How do I get a tilted equal sign for an equation?











up vote
3
down vote

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For example, this is what I am looking for:



enter image description here



Usually for limits where the value is 0 or 1 of a particular expression (e^x) in this case, you might see something like a tilted 45 degrees equal sign, and the value of that expression.



Not sure how to do it here.



Maybe something like a overbrace without the actual overbrace visible?










share|improve this question






















  • Why not this?
    – Werner
    3 hours ago










  • e^x is not equal to 1.
    – egreg
    3 hours ago










  • @Werner In general it is not true that lim_{xto0} Acdot B=(lim_{xto0} A)cdot (lim_{xto0} B). In this case it is, though, but I would certainly not want to typeset this as I find it somewhat misleading.
    – marmot
    3 hours ago










  • @egreg I thought this is precisely the reason why the OP does not want overbraces.
    – marmot
    3 hours ago










  • @marmot I find it wrong: it is not an equality and will confuse the students, rather than help them.
    – egreg
    3 hours ago















up vote
3
down vote

favorite












For example, this is what I am looking for:



enter image description here



Usually for limits where the value is 0 or 1 of a particular expression (e^x) in this case, you might see something like a tilted 45 degrees equal sign, and the value of that expression.



Not sure how to do it here.



Maybe something like a overbrace without the actual overbrace visible?










share|improve this question






















  • Why not this?
    – Werner
    3 hours ago










  • e^x is not equal to 1.
    – egreg
    3 hours ago










  • @Werner In general it is not true that lim_{xto0} Acdot B=(lim_{xto0} A)cdot (lim_{xto0} B). In this case it is, though, but I would certainly not want to typeset this as I find it somewhat misleading.
    – marmot
    3 hours ago










  • @egreg I thought this is precisely the reason why the OP does not want overbraces.
    – marmot
    3 hours ago










  • @marmot I find it wrong: it is not an equality and will confuse the students, rather than help them.
    – egreg
    3 hours ago













up vote
3
down vote

favorite









up vote
3
down vote

favorite











For example, this is what I am looking for:



enter image description here



Usually for limits where the value is 0 or 1 of a particular expression (e^x) in this case, you might see something like a tilted 45 degrees equal sign, and the value of that expression.



Not sure how to do it here.



Maybe something like a overbrace without the actual overbrace visible?










share|improve this question













For example, this is what I am looking for:



enter image description here



Usually for limits where the value is 0 or 1 of a particular expression (e^x) in this case, you might see something like a tilted 45 degrees equal sign, and the value of that expression.



Not sure how to do it here.



Maybe something like a overbrace without the actual overbrace visible?







math-mode stacking-symbols






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 3 hours ago









K Split X

15524




15524












  • Why not this?
    – Werner
    3 hours ago










  • e^x is not equal to 1.
    – egreg
    3 hours ago










  • @Werner In general it is not true that lim_{xto0} Acdot B=(lim_{xto0} A)cdot (lim_{xto0} B). In this case it is, though, but I would certainly not want to typeset this as I find it somewhat misleading.
    – marmot
    3 hours ago










  • @egreg I thought this is precisely the reason why the OP does not want overbraces.
    – marmot
    3 hours ago










  • @marmot I find it wrong: it is not an equality and will confuse the students, rather than help them.
    – egreg
    3 hours ago


















  • Why not this?
    – Werner
    3 hours ago










  • e^x is not equal to 1.
    – egreg
    3 hours ago










  • @Werner In general it is not true that lim_{xto0} Acdot B=(lim_{xto0} A)cdot (lim_{xto0} B). In this case it is, though, but I would certainly not want to typeset this as I find it somewhat misleading.
    – marmot
    3 hours ago










  • @egreg I thought this is precisely the reason why the OP does not want overbraces.
    – marmot
    3 hours ago










  • @marmot I find it wrong: it is not an equality and will confuse the students, rather than help them.
    – egreg
    3 hours ago
















Why not this?
– Werner
3 hours ago




Why not this?
– Werner
3 hours ago












e^x is not equal to 1.
– egreg
3 hours ago




e^x is not equal to 1.
– egreg
3 hours ago












@Werner In general it is not true that lim_{xto0} Acdot B=(lim_{xto0} A)cdot (lim_{xto0} B). In this case it is, though, but I would certainly not want to typeset this as I find it somewhat misleading.
– marmot
3 hours ago




@Werner In general it is not true that lim_{xto0} Acdot B=(lim_{xto0} A)cdot (lim_{xto0} B). In this case it is, though, but I would certainly not want to typeset this as I find it somewhat misleading.
– marmot
3 hours ago












@egreg I thought this is precisely the reason why the OP does not want overbraces.
– marmot
3 hours ago




@egreg I thought this is precisely the reason why the OP does not want overbraces.
– marmot
3 hours ago












@marmot I find it wrong: it is not an equality and will confuse the students, rather than help them.
– egreg
3 hours ago




@marmot I find it wrong: it is not an equality and will confuse the students, rather than help them.
– egreg
3 hours ago










1 Answer
1






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oldest

votes

















up vote
4
down vote













If you only want to draw one of those, this might be a bit of an overkill, but if you plan to do several annotations of that kind and wish to have access to more fancy features, this might be a reasonable way to go.



documentclass{article}
usepackage{amsmath}
usepackage{tikz}
usetikzlibrary{tikzmark}
begin{document}
[lim_{xto0}tikzmarknode{ex}{mathrm{e}^x}left(2+4xright)~=~2]
begin{tikzpicture}[overlay,remember picture]
path[red] ([xshift=1pt,yshift=1pt]ex.north east) -- ++(45:{width("=")*1pt-1pt})
node[midway,sloped]{$=$} node[above right=-2pt]{$1$};
end{tikzpicture}
end{document}


enter image description here



Arguably somewhat clearer alternatives include



documentclass{article}
usepackage{mathtools}
usepackage{tikz}
usetikzlibrary{tikzmark}
begin{document}
[lim_{xto0}tikzmarknode{ex}{mathrm{e}^x}left(2+4xright)~=~2]
begin{tikzpicture}[overlay,remember picture]
draw[red,->] ([xshift=1pt,yshift=1pt]ex.north east) --
++(45:{width("$scriptstyle xto0$")*1pt})
node[midway,sloped,above]{$scriptstyle xto0$} node[above right=-1pt]{$1$};
end{tikzpicture}
end{document}


enter image description here



which illustrates what I mean by "more fancy options".






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    1 Answer
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    up vote
    4
    down vote













    If you only want to draw one of those, this might be a bit of an overkill, but if you plan to do several annotations of that kind and wish to have access to more fancy features, this might be a reasonable way to go.



    documentclass{article}
    usepackage{amsmath}
    usepackage{tikz}
    usetikzlibrary{tikzmark}
    begin{document}
    [lim_{xto0}tikzmarknode{ex}{mathrm{e}^x}left(2+4xright)~=~2]
    begin{tikzpicture}[overlay,remember picture]
    path[red] ([xshift=1pt,yshift=1pt]ex.north east) -- ++(45:{width("=")*1pt-1pt})
    node[midway,sloped]{$=$} node[above right=-2pt]{$1$};
    end{tikzpicture}
    end{document}


    enter image description here



    Arguably somewhat clearer alternatives include



    documentclass{article}
    usepackage{mathtools}
    usepackage{tikz}
    usetikzlibrary{tikzmark}
    begin{document}
    [lim_{xto0}tikzmarknode{ex}{mathrm{e}^x}left(2+4xright)~=~2]
    begin{tikzpicture}[overlay,remember picture]
    draw[red,->] ([xshift=1pt,yshift=1pt]ex.north east) --
    ++(45:{width("$scriptstyle xto0$")*1pt})
    node[midway,sloped,above]{$scriptstyle xto0$} node[above right=-1pt]{$1$};
    end{tikzpicture}
    end{document}


    enter image description here



    which illustrates what I mean by "more fancy options".






    share|improve this answer



























      up vote
      4
      down vote













      If you only want to draw one of those, this might be a bit of an overkill, but if you plan to do several annotations of that kind and wish to have access to more fancy features, this might be a reasonable way to go.



      documentclass{article}
      usepackage{amsmath}
      usepackage{tikz}
      usetikzlibrary{tikzmark}
      begin{document}
      [lim_{xto0}tikzmarknode{ex}{mathrm{e}^x}left(2+4xright)~=~2]
      begin{tikzpicture}[overlay,remember picture]
      path[red] ([xshift=1pt,yshift=1pt]ex.north east) -- ++(45:{width("=")*1pt-1pt})
      node[midway,sloped]{$=$} node[above right=-2pt]{$1$};
      end{tikzpicture}
      end{document}


      enter image description here



      Arguably somewhat clearer alternatives include



      documentclass{article}
      usepackage{mathtools}
      usepackage{tikz}
      usetikzlibrary{tikzmark}
      begin{document}
      [lim_{xto0}tikzmarknode{ex}{mathrm{e}^x}left(2+4xright)~=~2]
      begin{tikzpicture}[overlay,remember picture]
      draw[red,->] ([xshift=1pt,yshift=1pt]ex.north east) --
      ++(45:{width("$scriptstyle xto0$")*1pt})
      node[midway,sloped,above]{$scriptstyle xto0$} node[above right=-1pt]{$1$};
      end{tikzpicture}
      end{document}


      enter image description here



      which illustrates what I mean by "more fancy options".






      share|improve this answer

























        up vote
        4
        down vote










        up vote
        4
        down vote









        If you only want to draw one of those, this might be a bit of an overkill, but if you plan to do several annotations of that kind and wish to have access to more fancy features, this might be a reasonable way to go.



        documentclass{article}
        usepackage{amsmath}
        usepackage{tikz}
        usetikzlibrary{tikzmark}
        begin{document}
        [lim_{xto0}tikzmarknode{ex}{mathrm{e}^x}left(2+4xright)~=~2]
        begin{tikzpicture}[overlay,remember picture]
        path[red] ([xshift=1pt,yshift=1pt]ex.north east) -- ++(45:{width("=")*1pt-1pt})
        node[midway,sloped]{$=$} node[above right=-2pt]{$1$};
        end{tikzpicture}
        end{document}


        enter image description here



        Arguably somewhat clearer alternatives include



        documentclass{article}
        usepackage{mathtools}
        usepackage{tikz}
        usetikzlibrary{tikzmark}
        begin{document}
        [lim_{xto0}tikzmarknode{ex}{mathrm{e}^x}left(2+4xright)~=~2]
        begin{tikzpicture}[overlay,remember picture]
        draw[red,->] ([xshift=1pt,yshift=1pt]ex.north east) --
        ++(45:{width("$scriptstyle xto0$")*1pt})
        node[midway,sloped,above]{$scriptstyle xto0$} node[above right=-1pt]{$1$};
        end{tikzpicture}
        end{document}


        enter image description here



        which illustrates what I mean by "more fancy options".






        share|improve this answer














        If you only want to draw one of those, this might be a bit of an overkill, but if you plan to do several annotations of that kind and wish to have access to more fancy features, this might be a reasonable way to go.



        documentclass{article}
        usepackage{amsmath}
        usepackage{tikz}
        usetikzlibrary{tikzmark}
        begin{document}
        [lim_{xto0}tikzmarknode{ex}{mathrm{e}^x}left(2+4xright)~=~2]
        begin{tikzpicture}[overlay,remember picture]
        path[red] ([xshift=1pt,yshift=1pt]ex.north east) -- ++(45:{width("=")*1pt-1pt})
        node[midway,sloped]{$=$} node[above right=-2pt]{$1$};
        end{tikzpicture}
        end{document}


        enter image description here



        Arguably somewhat clearer alternatives include



        documentclass{article}
        usepackage{mathtools}
        usepackage{tikz}
        usetikzlibrary{tikzmark}
        begin{document}
        [lim_{xto0}tikzmarknode{ex}{mathrm{e}^x}left(2+4xright)~=~2]
        begin{tikzpicture}[overlay,remember picture]
        draw[red,->] ([xshift=1pt,yshift=1pt]ex.north east) --
        ++(45:{width("$scriptstyle xto0$")*1pt})
        node[midway,sloped,above]{$scriptstyle xto0$} node[above right=-1pt]{$1$};
        end{tikzpicture}
        end{document}


        enter image description here



        which illustrates what I mean by "more fancy options".







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 2 hours ago

























        answered 3 hours ago









        marmot

        77.5k487164




        77.5k487164






























             

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