我的代码是:
\begin{align*}
\lim\limits_{x \to a} [f(x)g(x)-LM] &= \lim\limits_{x \to a}[h(x)M+i(x)L+h(x)i(x)]
\\
&= \lim\limits_{x \to a} h(x)M+ \lim\limits_{x \to a}i(x)L+ \lim\limits_{x \to a} h(x)i(x) \\
&= 0\\
\therefore \lim\limits_{x \to a} [f(x)g(x)-LM] &= \lim\limits_{x \to a} f(x)g(x) - LM = 0 \\
\therefore \lim\limits_{x \to a} f(x)g(x) &= LM
\end{align*}
我想:
我试过了:
\begin{align*}
\lim\limits_{x \to a} [f(x)g(x)-LM] &= \lim\limits_{x \to a}[h(x)M+i(x)L+h(x)i(x)]
\\
&= \lim\limits_{x \to a} h(x)M+ \lim\limits_{x \to a}i(x)L+ \lim\limits_{x \to a} h(x)i(x) \\
&= 0
\end{align*}
\begin{align*}
&\therefore \lim\limits_{x \to a} [f(x)g(x)-LM] = \lim\limits_{x \to a} f(x)g(x) - LM = 0 \\
&\therefore \lim\limits_{x \to a} f(x)g(x)= LM
\end{align*}
我更喜欢一种非手动且不繁琐的方式——谢谢。
答案1
alignat*{2}
如果您有两个对齐点(需要 3 &
),则可以使用它。
无关:您不必添加\limits
显示样式。
\documentclass{article}
\usepackage{amsmath, amssymb}
\begin{document}
\begin{alignat*}{2}
& \lim_{x \to a} [f(x)g(x)-LM] & & = \lim\limits_{x \to a}[h(x)M+i(x)L+h(x)i(x)]
\\
& & &= \lim_{x \to a} h(x)M+ \lim\limits_{x \to a}i(x)L+ \lim\limits_{x \to a} h(x)i(x) \\
& & &= 0\\
\therefore{} & \lim_{x \to a} [f(x)g(x)-LM] & &= \lim\limits_{x \to a} f(x)g(x) - LM = 0 \\
\therefore {}& \lim_{x \to a} f(x)g(x)= LM
\end{alignat*}
\end{document}