\sqrt symbol with a vertical part

Question

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}

Answer 1

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}