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align* 将方程式移出页面时出现问题

tag-icon • tag-icon math-mode align amsmath
align* 将方程式移出页面时出现问题

这是我想要排版成 LaTeX 的数学问题和答案：

我想用 LaTeX 排版的数学问题

延续

我正在使用align*环境来排版所有内容，这是我使用的代码：

\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsfonts}
\begin{document}
    \begin{align*}
        &\text{\textbf{Find the limit of} }\mathbf{\lim_{x\to\infty}\frac{x^3-7x^2+6x-17}{x^4-8}}\text{\textbf{:}}\\
        &=\lim_{x\to\infty}\frac{x^3-7x^2+6x-17}{x^4-8}\\
        &=\frac{(\infty)^3-7(\infty)-6(\infty)-17}{(\infty)^4-8}\\
        &\\
        &\text{Since this is an }\frac{\infty}{\infty}\text{ limit, we can use L`H}\hat{o}\text{pital's rule}\\
        &\frac{dy}{dx}(x^3-7x^2+6x-17) &&\frac{dy}{dx}(x^4-8) &\longrightarrow\text{ L`H}\hat{o}\text{pital's rule:}\\
        &=3x^2-14x+6                   && =4x^3               &\text{1st attempt}\\
        &\\
        &\text{Since this is still an }\frac{\infty}{\infty}\text{ limit, we can repeatedly use L`H}\hat{o}\text{pital's rule until we get to a result:}\\
        &\frac{dy}{dx}(3x^2-14x+6)     &&\frac{dy}{dx}(4x^3)  &\longrightarrow\text{ L`H}\hat{o}\text{pital's rule:}\\
        &=6x-14                        && =12x^2              &\text{2nd attempt}\\
        &\frac{dy}{dx}(6x-14)          &&\frac{dy}{dx}(12x^2) &\longrightarrow\text{ L`H}\hat{o}\text{pital's rule:}\\
        &=6                            && =24x                &\text{3rd attempt}\\
        &\\
        &\text{Now we have:}\\
        &\lim_{x\to\infty}\frac{6}{24x}\\
        &=\frac{6}{24(\infty)}\\
        &=\frac{6}{\infty}\\
        &=0
    \end{align*}
\end{document}

然而，问题在于我的方程式超出了页面范围，而且我不确定如何修复它。 在此处输入图片描述

笔记：这些数学图像来自旧的作业，而这项数学作业的 LaTeX 排版是我目前正在做的作业。

答案1

这个布局适合您吗？

笔记amsfonts：加载时无需加载amssymb– 后者已经为您完成了。

    \documentclass{article}
    \usepackage{mathtools}
    \usepackage{amssymb, bm}

    \begin{document}

        \begin{align*}
            \shortintertext{\bfseries{Find the limit of} $\displaystyle \bm{\lim_{x\to\infty}\frac{x^3-7x^2+6x-17}{x^4-8}} $ :}
            &=\lim_{x\to\infty}\frac{x^3-7x^2+6x-17}{x^4-8}\\
            &=\frac{(\infty)^3-7(\infty)-6(\infty)-17}{(\infty)^4-8}\\
            \intertext{Since this is an $\frac{\infty}{\infty}$, we can use L'Hôpital's rule}
            &\frac{dy}{dx}(x^3-7x^2+6x-17) &&\frac{dy}{dx}(x^4-8) &\longrightarrow\text{ L`Hôpital's rule:}\\
            &=3x^2-14x+6 && =4x^3 &\text{1st attempt}\\
            \intertext{Since this is still an $\frac{\infty}{\infty}$, we can repeatedly use L'Hôpital's rule until we get to a result:}
            &\frac{dy}{dx}(3x^2-14x+6) &&\frac{dy}{dx}(4x^3) &\longrightarrow\text{ L`Hôpital's rule:}\\
            &=6x-14 && =12x^2 &\text{2nd attempt}\\
            &\frac{dy}{dx}(6x-14) &&\frac{dy}{dx}(12x^2) &\longrightarrow\text{ L`Hôpital's rule:}\\
            &=6 && =24x &\text{3rd attempt}\\
            \intertext{Now we have:}
            &\lim_{x\to\infty}\frac{6}{24x}\\
            &=\frac{6}{24(\infty)}\\
            &=\frac{6}{\infty}\\
            &=0
        \end{align*}

        \end{document}

在此处输入图片描述

答案2

基于多重嵌套数学环境和左对齐方程的解决方案

在此处输入图片描述

\documentclass{article}
\usepackage{array}
\usepackage[fleqn]{mathtools}
\usepackage{amssymb}
\usepackage{bm}

\setlength{\parindent}{0pt}
\newcommand\prelen{\hspace{15pt}}
\newlength\eqskip   \setlength\eqskip{9pt} % skips at tries


\begin{document}
\textbf{Find the limit of} \(\displaystyle \mathbf{\bm{\lim}_{x\to\infty} \frac{x^3-7x^2+6x-17}{x^4-8}}\):

\begin{align*}
    &\lim_{x\to\infty}\frac{x^3-7x^2+6x-17}{x^4-8} \\
    &\prelen=\ \lim_{x\to\infty}
        \frac{(\infty)^3-7(\infty)-6(\infty)-17}{(\infty)^4-8} \\
    &\prelen=\ \frac{\infty}{\infty}
\end{align*}

Since this is an \(\displaystyle \frac{\infty}{\infty}\) limit, we can use L'Hôpital's rule:
\begin{align*}
    \begin{aligned}
        \frac{dy}{dx}\bigl(x^3-7x^2+6x-17\bigr) \\
        =\ 3x^2-14x+6
    \end{aligned}
        & \hspace{3em}
    \begin{aligned}   % {r @{\hspace{3em}} r}
        \frac{dy}{dx}\bigl(x^4-8\bigr) \\
        =\ 4x^3
    \end{aligned}
\end{align*}

Since this is still an \(\displaystyle \frac{\infty}{\infty}\) limit, we can repeatedly use L'Hôpital's rule until we get to a result:
\begin{align*}
    \begin{aligned}
        &\frac{dy}{dx}(3x^2-14x+6) \\
        &    \prelen=\ 6x - 14 \\[\eqskip]
        &\frac{dy}{dx}(6x - 14) \\
        &    \prelen=\ 6
    \end{aligned}
    &&  \begin{aligned}
            & \frac{dy}{dx}(4x^3)
                & \quad\longrightarrow
                &&& \smash{\begin{tabular}{l}
                                {L'Hôpital's rule} \\
                                (2nd try)
                            \end{tabular}} \\
            & \prelen=\ 12x^2 \\[\eqskip]
            & \frac{dy}{dx}(12x^2)
                & \quad\longrightarrow
                &&& \smash{\begin{tabular}{l}
                                {L'Hôpital's rule} \\
                                (3rd try)
                            \end{tabular}} \\
            & \prelen=\ 24x
        \end{aligned}             
\end{align*}

Now we have:
\begin{align*}
    \lim_{x\to\infty}\,\frac{6}{24x}\ &=\ \frac{6}{24(\infty)} \\
        & =\ \frac{6}{\infty} \\
        & =\ 0
\end{align*}

\end{document}

答案3

这里有很多可疑的代码，所以我最好只给你一个修正版本，而不是列出所有更改。简而言之，试图将整个结构强行合并为一个align会让你的生活变得非常困难，因为这不是环境设计的目的。我不会谈论将算术运算应用于无穷符号，除了这是一种可憎的行为。我会考虑创建一个宏\mathrm{d}并用它来代替微分运算符中的普通“d”。

\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsfonts}
\begin{document}
Find the limit of 
\(\displaystyle\mathbf{\lim_{x\to\infty}\frac{x^3-7x^2+6x-17}{x^4-8}}\):    
\begin{align*}
\lim_{x\to\infty}\frac{x^3-7x^2+6x-17}{x^4-8} 
  &= \lim_{x\to\infty}\frac{x^3-7x^2+6x-17}{x^4-8}\\
  &=\frac{(\infty)^3-7(\infty)-6(\infty)-17}{(\infty)^4-8}.
\end{align*}
Since this is an \(\infty/\infty\) limit, we can use L'H\^opital's rule:
\begin{align*}
&\frac{dy}{dx}(x^3-7x^2+6x-17) &\frac{dy}{dx}(x^4-8)&&\longrightarrow\text{L'H\^opital's rule:}\\
&    \quad =3x^2-14x+6         &\quad=4x^3          && \text{1st attempt}.
\end{align*}
Since this is still an \(\infty/\infty\) limit, we can repeatedly use L'H\^opital's rule until 
we get to a result:
\begin{align*}
&\frac{dy}{dx}(3x^2-14x+6) &\frac{dy}{dx}(4x^3)   &&\longrightarrow\text{L'H\^opital's rule:}\\
&\quad =6x-14              &\quad =12x^2          &&\text{2nd attempt}\\
&\frac{dy}{dx}(6x-14)      &\frac{dy}{dx}(12x^2)  &&\longrightarrow\text{L'H\^opital's rule:}\\
&\quad =6                  &\quad =24x            &&\text{3rd attempt}
\end{align*}
Now we have:
\begin{align*}
\lim_{x\to\infty}\frac{6}{24x}
        &=\frac{6}{24(\infty)}\\
        &=\frac{6}{\infty}\\
        &=0
\end{align*}
\end{document}

方程式

答案4

我会这样写你的作业：

在此处输入图片描述

\documentclass{article}
\usepackage{mathtools}
\usepackage{amssymb}

\begin{document}
\noindent\textbf{Find the limit of $\mathbf{\lim\limits_{x\to\infty}\frac{x^3-7x^2+6x-17}{x^4-8}}$ :}    
    \begin{align*}
\lim_{x\to\infty}\frac{x^3 - 7x^2+6x - 17}{x^4 - 8}
        & = \frac{(\infty)^3 - 7(\infty)-6(\infty)-17}{(\infty)^4 - 8}      \\
\intertext{Since this is an $\frac{\infty}{\infty}$ limit, we can use L'Hôpital's rule}
\frac{dy}{dx}(x^3 - 7x^2+6x - 17)     
        & = 3x^2 - 14x + 6 
                &  \longrightarrow\quad \parbox[t]{7em}{L'Hôpital's rule:\\
                                                        1st attempt}         \\
\frac{dy}{dx}(x^4 - 8)         
        & = 4x^3            
\intertext{Since this is still an $\frac{\infty}{\infty}$ limit, we can repeatedly use L'Hôpital's rule rule until we get the final result:}
\frac{dy}{dx}(3x^2-14x+6)
        & = 6x - 14       
                &  \longrightarrow\quad \parbox[t]{7em}{L'Hôpital's rule:\\
                                                        2nd attempt}         \\
\frac{dy}{dx}(4x^3)  
        & =12x^2                                                                        
\shortintertext{and}
\frac{dy}{dx}(6x-14)    
        & = 6   
                &  \longrightarrow\quad \parbox[t]{7em}{L'Hôpital's rule:\\
                                                        3rd attempt}        \\
\frac{dy}{dx}(12x^2) 
        & =24x                                                              \\
\intertext{Now we have:}
\lim_{x\to\infty}\frac{6}{24x}
        & = \frac{6}{24(\infty)} = \frac{6}{\infty}                         \\
        & = \boxed{ 0 }
    \end{align*}
\end{document}

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