Align the system of equations

Question

Consider:

\documentclass[a4paper]{article}
\usepackage{amsmath,amssymb}
\usepackage{amsfonts}
\usepackage{mathtools}
\begin{document}
\[
\begin{array}{llr}
\begin{rcases*}
    $\lim\limits_{x\to2^+}f(x)&=\lim\limits_{x\to2^+}\sqrt{5x-1}&=3$\\
    $\lim\limits_{x\to2^-}f(x)&=\lim\limits_{x\to2^-}(3x^2-4x-1)&=3$\\
    $f(2)&=\sqrt{5\cdot2-1}&=3$
\end{rcases*}\Leftrightarrow \displaystyle \lim\limits_{x\to2^-}f(x)=f(2)=\lim\limits_{x\to2^+}f(x)
\end{array}
\]
\end{document}

I was getting:

But I want the result like this:

What should I do?

Answer 1

\documentclass[a4paper]{article}
\usepackage{amsmath,amssymb}
\usepackage{mathtools}
\begin{document}  
    \[\def\arraystretch{2}\arraycolsep=1.6pt
    \left.\begin{array}{llr}
        \lim\limits_{x\to2^+}f(x) &= \lim\limits_{x\to2^+}\sqrt{5x-1} &= 3\\
        \lim\limits_{x\to2^-}f(x) &= \lim\limits_{x\to2^-}(3x^2-4x-1) &= 3\\
        f(2)                     &= \sqrt{5\cdot2-1}                  &=3
    \end{array}\right\}
    \Leftrightarrow \displaystyle \lim\limits_{x\to2^-}f(x)=f(2)=\lim\limits_{x\to2^+}f(x)
    \]
\end{document}

Answer 2

I have three proposals, none of which uses cases or variants, which are not suited for the job.

\documentclass[a4paper]{article}
\usepackage{amsmath,amssymb}
\usepackage{mathtools}
\usepackage{array}

\begin{document}  

\[
\left.
\begin{alignedat}{2}
  \lim_{x\to2^+}f(x) &= \lim_{x\to2^+}\sqrt{5x-1}&&=3\\
  \lim_{x\to2^-}f(x) &= \lim_{x\to2^-}(3x^2-4x-1)&&=3\\
  f(2)&=\sqrt{5\cdot2-1}&&=3
\end{alignedat}
\right\}\Leftrightarrow \lim_{x\to2^-}f(x)=f(2)=\lim_{x\to2^+}f(x)
\]

\[
\left.
\renewcommand{\arraystretch}{1.5}
\setlength{\arraycolsep}{0pt}
\newcolumntype{A}{>{\displaystyle}l}
\newcolumntype{B}{>{\displaystyle{}}l}
\begin{array}{ABB}
  \lim_{x\to2^+}f(x) &= \lim_{x\to2^+}\sqrt{5x-1}&=3\\
  \lim_{x\to2^-}f(x) &= \lim_{x\to2^-}(3x^2-4x-1)&=3\\
  f(2)&=\sqrt{5\cdot2-1}&=3
\end{array}
\right\}\Leftrightarrow \lim_{x\to2^-}f(x)=f(2)=\lim_{x\to2^+}f(x)
\]

\[
\left.
\begin{aligned}
&\! \lim_{x\to2^+}f(x) = \lim_{x\to2^+}\sqrt{5x-1}=3 \\
&\! \lim_{x\to2^-}f(x) = \lim_{x\to2^-}(3x^2-4x-1)=3 \\
&   f(2)=\sqrt{5\cdot2-1}=3
\end{aligned}
\right\}\lim_{x\to2^-}f(x)=f(2)=\lim_{x\to2^+}f(x)
\]

\end{document}

I'd much prefer the last one: the three parts left to the brace are not really connected to one another and aligning the equals signs is not needed and makes for bad blanks. The \Leftrightarrow symbol is mathematically wrong.

Answer 3

You're not making good use of the rcases* machinery; I would ditch it in favor a simple \left. ... \right\} wrapper. To format the material to left of the tall curly brace, I suggest you employ an alignedat{3} environment. This will allow you to get rid of all \displaystyle and \limits directives.

Optionally, you could use \mathclap directives to get rid of the whitespace around the \lim terms. The solutions without and with the \mathclap idea are shown in the following screenshot.

\documentclass[a4paper]{article}
\usepackage{%%amsmath, %% amsmath is loaded automatically by mathtools
            amssymb}
%%\usepackage{amsfonts} %% amsfonts is loaded automatically by amssymb
\usepackage{mathtools} % for \mathclap macro
\begin{document}  

\noindent
Basic solution:
\[
\left.
\begin{alignedat}{3}
    &\lim_{x\to2^+}f(x)&&=\lim_{x\to2^+}\sqrt{5x-1}&&=3\\
    &\lim_{x\to2^-}f(x)&&=\lim_{x\to2^-}(3x^2-4x-1)&&=3\\
    &f(2)              &&=\sqrt{5\cdot2-1}         &&=3
\end{alignedat}
\,\right\}
\Leftrightarrow  \lim_{x\to2^-}f(x)=f(2)=\lim_{x\to2^+}f(x)
\]


\bigskip\noindent
With \verb+\mathclap+:
\[
\left.
\begin{alignedat}{3}
    &\lim_{\mathclap{x\to2^+}}f(x)&&=\lim_{\mathclap{x\to2^+}}\sqrt{5x-1}&&=3\\
    &\lim_{\mathclap{x\to2^-}}f(x)&&=\lim_{\mathclap{x\to2^-}}(3x^2-4x-1)&&=3\\
    &f(2)                         &&=\sqrt{5\cdot2-1}                    &&=3
\end{alignedat}
\,\right\}
\Leftrightarrow  \lim_{\mathclap{x\to2^-}}f(x)=f(2)=\lim_{\mathclap{x\to2^+}}f(x)
\]
\end{document}