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












7














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


















  • 1




    Why not this?
    – Werner
    Nov 23 at 22:24






  • 5




    e^x is not equal to 1.
    – egreg
    Nov 23 at 22:32






  • 1




    @marmot I find it wrong: it is not an equality and will confuse the students, rather than help them.
    – egreg
    Nov 23 at 22:35










  • @marmot: Sure, there are conditions imposed on A and B - they both need to have their own respective limits, but that is besides the point here. Do you find the product of the limits is the limit of the products confusing? I find the =1 coloured, slanted superscript confusing.
    – Werner
    Nov 23 at 22:36






  • 2




    Please ignore whether the math is correct or not correct. This is simply what I have seen around in class, where if we are evaluating the limit at x=0, then clearly e^0 = 1, this is why I have the = 1
    – K Split X
    Nov 24 at 1:09
















7














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


















  • 1




    Why not this?
    – Werner
    Nov 23 at 22:24






  • 5




    e^x is not equal to 1.
    – egreg
    Nov 23 at 22:32






  • 1




    @marmot I find it wrong: it is not an equality and will confuse the students, rather than help them.
    – egreg
    Nov 23 at 22:35










  • @marmot: Sure, there are conditions imposed on A and B - they both need to have their own respective limits, but that is besides the point here. Do you find the product of the limits is the limit of the products confusing? I find the =1 coloured, slanted superscript confusing.
    – Werner
    Nov 23 at 22:36






  • 2




    Please ignore whether the math is correct or not correct. This is simply what I have seen around in class, where if we are evaluating the limit at x=0, then clearly e^0 = 1, this is why I have the = 1
    – K Split X
    Nov 24 at 1:09














7












7








7







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













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




math-mode stacking-symbols






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 23 at 21:59









K Split X

18224




18224








  • 1




    Why not this?
    – Werner
    Nov 23 at 22:24






  • 5




    e^x is not equal to 1.
    – egreg
    Nov 23 at 22:32






  • 1




    @marmot I find it wrong: it is not an equality and will confuse the students, rather than help them.
    – egreg
    Nov 23 at 22:35










  • @marmot: Sure, there are conditions imposed on A and B - they both need to have their own respective limits, but that is besides the point here. Do you find the product of the limits is the limit of the products confusing? I find the =1 coloured, slanted superscript confusing.
    – Werner
    Nov 23 at 22:36






  • 2




    Please ignore whether the math is correct or not correct. This is simply what I have seen around in class, where if we are evaluating the limit at x=0, then clearly e^0 = 1, this is why I have the = 1
    – K Split X
    Nov 24 at 1:09














  • 1




    Why not this?
    – Werner
    Nov 23 at 22:24






  • 5




    e^x is not equal to 1.
    – egreg
    Nov 23 at 22:32






  • 1




    @marmot I find it wrong: it is not an equality and will confuse the students, rather than help them.
    – egreg
    Nov 23 at 22:35










  • @marmot: Sure, there are conditions imposed on A and B - they both need to have their own respective limits, but that is besides the point here. Do you find the product of the limits is the limit of the products confusing? I find the =1 coloured, slanted superscript confusing.
    – Werner
    Nov 23 at 22:36






  • 2




    Please ignore whether the math is correct or not correct. This is simply what I have seen around in class, where if we are evaluating the limit at x=0, then clearly e^0 = 1, this is why I have the = 1
    – K Split X
    Nov 24 at 1:09








1




1




Why not this?
– Werner
Nov 23 at 22:24




Why not this?
– Werner
Nov 23 at 22:24




5




5




e^x is not equal to 1.
– egreg
Nov 23 at 22:32




e^x is not equal to 1.
– egreg
Nov 23 at 22:32




1




1




@marmot I find it wrong: it is not an equality and will confuse the students, rather than help them.
– egreg
Nov 23 at 22:35




@marmot I find it wrong: it is not an equality and will confuse the students, rather than help them.
– egreg
Nov 23 at 22:35












@marmot: Sure, there are conditions imposed on A and B - they both need to have their own respective limits, but that is besides the point here. Do you find the product of the limits is the limit of the products confusing? I find the =1 coloured, slanted superscript confusing.
– Werner
Nov 23 at 22:36




@marmot: Sure, there are conditions imposed on A and B - they both need to have their own respective limits, but that is besides the point here. Do you find the product of the limits is the limit of the products confusing? I find the =1 coloured, slanted superscript confusing.
– Werner
Nov 23 at 22:36




2




2




Please ignore whether the math is correct or not correct. This is simply what I have seen around in class, where if we are evaluating the limit at x=0, then clearly e^0 = 1, this is why I have the = 1
– K Split X
Nov 24 at 1:09




Please ignore whether the math is correct or not correct. This is simply what I have seen around in class, where if we are evaluating the limit at x=0, then clearly e^0 = 1, this is why I have the = 1
– K Split X
Nov 24 at 1:09










1 Answer
1






active

oldest

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12














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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    1 Answer
    1






    active

    oldest

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    active

    oldest

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    active

    oldest

    votes









    12














    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




























      12














      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


























        12












        12








        12






        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 Nov 23 at 22:47

























        answered Nov 23 at 22:12









        marmot

        84.6k495179




        84.6k495179






























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