# \sqrt symbol with a vertical part

It seems that for a given height, the \sqrt symbol "switches" it's type, and it becomes vertical at some part, like here:

But I want it to be like the second one. What is the solution for this?

EDIT: I've cut this line out of my .tex file. It looks like this:

\documentclass[12pt]{article}

\usepackage{enumerate}
\usepackage{lmodern}
\usepackage{amsmath}

\begin{document}

\begin{enumerate}[1.]

\item[\textbf{MO:}] {}

\begin{enumerate}[a)]

\item{
$$\lim \limits_{k \to \infty} \sqrt{\frac{1+(-1)^k}{3+\dfrac{1}{k^2}+\dfrac{8}{k^3}}}= \sqrt{\dfrac{1+1}{3+0+0}}=\sqrt{\dfrac{2}{3}}$$}

\end{enumerate}

\end{enumerate}

\end{document}

• Please add a minimal working example (MWE) that illustrates your problem. It will be much easier for us to reproduce your situation and find out what the issue is when we see compilable code, starting with \documentclass{...} and ending with \end{document}. – jub0bs Aug 12 '13 at 19:41
• I've done it, hope it gives enough information about it. – Sphery Aug 12 '13 at 19:52
• the computer modern font (the default font with (la)tex) has several sizes of radical signs, but after a certain point, it can't enlarge them any more, and instead uses the version with a vertical part that can be extended ad infinitum with vertical line segments. you've exceeded the size limit. – barbara beeton Aug 12 '13 at 19:54
• Your code, as it is, is not compilable. Please load the required packages. Also, refrain from using  for delimiting display math; see tex.stackexchange.com/questions/503/why-is-preferable-to for more detail. – jub0bs Aug 12 '13 at 19:58
• Unrelated note: \item does not take a mandatory argument, so you do not have to write \item{item text}, it is sufficient with \item item text. – Torbjørn T. Aug 13 '13 at 0:46

## 1 Answer

EDITED to provide second cut (with optional argument for fine-tuning the vertical height of the scaled root symbol). What I do here is take a standard size sqrt sign and stretch it vertically to match the height of its argument, using the scalerel package. The fine-tuning of the height is done with the stackengine package.

\documentclass{article}
\usepackage{amsmath}
\usepackage{stackengine}[2013-10-15]
\usepackage{scalerel}
\newcommand\mysqrt[2][0pt]{\stretchrel{\sqrt{}}{\addstackgap%
[#1]{$\displaystyle\overline{#2}$}}}
\begin{document}
$\lim \limits_{k \to \infty} \mysqrt[1pt]{\frac{1+(-1)^k}{3+\dfrac{1}{k^2}+\dfrac{8}{k^3}}}= \sqrt{\dfrac{1+1}{3+0+0}}=\sqrt{\dfrac{2}{3}}$
$\mysqrt[.4pt]{\frac{\frac{x}{y}}{\frac{a}{b}}}$
\end{document}


• This also seems to be a solution for my integral notation problems, as you generally provided a solution! Thank you! – Sphery Aug 14 '13 at 11:41
• @Sphery Glad to help. Note that \stretchrel can take an optional argument that specifies the minimum aspect ratio (in %), so that if the vertical stretch is too large, horizontal expansion occurs. Thus, \stretchrel[400]{...} limits the stretched aspect ratio of the root sign to 4.00 – Steven B. Segletes Aug 14 '13 at 12:43
• @Sphery Please see revised answer. – Steven B. Segletes Oct 20 '13 at 2:11