我已阅读了几篇有类似错误的帖子，但找不到我的错误在哪里。
我有以下代码用于在alignat*环境中的顺序不等式旁边添加注释。不幸的是，每次我使用\cm命令时，latex 都会给出错误，指出缺失的数字被视为零。我想我的某个长度命令中可能缺少一些参数，但我不知道在哪里。提前致谢。
\newlength{\argwidth}
\newlength{\whatsleft}
\newcommand{\measureremainder}[1]{%
\begin{tikzpicture}[overlay,remember picture]
\path (current page.north west) ++(\hoffset, -\voffset)
node[anchor=north west, shape=rectangle, inner sep=0, minimum width=\paperwidth, minimum height=\paperheight]
(pagearea) {};
\path (pagearea.north west) ++(1in+\currentsidemargin,-1in-\topmargin-\headheight-\headsep)
node[anchor=north west, shape=rectangle, inner sep=0, minimum width=\textwidth, minimum height=\textheight]
(textarea) {};
\path let \p0 = (0,0), \p1 = (textarea.east) in
[/utils/exec={\pgfmathsetlength#1{\x1-\x0}\global#1=#1}];
\end{tikzpicture}%
}
\newcommand{\cm}[1]{&\quad&\hspace{1.3mm}\measureremainder{\whatsleft}\settowidth{\argwidth}{#1}\ifthenelse{\lengthtest{\argwidth < \whatsleft}}{\text{{#1}}}{\begin{minipage}[c]{\whatsleft} \raggedright {#1} \vspace{2mm}\end{minipage}}}
例如，我将按如下方式使用它：
\documentclass[12pt]{report}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage[margins = 1.0in]{geometry}
\usepackage{MnSymbol}
\usepackage{tikz}
\usepackage{mathtools}
\usepackage{calc}
\usepackage{tkz-fct}
\usepackage{ifthen}
\usepackage{xifthen}
\newlength{\whatsleft}
\newlength{\argwidth}
\newcommand{\measureremainder}[1]{%
\begin{tikzpicture}[overlay,remember picture]
\path (current page.north west) ++(\hoffset, -\voffset)
node[anchor=north west, shape=rectangle, inner sep=0, minimum width=\paperwidth, minimum height=\paperheight]
(pagearea) {};
\path (pagearea.north west) ++(1in+\currentsidemargin,-1in-\topmargin-\headheight-\headsep)
node[anchor=north west, shape=rectangle, inner sep=0, minimum width=\textwidth, minimum height=\textheight]
(textarea) {};
\path let \p0 = (0,0), \p1 = (textarea.east) in
[/utils/exec={\pgfmathsetlength#1{\x1-\x0}\global#1=#1}];
\end{tikzpicture}%
}
\newcommand{\cm}[1]{&\quad &\hspace{1.3mm}\measureremainder{\whatsleft}\settowidth{\argwidth}{#1}\ifthenelse{\lengthtest{\argwidth < \whatsleft}}{\text{{#1} }}{\begin{minipage}[c]{\whatsleft} \raggedright {#1} \vspace{2mm}\end{minipage}}}
\begin{document}
\begin{alignat*}{2}
\lim_{x \to 3} x^2 - 4x + 3 &= \lim_{x \to 3} x^2 - \lim_{x \to 3} 4x + \lim_{x \to 3} 3 \cm{Using the sum rule} \\
&= \left(\lim_{x \to 3} x\right)\left(\lim_{x \to 3} x\right) - \left(4\lim_{x \to 3} x\right) + \lim_{x \to 3} 3 \cm{Using the product rule} \\
&= 3 \cdot 3 - 4 \cdot 3 + 3 \\
&= 0 \end{alignat*}
\end{document}
在此先感谢您的帮助。
答案1
错误发生在\lim之后的第一个位置\begin{alignat*}。此环境需要一个参数，即“方程列”的数量。由于\lim不是数字，TeX 会抛出错误。
除非您重新定义了\(和\)，否则它们会引起更多麻烦，因为它们通常用于内联数学模式。因此，我用普通括号替换了它们。我不认为，尽管有酸橙的索引，但这里需要\left和\right。在我看来，普通括号在这里看起来更好。
完整示例（需要运行两次 LaTeX）：
\documentclass{article}
\usepackage{ifthen}
\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{calc}
\newlength{\currentsidemargin}
\newlength{\argwidth}
\newlength{\whatsleft}
\newcommand{\measureremainder}[1]{%
\begin{tikzpicture}[overlay,remember picture]
\path
(current page.north west) ++(\hoffset, -\voffset)
node[
anchor=north west,
shape=rectangle,
inner sep=0,
minimum width=\paperwidth,
minimum height=\paperheight
] (pagearea) {}
;
\path
(pagearea.north west)
++(1in+\currentsidemargin,-1in-\topmargin-\headheight-\headsep)
node[
anchor=north west,
shape=rectangle,
inner sep=0,
minimum width=\textwidth,
minimum height=\textheight
] (textarea) {}
;
\path
let \p0 = (0,0),
\p1 = (textarea.east)
in
[/utils/exec={\pgfmathsetlength#1{\x1-\x0}\global#1=#1}]
;
\end{tikzpicture}%
}
\newcommand{\cm}[1]{%
&\quad
&\hspace{1.3mm}%
\measureremainder{\whatsleft}%
\settowidth{\argwidth}{#1}%
\ifthenelse{\lengthtest{\argwidth < \whatsleft}}{%
\text{#1}
}{%
\begin{minipage}[c]{\whatsleft}
\raggedright
#1\par
\vspace{2mm}%
\end{minipage}%
}%
}
\begin{document}
\begin{alignat*}{2}
\lim_{x \to 3} (x^2 - 4x + 3) &= \lim_{x \to 3} x^2 - \lim_{x \to 3} 4x +
\lim_{x \to 3} 3 \cm{Using the sum rule} \\
&= (\lim_{x \to 3} x)
(\lim_{x \to 3} x)
- 4(\lim_{x \to 3} x) +
\lim_{x \to 3} 3 \cm{Using the product rule} \\
&= 3 \cdot 3 - 4 \cdot 3 + 3 \\
&= 0
\end{alignat*}
\end{document}
