How do I left-align selective equation blocks, i.e. not centered ?

I know how to do it for ALL equation blocks in the document setting


in the preamble, but I only want to do this for some equation blocks.



De raaklijn aan $(x_{0},f(x_{0}))$ wordt algemeen bepaald met :
\textnormal{$\begin{cases}z=f(x)\text{ is continu in }x_{0}\text{.}\\
f'(x_{0})=0\text{ : }z=f(x_{0})\text{ is de horizontale raaklijn.}\\
\lim_{x\to x_{0}} f'(x)=\pm\infty\text{ : }x=x_{0}\text{ is de verticale raaklijn.}\\
f'(x_{0})\in\mathbb{R}_{0}\text{ : }z-f(x_{0})=f'(x_{0})\cdot(x-x_{0})\text{ is de raaklijn.\hspace{-12pt}}\end{cases}$}\end{empheq}

  • Place '&' at the beginning of each line inside the empheq environment and do not use the cases environment. This is the easiest way to get what you want. – Daniel F Jul 17 '15 at 16:23

Don't use align*.

De raaklijn aan $(x_{0},f(x_{0}))$ wordt algemeen bepaald met:
z=f(x) \text{ is continu in } x_{0}.
f'(x_{0})=0 \text{ : } z=f(x_{0}) \text{ is de horizontale raaklijn.}
\lim_{x\to x_{0}} f'(x)=\pm\infty \text{ : } x=x_{0} \text{ is de verticale raaklijn.}
f'(x_{0})\in\mathbb{R}_{0} \text{ : } z-f(x_{0})=f'(x_{0})\cdot(x-x_{0})
  \text{ is de raaklijn.}

enter image description here

| improve this answer | |
  • What does \kern-\nulldelimiterspace\kern-2\arraycolsep} do in this ? – Petoetje59 Oct 5 '12 at 14:28
  • @Petoetje59 It's a substitute of your \hspace{-12pt} based on document parameters, rather than "calculation by eye". – egreg Oct 5 '12 at 14:45
  • Not allowing you to get the unsung hero badge! ;) +1. – user11232 Oct 5 '12 at 14:48
  • @HarishKumar Darn! – egreg Oct 5 '12 at 14:49
  • How come \begin{align*} doesn't work but flushleft does? – Carlos - the Mongoose - Danger Sep 1 '15 at 8:18

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