# [Tex/LaTex] \sqrt symbol with a vertical part

math-modesymbols

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}


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}
[#1]{$\displaystyle\overline{#2}$}}}
$\lim \limits_{k \to \infty} \mysqrt[1pt]{\frac{1+(-1)^k}{3+\dfrac{1}{k^2}+\dfrac{8}{k^3}}}= \sqrt{\dfrac{1+1}{3+0+0}}=\sqrt{\dfrac{2}{3}}$
$\mysqrt[.4pt]{\frac{\frac{x}{y}}{\frac{a}{b}}}$ 