6

I am trying to put an underbrace underneath a the innermost of multiple square root signs. Unfortunately, I can't get the brace to line up underneath the entire line. Here's the relevant code and a picture of the output:

\documentclass{article}
\usepackage[utf8]{inputenc}
\begin{document}
\begin{align*}
\lim_{x\to0}f(f(f(x))) &= f\left(f\left(\lim_{x\to0}f(x)\right)\right) \\
&= \sqrt{\sqrt{\underbrace{\sqrt{1+0}}_{\text{Because $f(x)$ is continuous}}}}
\end{align*}

enter image description here

It should be pretty clear what I'm talking about from the picture.

It would also be cool if I could get the two outermost square root signs to be the same, but it's not a huge deal. The underbrace is the main issue here.

Thanks in advance!

NNN
  • 230

2 Answers2

7

Three suggestions:

  • Encase the text below the underbrace in a \mathclap directive so that its width doesn't affect the widths of the outermost and middle square-root symbols,

  • Use a \smash[b]{...} directive to ensure that the outer surds have the same (zero) depth, and

  • Insert a \\ directive at the end of the second line to ensure enough whitespace below the equation. (The \smash directive has "hidden" the underbrace material from TeX's ordinary spacing calculations.)

enter image description here

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{mathtools}
\begin{document}
\begin{align*}
\lim_{x\to0}f(f(f(x))) 
&= f\Bigl(f\bigl(\,\lim_{x\to0}f(x)\bigr)\Bigr) \\
&= \sqrt{\sqrt{\smash[b]{\underbrace{\sqrt{1+0}}_{%
   \mathclap{\text{because $f(x)$ is continuous}}}}}}\\ % <-- note the "\\"
\end{align*}
Some text following this two-line displayed equation.
\end{document}
Mico
  • 506,678
5

You can use a combination of \smashing the and \vphantom to ensure a proper vertical spacing and \mathclap to remove the horizontal intrusion of the under-script.

enter image description here

\documentclass{article}

\usepackage{mathtools,lipsum}

\begin{document}

\lipsum*[1]
\begin{align*}
  \lim_{x \to 0} f(f(f(x))) &= f\bigl( f\bigl( \lim_{x \to 0} f(x) \bigr)\bigr) \\
                            &= \sqrt{\sqrt{\underbrace{\sqrt{1+0}}_{\text{Because $f(x)$ is continuous}}}} \\
                            &= \sqrt{\sqrt{
                                 \vphantom{\sqrt{1}}
                                 \smash{\underbrace{\sqrt{1+0}}_{\mathclap{\text{Because $f(x)$ is continuous}}}}
                               }}
                               \vphantom{\underbrace{\sqrt{1}}_{\text{Because $f(x)$ is continuous}}}
\end{align*}
\lipsum[1]

\end{document}

An additional \vphantom at the end ensures the appropriate vertical spacing after the align is maintained if content follows it.

Werner
  • 603,163