# Make enumerate start from a new line [duplicate]

I'd like to put enumerate on a new line, but cannot use \newline because there is no line to end!
Here is the code:

\newtheorem*{quoz}{Teoremi del quoziente}
\begin{quoz} % \newline here does not work
\begin{enumerate}
\item Se $\lim\limits_{x \to \alpha} |f(x)| = +\infty$ allora $\lim\limits_{x \to \alpha} \frac1{f(x)} = 0$.
\item Se $\lim\limits_{x \to \alpha} f(x) = 0$ allora $\lim\limits_{x \to \alpha} \frac1{|f(x)|} = +\infty$.
\item Se $\lim\limits_{x \to \alpha} f(x) = l$ allora $\lim\limits_{x \to \alpha} \frac1{f(x)} = \frac1l$.
\item Se $\lim\limits_{x \to \alpha} f(x) = l$ e $\lim\limits_{x \to \alpha} g(x) = m \neq 0$, allora $\lim\limits_{x \to \alpha} \frac{f(x)}{g(x)} = \frac lm$.
\end{enumerate}
\end{quoz}


Currently the problem is that the first point starts where the theorem's title is. I find it kind of ugly. Is there a way to make it start after a new line?

I tried doing

\begin{quoz}\newline
\begin{enumerate}
%...


But that adds too much space and feels like a hack. Is there a better way?

EDIT: I discovered that for some weird reason this works

\begin{quoz}
\begin{enumerate}
%...


I don't get why because  should be inline... Nevertheless, is there a better way?

• Does \leavevmode do? Feb 2, 2015 at 13:03
• @Manuel: I get undefined control sequence. Is it defined in some package? Feb 2, 2015 at 13:05
• Perhaps you mistyped it. \leavevmode is basic TeX (no package required). Feb 2, 2015 at 13:08
• \leavevmode should work. As anything that starts a paragraph like \ , ~, \mbox{}, ... Feb 2, 2015 at 13:09
• question answered here: Is it possible to use the enumerate/itemize environment within the proof/remark environment in a nice way?. the ams author faq item mentioned there has even more information. Feb 2, 2015 at 13:58

Here's a solution that also teaches you how not to define a new unnumbered theorem each time. The *-version of genthm does a break, in case you want to start with a list. However, I find this bad style and I'd start the theorem by stating the general assumptions we're making on the functions.

\documentclass{article}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage[italian]{babel}

\usepackage{amsthm}

\newtheorem*{nonamethm}{\nonamethmname}
\newcommand{\nonamethmname}{}

\newenvironment{genthm}
{\renewcommand{\nonamethmname}{#1}\nonamethm}
{\endnonamethm}
\newenvironment{genthm*}
{\renewcommand{\nonamethmname}{#1}\nonamethmcheck}
{\endnonamethm}

\newcommand\nonamethmcheck[]{%
\if\relax\detokenize{#1}\relax
\nonamethm\relax
\else
\nonamethm[#1]%
\fi
\mbox{}%
}

\begin{document}

\begin{genthm*}{Teoremi del quoziente}
\begin{enumerate}
\item Se $\lim\limits_{x \to \alpha} |f(x)| = +\infty$,
allora $\lim\limits_{x \to \alpha} \frac{1}{f(x)} = 0$.

\item Se $\lim\limits_{x \to \alpha} f(x) = 0$,
allora $\lim\limits_{x \to \alpha} \frac1{|f(x)|} = +\infty$.

\item Se $\lim\limits_{x \to \alpha} f(x) = l \neq 0$,
allora $\lim\limits_{x \to \alpha} \frac1{f(x)} = \frac{1}{l}$.

\item Se $\lim\limits_{x \to \alpha} f(x) = l$ e $\lim\limits_{x \to \alpha} g(x) = m \neq 0$,
allora $\lim\limits_{x \to \alpha} \frac{f(x)}{g(x)} = \frac{l}{m}$.
\end{enumerate}
\end{genthm*}

\begin{genthm}{Teoremi del quoziente}
Siano $f$ e $g$ funzioni definite in un intorno bucato di~$\alpha$.
\begin{enumerate}
\item Se $\lim\limits_{x \to \alpha} |f(x)| = +\infty$,
allora $\lim\limits_{x \to \alpha} \frac{1}{f(x)} = 0$.

\item Se $\lim\limits_{x \to \alpha} f(x) = 0$,
allora $\lim\limits_{x \to \alpha} \frac1{|f(x)|} = +\infty$.

\item Se $\lim\limits_{x \to \alpha} f(x) = l \neq 0$,
allora $\lim\limits_{x \to \alpha} \frac1{f(x)} = \frac{1}{l}$.

\item Se $\lim\limits_{x \to \alpha} f(x) = l$ e $\lim\limits_{x \to \alpha} g(x) = m \neq 0$,
allora $\lim\limits_{x \to \alpha} \frac{f(x)}{g(x)} = \frac{l}{m}$.
\end{enumerate}
\end{genthm}

\end{document}


The optional argument (coming after the mandatory argument) is supported for both environments. • This is very useful, I'll do like you suggested. Thanks again! Feb 2, 2015 at 15:18