \sum _{k=0}^{n }{\frac {x^{k}}{k!}}+o\bigl(x^{n}\bigr)  &
\mathrm{e}^{x}\eqx  1+{\frac {x}{1!}}+{\frac {x^{2}}{2!}}+{\frac {x^{3}}{3!}}+{\frac {x^{4}}{4!}}+{\frac {x^{5}}{5!}}+o\bigl(x^{5}\bigr) &  & 
\mathrm{R}=+\infty,\ \forall x\in \mathbb{R}\quad 
\mathrm{e}^{x}=\sum _{n=0}^{+\infty }{\frac {x^{n}}{n!}} \\   
\hline
$\ch x$   & 
\ch x\eqx \sum _{k=0}^{n }{\frac {x^{2k}}{(2k)!}}+o\bigl(x^{2n}\bigr) & 
\ch x\eqx  1+{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}+o\bigl(x^{5}\bigr) &  & 
\mathrm{R}=+\infty,\ \forall x\in \mathbb{R},\quad 
\ch x=\sum _{n=0}^{+{\infty }}{\frac {x^{2n}}{(2n)!}} \\   
\hline
$\sh x$   & 
\sh x\eqx \sum _{k=0}^{n}\frac {x^{2k+1}}{(2k+1)!} +o\bigl(x^{2n+1}\bigr) &
\sh x\eqx  x+{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}+o\bigl(x^{5}\bigr) &  &
\mathrm{R}=+\infty,\ \forall x\in \mathbb{R},\quad 
\sh x=\sum _{n=0}^{+{\infty }}{\frac {x^{2n+1}}{(2n+1)!}} \\   
\hline
$\cos x$ & 
\cos x\eqx \sum _{k=0}^{n }{(-1)^k\frac {x^{2k}}{(2k)!}}+o\bigl(x^{2n}\bigr)  & 
\cos x\eqx 1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}+o\bigl(x^{5}\bigr) &
\begin{aligned}[c] 
   \cos x-1 &\underset{\scriptscriptstyle x\to0}{\sim} -{\frac {x^{2}}{2!}}\\
   1-\cos x &\underset{\scriptscriptstyle x\to0}{\sim} {\frac {x^{2}}{2!}} 
\end{aligned} &
\mathrm{R}=+\infty,\ \forall x\in \mathbb{R},\quad 
\cos x=\sum _{n=0}^{+{\infty }}{(-1)^n\frac {x^{2n}}{(2n)!}} \\   
\hline
$\sin x$   & 
\sin x\eqx  \sum _{k=0}^{n }{(-1)^k\frac {x^{2k+1}}{(2k+1)!}}+o\bigl(x^{2n+1}\bigr)   & 
\sin x\eqx   x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}+o\bigl(x^{5}\bigr) &  &
\mathrm{R}=+\infty,\ \forall x\in \mathbb{R},\quad 
\sin x=\sum _{n=0}^{+{\infty }}{(-1)^n\frac {x^{2n+1}}{(2n+1)!}} \\   
\hline
$\arctan x$   & \arctan x\eqx  \sum _{k=0}^{n }{(-1)^k\frac {x^{2k+1}}{(2k+1)}}+o\bigl(x^{2n+1}\bigr) & 
\arctan x\eqx   x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}+o\bigl(x^{5}\bigr) &  & 
\mathrm{R}=1,\  \forall x\in [-1;1](x\not =\pm i),\quad 
\arctan x=\sum _{n=0}^{+{\infty }}{(-1)^n\frac {x^{2n+1}}{(2n+1)}} \\   
\hline
$\arctanh x$   &
\arctanh x\eqx \sum _{k=0}^{n }{\frac {x^{2k+1}}{(2k+1)}}+o\bigl(x^{2n+1}\bigr)  &  
\arctanh x\eqx  x+{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}+o\bigl(x^{5}\bigr) &  & 
\mathrm{R}=1,\  \forall x\in \mathopen]-1;1[,\quad 
\arctanh x=\sum _{n=0}^{+{\infty }}{\frac {x^{2n+1}}{(2n+1)}} \\   
\hline
\end{tabular}
\end{sidewaystable}
\thispagestyle{empty}
\end{document}

Answer

无需使用adjustbox环境。最好使用\scriptsize指令并降低的值\setcellgapes。

在下面的代码中，我还清理并简化了代码的各个部分。最大的变化应该在最后一列中显而易见，因为用替换了\rm R。\mathrm{R}我还为用于表示开区间或半开区间的左侧边缘的\mathopen情况提供了前缀。（默认情况下，的数学代码是；这种区别会影响符号的解释方式：它可以是二元运算符或一元运算符。）]]mathclose-

\documentclass[12pt,a4paper]{report}
 %This code was written by Mohcine
\usepackage[margin=1cm]{geometry}
\usepackage[french]{babel}
\usepackage{fontspec}
\usepackage{makecell}
\usepackage{rotating}
\setcellgapes{4pt}
\usepackage{amsthm,amssymb,amsfonts,mathtools}
\DeclareMathOperator{\arctanh}{arctanh}
\DeclareMathOperator{\ch}{ch}
\DeclareMathOperator{\sh}{sh}
\newcommand{\eqx}{\mathrel{\underset{\scriptscriptstyle x\to0}{=}}}

\begin{document}
\begin{sidewaystable}
\centering
\makegapedcells
\scriptsize
\begin{tabular}{ | c | *{4}{>{$\displaystyle}c<{$}|}}
\hline
Mohcine    
&   \multicolumn{2}{c|}{Développement limité en $0$ de fonctions usuelles }    & 
\text{Equivalents} & 
\text{Développements en série entière usuels} \\ 
\hline
Théorème &   
\multicolumn{2}{c|}{$\displaystyle f(x)\underset{\scriptscriptstyle x\to a}{=} \sum_{k=0}^{n}a_k(x-1)^{k}+o\bigl((x-a)^{n}\bigr)$}
&  &   
f(x)=\sum_{n=0}^{+\infty} a_n(x)^{n} \\ 
\hline
Formulaire        &   
\multicolumn{2}{c|}{$\displaystyle f(x)\underset{\scriptscriptstyle x\to a}{=}\sum_{k=0}^{n}a_k(x-1)^{k}+o\bigl((x-a)^{n}\bigr)$}
& &  
f(x)=\sum_{n\geq 0} a_k(x-1)^{k} \\ 
\hline
Ordre   &   DL_{n}(0)   &   DL_{5}(0)   &  & \\ 
\hline
$\dfrac{1}{1-x}$   &  
\frac{1}{1-x}\eqx \sum_{k=0}^{n} {x^k} +o\bigl(x^{n}\bigr) &  
\frac{1}{1-x}\eqx 1+x+x^2+x^3+x^4+x^5+o\bigl(x^{5}\bigr) &  & 
\mathrm{R}=1,\ \forall x\in \mathopen]-1;1[ \quad 
\frac{1}{1-x}=\sum_{n=0}^{+\infty} {x^n} \\   
\hline
$\dfrac{1}{1+x}$   &  
\frac{1}{1+x}\eqx \sum_{k=0}^{n} (-1)^{k}{x^k} +o\bigl(x^{n}\bigr) &  
\frac{1}{1+x}\eqx  1-x+x^2-x^3+x^4-x^5+o\bigl(x^{5}\bigr) &  & 
\mathrm{R}=1,\ \forall x\in \mathopen]-1;1[\quad 
\frac{1}{1+x}=\sum_{n=0}^{+\infty} (-1)^{n}{x^n} \\   
\hline
$-\ln(1-x)$   &   
-\ln(1-x) \eqx \sum_{k=0}^{n} \frac{x^{k+1}}{k+1} +o\bigl(x^{n}\bigr)  
&  
-\ln(1-x)\eqx  x+{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}+{\frac {x^{4}}{4}}+{\frac {x^{5}}{5}}+o\bigl(x^{5}\bigr) & & 
\mathrm{R}=1,\ \forall x\in [-1;1[\quad 
-\ln(1-x)=\sum_{n=0}^{+\infty} \frac{x^{n+1}}{n+1} \\   
\hline
$\ln(1+x)$   &   
\ln(1+x)\eqx \sum_{k=0}^{n} (-1)^{k}\frac{x^{k+1}}{k+1} +o\bigl(x^{n}\bigr)   &  
\ln(1+x)\eqx  x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-{\frac {x^{4}}{4}}+{\frac {x^{5}}{5}}+o\bigl(x^{5}\bigr)  &  & 
\mathrm{R}=1,\ \forall x\in \mathopen]-1;1]\quad 
\ln(1+x)=\sum_{n=0}^{+\infty}(-1)^{n}\frac{x^{n+1}}{n+1}\\   
\hline
& \multicolumn{4}{c|}{On en déduit la somme de la série harmonique alternée $\displaystyle\sum_{n=1}^{+\infty} \frac{(-1)^{n-1}}{n}=\ln(2)$} \\  
\hline
$\exp(x)$   & 
\mathrm{e}^{x}\eqx \sum _{k=0}^{n }{\frac {x^{k}}{k!}}+o\bigl(x^{n}\bigr)  &
\mathrm{e}^{x}\eqx  1+{\frac {x}{1!}}+{\frac {x^{2}}{2!}}+{\frac {x^{3}}{3!}}+{\frac {x^{4}}{4!}}+{\frac {x^{5}}{5!}}+o\bigl(x^{5}\bigr) &  & 
\mathrm{R}=+\infty,\ \forall x\in \mathbb{R}\quad 
\mathrm{e}^{x}=\sum _{n=0}^{+\infty }{\frac {x^{n}}{n!}} \\   
\hline
$\ch x$   & 
\ch x\eqx \sum _{k=0}^{n }{\frac {x^{2k}}{(2k)!}}+o\bigl(x^{2n}\bigr) & 
\ch x\eqx  1+{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}+o\bigl(x^{5}\bigr) &  & 
\mathrm{R}=+\infty,\ \forall x\in \mathbb{R},\quad 
\ch x=\sum _{n=0}^{+{\infty }}{\frac {x^{2n}}{(2n)!}} \\   
\hline
$\sh x$   & 
\sh x\eqx \sum _{k=0}^{n}\frac {x^{2k+1}}{(2k+1)!} +o\bigl(x^{2n+1}\bigr) &
\sh x\eqx  x+{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}+o\bigl(x^{5}\bigr) &  &
\mathrm{R}=+\infty,\ \forall x\in \mathbb{R},\quad 
\sh x=\sum _{n=0}^{+{\infty }}{\frac {x^{2n+1}}{(2n+1)!}} \\   
\hline
$\cos x$ & 
\cos x\eqx \sum _{k=0}^{n }{(-1)^k\frac {x^{2k}}{(2k)!}}+o\bigl(x^{2n}\bigr)  & 
\cos x\eqx 1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}+o\bigl(x^{5}\bigr) &
\begin{aligned}[c] 
   \cos x-1 &\underset{\scriptscriptstyle x\to0}{\sim} -{\frac {x^{2}}{2!}}\\
   1-\cos x &\underset{\scriptscriptstyle x\to0}{\sim} {\frac {x^{2}}{2!}} 
\end{aligned} &
\mathrm{R}=+\infty,\ \forall x\in \mathbb{R},\quad 
\cos x=\sum _{n=0}^{+{\infty }}{(-1)^n\frac {x^{2n}}{(2n)!}} \\   
\hline
$\sin x$   & 
\sin x\eqx  \sum _{k=0}^{n }{(-1)^k\frac {x^{2k+1}}{(2k+1)!}}+o\bigl(x^{2n+1}\bigr)   & 
\sin x\eqx   x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}+o\bigl(x^{5}\bigr) &  &
\mathrm{R}=+\infty,\ \forall x\in \mathbb{R},\quad 
\sin x=\sum _{n=0}^{+{\infty }}{(-1)^n\frac {x^{2n+1}}{(2n+1)!}} \\   
\hline
$\arctan x$   & \arctan x\eqx  \sum _{k=0}^{n }{(-1)^k\frac {x^{2k+1}}{(2k+1)}}+o\bigl(x^{2n+1}\bigr) & 
\arctan x\eqx   x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}+o\bigl(x^{5}\bigr) &  & 
\mathrm{R}=1,\  \forall x\in [-1;1](x\not =\pm i),\quad 
\arctan x=\sum _{n=0}^{+{\infty }}{(-1)^n\frac {x^{2n+1}}{(2n+1)}} \\   
\hline
$\arctanh x$   &
\arctanh x\eqx \sum _{k=0}^{n }{\frac {x^{2k+1}}{(2k+1)}}+o\bigl(x^{2n+1}\bigr)  &  
\arctanh x\eqx  x+{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}+o\bigl(x^{5}\bigr) &  & 
\mathrm{R}=1,\  \forall x\in \mathopen]-1;1[,\quad 
\arctanh x=\sum _{n=0}^{+{\infty }}{\frac {x^{2n+1}}{(2n+1)}} \\   
\hline
\end{tabular}
\end{sidewaystable}
\thispagestyle{empty}
\end{document}

Question 2

按照此表格编码在您的文档中：我刚刚使用sidewaystable包rotating

\begin{sidewaystable}
  %\resizebox{\textwidth}{!}{%
    \begin{adjustbox}{width=1\textwidth}
\makegapedcells
    \begin{tabular}{ | c | *{5}{>{$\displaystyle}c<{$}|}}
    \hline
    Mohcine    &   \multicolumn{2}{c|}{Développement limité en 0 de fonctions usuelles }    & \text{ Equivalents} & \text{ Développements en série entière usuels} \\ \hline
            Théorème         &   \multicolumn{2}{c|}{$\displaystyle
            f(x)\underset{{x\to a}}{=}\sum_{k=0}^{n}a_k(x-1)^{k}+o((x-a)^{n})$}
                                                &  &   f(x)=\sum_{n=0}^{+\infty} a_n(x)^{n}     \\
    \hline
Formulaire        &   \multicolumn{2}{c|}{$\displaystyle
            f(x)\underset{{x\to a}}{=}\sum_{k=0}^{n}a_k(x-1)^{k}+o((x-a)^{n})$}
                                               & &  f(x)=\sum_{n\geq 0} a_k(x-1)^{k}    \\
    \hline
Ordre   &   DL_{n}(0)   &   DL_{5}(0)           &  &                                      \\
    \hline
        $\dfrac{1}{1-x}$   &  \dfrac{1}{1-x}\underset{{x\to 0}}{=}\sum_{k=0}^{n} {x^k} +o(x^{n})
                        &  \dfrac{1}{1-x}\underset{{x\to 0}}{=}1+x+x^2+x^3+x^4+x^5+o(x^{5}) &  & \rm{R}=1,\ \forall x\in ]-1;1[\quad \dfrac{1}{1-x}=\sum_{n=0}^{+\infty} {x^n}       \\
    \hline
            $\dfrac{1}{1+x}$   &  \dfrac{1}{1+x}\underset{{x\to 0}}{=}\sum_{k=0}^{n} (-1)^{k}{x^k} +o(x^{n})
                        &  \dfrac{1}{1+x}\underset{{x\to 0}}{=} 1-x+x^2-x^3+x^4-x^5+o(x^{5}) &  & \rm{R}=1,\ \forall x\in ]-1;1[\quad \dfrac{1}{1+x}=\sum_{n=0}^{+\infty} (-1)^{n}{x^n}       \\
    \hline
                $-\ln(1-x)$   &   -\ln(1-x)\underset{{x\to 0}}{=}\sum_{k=0}^{n} \dfrac{x^{k+1}}{k+1} +o(x^{n})
                        &  -\ln(1-x)\underset{{x\to 0}}{=} x+{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}+{\frac {x^{4}}{4}}+{\frac {x^{5}}{5}}+o(x^{5})  &  & \rm{R}=1,\ \forall x\in [-1;1[\quad -\ln(1-x)=\sum_{n=0}^{+\infty} \dfrac{x^{n+1}}{n+1}      \\
    \hline
            $\ln(1+x)$   &   \ln(1+x)\underset{{x\to 0}}{=}\sum_{k=0}^{n} (-1)^{k}\dfrac{x^{k+1}}{k+1} +o(x^{n})
                        &  \ln(1+x)\underset{{x\to 0}}{=} x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-{\frac {x^{4}}{4}}+{\frac {x^{5}}{5}}+o(x^{5})  &  &
                                                {\displaystyle\rm{R}=1,\ \forall x\in ]-1;1]\quad \ln(1+x)=\sum_{n=0}^{+\infty} (-1)^{n}\dfrac{x^{n+1}}{n+1} }     \\
    \hline
                & \multicolumn{4}{c|}{ On en déduit la somme de la série harmonique alternée  $\displaystyle\quad  \sum_{n=1}^{+\infty} \dfrac{(-1)^{n-1}}{n}=\ln(2)$ } \\
        \hline

$\exp(x)$   &   {\displaystyle {\rm {e}}^{x}\underset{{x\to 0}}{=}\sum _{k=0}^{n }{\frac {x^{k}}{k!}}+o(x^{n}) }
                        &  {\displaystyle {\rm {e}}^{x}\underset{{x\to 0}}{=} 1+{\frac {x}{1!}}+{\frac {x^{2}}{2!}}+{\frac {x^{3}}{3!}}+{\frac {x^{4}}{4!}}+{\frac {x^{5}}{5!}}+o(x^{5})} &  & {\displaystyle \rm{R}=+\infty,\ \forall x\in \mathbb{R}\quad {\rm {e}}^{x}=\sum _{n=0}^{+\infty }{\frac {x^{n}}{n!}}}        \\
    \hline
        $\operatorname {ch} \,x$   &   {\displaystyle \operatorname {ch} \,x\underset{{x\to 0}}{=}\sum _{k=0}^{n }{\frac {x^{2k}}{(2\,k)!}}+o(x^{2n}) }
                        &  {\displaystyle \operatorname {ch} \,x\underset{{x\to 0}}{=} 1+{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}+o(x^{5})} &  & {\displaystyle \rm{R}=+\infty,\ \forall x\in \mathbb{R}\quad  ,\,\operatorname{ch} \,x=\sum _{n=0}^{+{\infty }}{\frac {x^{2n}}{(2\,n)!}}}        \\
    \hline
            $\operatorname {sh} \,x$   &   {\displaystyle \operatorname {sh} \,x\underset{{x\to 0}}{=}\sum _{k=0}^{n }{\frac {x^{2k+1}}{(2\,k+1)!}}+o(x^{2n+1}) }
                        &  {\displaystyle \operatorname {sh} \,x\underset{{x\to 0}}{=} x+{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}+o(x^{5})} &  & {\displaystyle \rm{R}=+\infty,\ \forall x\in \mathbb{R}\quad  ,\,\operatorname{sh} \,x=\sum _{n=0}^{+{\infty }}{\frac {x^{2n+1}}{(2\,n+1)!}}}        \\
    \hline
                        $\cos \,x$   &   {\displaystyle \cos \,x\underset{{x\to 0}}{=} \sum _{k=0}^{n }{(-1)^k\frac {x^{2k}}{(2\,k)!}}+o(x^{2n}) }
                        &  {\displaystyle \cos \,x\underset{{x\to 0}}{=}  1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}+o(x^{5})}
                                                & {\displaystyle\begin{aligned}[c] \cos \,x-1 &\underset{{x\to 0}}{\sim} -{\frac {x^{2}}{2!}}\\
1-\cos \,x &\underset{{x\to 0}}{\sim} {\frac {x^{2}}{2!}} \end{aligned}}   &
                                                {\displaystyle \rm{R}=+\infty,\ \forall x\in \mathbb{R}\quad  ,\,\cos \,x=\sum _{n=0}^{+{\infty }}{(-1)^n\frac {x^{2n}}{(2\,n)!}}}        \\
    \hline
            $\sin \,x$   &   {\displaystyle \sin \,x\underset{{x\to 0}}{=} \sum _{k=0}^{n }{(-1)^k\frac {x^{2k+1}}{(2\,k+1)!}}+o(x^{2n+1}) }
                        &  {\displaystyle \sin \,x\underset{{x\to 0}}{=}  x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}+o(x^{5})} &  & {\displaystyle \rm{R}=+\infty,\ \forall x\in \mathbb{R}\quad  ,\,\sin \,x=\sum _{n=0}^{+{\infty }}{(-1)^n\frac {x^{2n+1}}{(2\,n+1)!}}}        \\
    \hline
        $\arctan \,x$   &   {\displaystyle \arctan \,x\underset{{x\to 0}}{=} \sum _{k=0}^{n }{(-1)^k\frac {x^{2k+1}}{(2\,k+1)}}+o(x^{2n+1}) }
                        &  {\displaystyle \arctan  \,x\underset{{x\to 0}}{=}  x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}+o(x^{5})} &  & {\displaystyle \rm{R}=1,\  \forall x\in [-1;1](x\not =\pm i)\quad  ,\,\arctan  \,x=\sum _{n=0}^{+{\infty }}{(-1)^n\frac {x^{2n+1}}{(2\,n+1)}}}        \\
    \hline
            $\operatorname {arctanh} \,x$   &   {\displaystyle \operatorname {arctanh} \,x\underset{{x\to 0}}{=}\sum _{k=0}^{n }{\frac {x^{2k+1}}{(2\,k+1)}}+o(x^{2n+1}) }
                        &  {\displaystyle \operatorname {arctanh} \,x\underset{{x\to 0}}{=} x+{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}+o(x^{5})} &  & {\displaystyle \rm{R}=1,\  \forall x\in ]-1;1[\quad  ,\,\operatorname {arctanh} \,x=\sum _{n=0}^{+{\infty }}{\frac {x^{2n+1}}{(2\,n+1)}}}        \\
    \hline
                    \end{tabular}
    %}
    \end{adjustbox}
 \end{sidewaystable}

\usepackage{rotating}使用前言部分的编码。

Answer

按照此表格编码在您的文档中：我刚刚使用sidewaystable包rotating

\begin{sidewaystable}
  %\resizebox{\textwidth}{!}{%
    \begin{adjustbox}{width=1\textwidth}
\makegapedcells
    \begin{tabular}{ | c | *{5}{>{$\displaystyle}c<{$}|}}
    \hline
    Mohcine    &   \multicolumn{2}{c|}{Développement limité en 0 de fonctions usuelles }    & \text{ Equivalents} & \text{ Développements en série entière usuels} \\ \hline
            Théorème         &   \multicolumn{2}{c|}{$\displaystyle
            f(x)\underset{{x\to a}}{=}\sum_{k=0}^{n}a_k(x-1)^{k}+o((x-a)^{n})$}
                                                &  &   f(x)=\sum_{n=0}^{+\infty} a_n(x)^{n}     \\
    \hline
Formulaire        &   \multicolumn{2}{c|}{$\displaystyle
            f(x)\underset{{x\to a}}{=}\sum_{k=0}^{n}a_k(x-1)^{k}+o((x-a)^{n})$}
                                               & &  f(x)=\sum_{n\geq 0} a_k(x-1)^{k}    \\
    \hline
Ordre   &   DL_{n}(0)   &   DL_{5}(0)           &  &                                      \\
    \hline
        $\dfrac{1}{1-x}$   &  \dfrac{1}{1-x}\underset{{x\to 0}}{=}\sum_{k=0}^{n} {x^k} +o(x^{n})
                        &  \dfrac{1}{1-x}\underset{{x\to 0}}{=}1+x+x^2+x^3+x^4+x^5+o(x^{5}) &  & \rm{R}=1,\ \forall x\in ]-1;1[\quad \dfrac{1}{1-x}=\sum_{n=0}^{+\infty} {x^n}       \\
    \hline
            $\dfrac{1}{1+x}$   &  \dfrac{1}{1+x}\underset{{x\to 0}}{=}\sum_{k=0}^{n} (-1)^{k}{x^k} +o(x^{n})
                        &  \dfrac{1}{1+x}\underset{{x\to 0}}{=} 1-x+x^2-x^3+x^4-x^5+o(x^{5}) &  & \rm{R}=1,\ \forall x\in ]-1;1[\quad \dfrac{1}{1+x}=\sum_{n=0}^{+\infty} (-1)^{n}{x^n}       \\
    \hline
                $-\ln(1-x)$   &   -\ln(1-x)\underset{{x\to 0}}{=}\sum_{k=0}^{n} \dfrac{x^{k+1}}{k+1} +o(x^{n})
                        &  -\ln(1-x)\underset{{x\to 0}}{=} x+{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}+{\frac {x^{4}}{4}}+{\frac {x^{5}}{5}}+o(x^{5})  &  & \rm{R}=1,\ \forall x\in [-1;1[\quad -\ln(1-x)=\sum_{n=0}^{+\infty} \dfrac{x^{n+1}}{n+1}      \\
    \hline
            $\ln(1+x)$   &   \ln(1+x)\underset{{x\to 0}}{=}\sum_{k=0}^{n} (-1)^{k}\dfrac{x^{k+1}}{k+1} +o(x^{n})
                        &  \ln(1+x)\underset{{x\to 0}}{=} x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-{\frac {x^{4}}{4}}+{\frac {x^{5}}{5}}+o(x^{5})  &  &
                                                {\displaystyle\rm{R}=1,\ \forall x\in ]-1;1]\quad \ln(1+x)=\sum_{n=0}^{+\infty} (-1)^{n}\dfrac{x^{n+1}}{n+1} }     \\
    \hline
                & \multicolumn{4}{c|}{ On en déduit la somme de la série harmonique alternée  $\displaystyle\quad  \sum_{n=1}^{+\infty} \dfrac{(-1)^{n-1}}{n}=\ln(2)$ } \\
        \hline

$\exp(x)$   &   {\displaystyle {\rm {e}}^{x}\underset{{x\to 0}}{=}\sum _{k=0}^{n }{\frac {x^{k}}{k!}}+o(x^{n}) }
                        &  {\displaystyle {\rm {e}}^{x}\underset{{x\to 0}}{=} 1+{\frac {x}{1!}}+{\frac {x^{2}}{2!}}+{\frac {x^{3}}{3!}}+{\frac {x^{4}}{4!}}+{\frac {x^{5}}{5!}}+o(x^{5})} &  & {\displaystyle \rm{R}=+\infty,\ \forall x\in \mathbb{R}\quad {\rm {e}}^{x}=\sum _{n=0}^{+\infty }{\frac {x^{n}}{n!}}}        \\
    \hline
        $\operatorname {ch} \,x$   &   {\displaystyle \operatorname {ch} \,x\underset{{x\to 0}}{=}\sum _{k=0}^{n }{\frac {x^{2k}}{(2\,k)!}}+o(x^{2n}) }
                        &  {\displaystyle \operatorname {ch} \,x\underset{{x\to 0}}{=} 1+{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}+o(x^{5})} &  & {\displaystyle \rm{R}=+\infty,\ \forall x\in \mathbb{R}\quad  ,\,\operatorname{ch} \,x=\sum _{n=0}^{+{\infty }}{\frac {x^{2n}}{(2\,n)!}}}        \\
    \hline
            $\operatorname {sh} \,x$   &   {\displaystyle \operatorname {sh} \,x\underset{{x\to 0}}{=}\sum _{k=0}^{n }{\frac {x^{2k+1}}{(2\,k+1)!}}+o(x^{2n+1}) }
                        &  {\displaystyle \operatorname {sh} \,x\underset{{x\to 0}}{=} x+{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}+o(x^{5})} &  & {\displaystyle \rm{R}=+\infty,\ \forall x\in \mathbb{R}\quad  ,\,\operatorname{sh} \,x=\sum _{n=0}^{+{\infty }}{\frac {x^{2n+1}}{(2\,n+1)!}}}        \\
    \hline
                        $\cos \,x$   &   {\displaystyle \cos \,x\underset{{x\to 0}}{=} \sum _{k=0}^{n }{(-1)^k\frac {x^{2k}}{(2\,k)!}}+o(x^{2n}) }
                        &  {\displaystyle \cos \,x\underset{{x\to 0}}{=}  1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}+o(x^{5})}
                                                & {\displaystyle\begin{aligned}[c] \cos \,x-1 &\underset{{x\to 0}}{\sim} -{\frac {x^{2}}{2!}}\\
1-\cos \,x &\underset{{x\to 0}}{\sim} {\frac {x^{2}}{2!}} \end{aligned}}   &
                                                {\displaystyle \rm{R}=+\infty,\ \forall x\in \mathbb{R}\quad  ,\,\cos \,x=\sum _{n=0}^{+{\infty }}{(-1)^n\frac {x^{2n}}{(2\,n)!}}}        \\
    \hline
            $\sin \,x$   &   {\displaystyle \sin \,x\underset{{x\to 0}}{=} \sum _{k=0}^{n }{(-1)^k\frac {x^{2k+1}}{(2\,k+1)!}}+o(x^{2n+1}) }
                        &  {\displaystyle \sin \,x\underset{{x\to 0}}{=}  x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}+o(x^{5})} &  & {\displaystyle \rm{R}=+\infty,\ \forall x\in \mathbb{R}\quad  ,\,\sin \,x=\sum _{n=0}^{+{\infty }}{(-1)^n\frac {x^{2n+1}}{(2\,n+1)!}}}        \\
    \hline
        $\arctan \,x$   &   {\displaystyle \arctan \,x\underset{{x\to 0}}{=} \sum _{k=0}^{n }{(-1)^k\frac {x^{2k+1}}{(2\,k+1)}}+o(x^{2n+1}) }
                        &  {\displaystyle \arctan  \,x\underset{{x\to 0}}{=}  x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}+o(x^{5})} &  & {\displaystyle \rm{R}=1,\  \forall x\in [-1;1](x\not =\pm i)\quad  ,\,\arctan  \,x=\sum _{n=0}^{+{\infty }}{(-1)^n\frac {x^{2n+1}}{(2\,n+1)}}}        \\
    \hline
            $\operatorname {arctanh} \,x$   &   {\displaystyle \operatorname {arctanh} \,x\underset{{x\to 0}}{=}\sum _{k=0}^{n }{\frac {x^{2k+1}}{(2\,k+1)}}+o(x^{2n+1}) }
                        &  {\displaystyle \operatorname {arctanh} \,x\underset{{x\to 0}}{=} x+{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}+o(x^{5})} &  & {\displaystyle \rm{R}=1,\  \forall x\in ]-1;1[\quad  ,\,\operatorname {arctanh} \,x=\sum _{n=0}^{+{\infty }}{\frac {x^{2n+1}}{(2\,n+1)}}}        \\
    \hline
                    \end{tabular}
    %}
    \end{adjustbox}
 \end{sidewaystable}

\usepackage{rotating}使用前言部分的编码。

Question 3

使用\resizebox{.96\linewidth}{!}{...}和删除table环境

.96 值确保表格不会移动到下一页。

\linewidth应该在横向模式下使用，而不是\textwidth

Answer

使用\resizebox{.96\linewidth}{!}{...}和删除table环境

.96 值确保表格不会移动到下一页。

\linewidth应该在横向模式下使用，而不是\textwidth

缩放包含合并单元格的表格以适合整个页面

答案1

答案2

答案3

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