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I just discovered how to use flalign* to left justify my multi-line equation, but I would like it to start immediately after the a) lim... in my enumerate without going to the next line.

\begin{enumerate}
  \item \begin{flalign*}
          L&=\lim\limits_{n\to\infty}\frac{n}{n^2-1}&\\
           &=\lim\limits_{n\to\infty}\frac{n}{n^2-1}\frac{(\frac{1}{n^2})}{(\frac{1}{n^2})}‌​\\
           &=\lim\limits_{n\to\infty}\frac{\frac{n}{n^2}}{\frac{n^2}{n^2}-\frac{1}{n^2}}\\ 
           &=\lim\limits_{n\to\infty}\frac{\frac{1}{n}}{1-\frac{1}{n^2}}\\
           &=\frac{0}{1-0}\\
           &=0 
        \end{flalign*}
   \item

It always goes to the next line. How do I move it up?

Stephen
  • 14,890
Euler Fan
  • 21
  • If you are not using a multi-line equation, you should just use $<eqn>$, rather than \begin{flalign*}<eqn>\end{flalign*}. – Werner Jan 29 '14 at 17:29
  • 2
    Please show a complete small document that shows what you are trying to do. why enumerate rather than using a numbered math alignment? – David Carlisle Jan 29 '14 at 17:30
  • Welcome to TeX.SE. Perhaps this question will solve your issue: Vertical alignment of align* in enumerate. – Peter Grill Jan 29 '14 at 18:08
  • I am using a multi-line equation. I would just like this to start with a) lim... and so on without going to the next line. \begin{enumerate} \item \begin{flalign} L&=\lim\limits_{n\to\infty}\frac{n}{n^2-1}&\ &=\lim\limits_{n\to\infty}\frac{n}{n^2-1}\frac{(\frac{1}{n^2})}{(\frac{1}{n^2})}\ &=\lim\limits_{n\to\infty}\frac{\frac{n}{n^2}}{\frac{n^2}{n^2}-\frac{1}{n^2}}\ &=\lim\limits_{n\to\infty}\frac{\frac{1}{n}}{1-\frac{1}{n^2}}\ &=\frac{0}{1-0}\ &=0 \end{flalign} \item – Euler Fan Jan 29 '14 at 18:17
  • The first line in the flalign* includes a second & - is that intentional? – Stephen Jan 29 '14 at 19:07
  • As per Vertical alignment of align* in enumerate, just using \item $\begin{aligned}[t] L&=\lim\limits_{n\to\infty}\frac{n}{n^2-1}&\\ &=\lim\limits_{n\to\infty}\frac{n}{n^2-1}\frac{(\frac{1}{n^2})}{(\frac{1}{n^2})}‌​\\ &=\lim\limits_{n\to\infty}\frac{\frac{n}{n^2}}{\frac{n^2}{n^2}-\frac{1}{n^2}}\\ &=\lim\limits_{n\to\infty}\frac{\frac{1}{n}}{1-\frac{1}{n^2}}\\ &=\frac{0}{1-0}\\ &=0 \end{aligned}$ seems to work. – Peter Grill Jan 29 '14 at 19:16

0 Answers0