Conditional limit

Question

I want to write ``conditional limit'' in math mode, which contains a right arrow, a down arrow and three parameters (the expression tends to A while B goes to C). I have tried to solve it with \mkern, but it's not the best.

MWE:

\documentclass{article}

\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage[english]{babel}

\usepackage{amsmath}

\def\limdownarrow{\qopname\relax m{\downarrow}}

\begin{document}

The expression goes to zero, while $\epsilon$ goes to zero:
$$\lim\limits_{\epsilon\to0} \left( F(x+\epsilon) - F(x-\epsilon) \right) = 0$$
Which could be written as:
$$F(x+\epsilon) - F(x-\epsilon) \to \mkern-19mu\limdownarrow_0^{\epsilon} 0 $$

\end{document}

Creates:

Is there any ``bulit-in'' operator for this quest?

Answer 1

Never seen such a notation.

\documentclass{article}
\usepackage{amsmath}

\makeatletter
\newcommand{\tendstoas}[2]{\mathrel{\tendstoas@{#1}{#2}}}
\newcommand{\tendstoas@}[2]{%
  \mathop{%
    \vphantom{\big\downarrow}%
    \ooalign{$\m@th\longrightarrow$\cr\hfil$\m@th\big\downarrow$\hfil}%
  }\limits^{#1}_{#2}%
}
\makeatother

\begin{document}

The expression goes to zero, while $\epsilon$ goes to zero:
\[
\lim_{\epsilon\to0} \left( F(x+\epsilon) - F(x-\epsilon) \right) = 0
\]
Which could be written as:
\[
F(x+\epsilon) - F(x-\epsilon) \tendstoas{\epsilon}{0} 0
\]

\end{document}

Notes: \limits is redundant in display math mode; $$ should never be used in LaTeX.

Alternatives:

Which could be written as:
\[
F(x+\epsilon) - F(x-\epsilon) \xrightarrow[\epsilon\to0]{} 0
\]
Which could be written as:
\[
F(x+\epsilon) - F(x-\epsilon) \xrightarrow{\epsilon\to0} 0
\]

Following the suggestion of Sebastiano to use a symbol from the STIX2 fonts, namely \tosa ⤩, but rotating it 45 degrees, here's a variant:

\DeclareFontFamily{U}{stix2sf}{}
\DeclareFontShape{U}{stix2sf}{m}{it}{<-> stix2-mathsfit}{}
\newcommand{\tendstoas}[2]{%
  \mathrel{%
    \mathop{%
      \rotatebox[origin=c]{45}{\usefont{U}{stix2sf}{m}{it}\symbol{"C2}}%
    }\limits^{#1}_{#2}%
  }%
}

Answer 2

\documentclass{article}
\usepackage{stackengine}
\newcommand\condlimit[2]{\mathrel{\ensurestackMath{%
  \stackinset{c}{-1.7pt}{c}{-.4pt}{\mathop{\downarrow}\limits^{#1}_{#2}}%
  {\rightarrow}}}}
\begin{document}
\[
1 + F(x+\epsilon) - F(x-\epsilon)\condlimit{0}{\epsilon} 1
\]
\end{document}

Answer 3

Alternative solution:

\newbox\xtobox
\setbox\xtobox=\hbox{\mathsurround=0pt$\longrightarrow$}
\setbox\xtobox=\hbox to\wd\xtobox{\rlap{\box\xtobox}\hss\lower.2ex\hbox{$\downarrow$}\hss}
\def\xto#1#2{\mathop{\copy\xtobox}\limits_{#1}^{#2}}

test:
$F(x+\epsilon) - F(x-\epsilon) \xto{\epsilon}{0} 0$

Differences from egregs solution: (1) The vertical arrow is not \big but it is smaller and lowered. This takes less vertical space in the math formula. (2) The operator box is pre-calculated in the preamble of the document, not calculated in every usage.