Xjyuk

How is it possible to erase the line (keep it short) that goes up to infinity and use an infinity symbol to indicate so? (Do not touch the rest of the plot and the inserted numbers please if possible.)

documentclassstandalone
usepackageexpl3
ExplSyntaxOn
int_zero_new:N g__prg_map_int
ExplSyntaxOff
usepackageamsmath
usepackageamssymb
usepackagegraphicx
usepackagetikz
usetikzlibraryarrows,shapes,positioning
usepackagepgfplots
usetikzlibrarycalc
begindocument
newcommandddfrac[2]fracdisplaystyle #1displaystyle #2
pagestyleempty
centering
begintikzpicture[yscale=1,xscale=0.05]
 draw[thick,->,>=stealth] (-10,0) -- (45,0);
 draw[thick] (0,-2) -- (0,5);
 draw [smooth,very thick,domain=0.1:40,black] plot (x,((1*exp(-x/10))/(1-1*exp(-x/10)))+((0.00001*exp(-x/10))/((1-1*exp(-x/10))*(1-1*exp(-x/10)))));
 endtikzpicture
 enddocument 

edit: it seems that my solution is almost the same as from @marmoth (+1). In time of uploading mine, I was not aware about this.

I understand that you looking for something like this:

With use of clip function:

documentclass[margin=3mm]standalone
usepackagepgfplots
usetikzlibraryarrows,
 calc,
 positioning,
 shapes
begindocument
newcommandddfrac[2]fracdisplaystyle #1displaystyle #2
pagestyleempty
centering
begintikzpicture[yscale=1, xscale=0.1]
clip (-1,-1.75) rectangle (50,5);
 draw[-stealth] (-1,0) -- (45,0);
 draw[-stealth] (0,-1.75) -- (0,5);
 draw [domain=0.1:40, smooth,very thick] plot (x,((1*exp(-x/10))/(1-1*exp(-x/10)))+((0.00001*exp(-x/10))/((1-1*exp(-x/10))*(1-1*exp(-x/10)))));
 draw [domain=0.1:40, samples=51,red] plot (x,((1*exp(-x/10))/(1-1*exp(-x/10)))+((0.00001*exp(-x/10))/((1-1*exp(-x/10))*(1-1*exp(-x/10)))));
node[below right] at (1,5.1) $uparrowinfty$;
 endtikzpicture
 enddocument

Selection smooth in your draw plot options is wrong. Correct result is when you replace it with selected number of samples. for example with samples=51. See red curve.

I am assuming that you do not want to use pgfplots even though you are loading it. In that case you could clip and intersect.

documentclass[tikz]standalone
usetikzlibrarypositioning,intersections
begindocument
begintikzpicture[yscale=1,xscale=0.1]
 draw[thick,->,>=stealth] (-10,0) -- (45,0);
 draw[thick] (0,-2) -- (0,5);
 beginscope
 clip[name path global=box] (-10,-2) rectangle (45,5);
 draw[smooth,very thick,domain=0.1:40,samples=101,black,name path global=curve] 
 plot (x,((1*exp(-x/10))/(1-1*exp(-x/10)))+((0.00001*exp(-x/10))/((1-1*exp(-x/10))*(1-1*exp(-x/10)))));
 endscope
 draw[-stealth,very thick,name intersections=of=box and curve] 
 ([yshift=-0.1pt]intersection-1) -- ++(0,0.5) node[midway,right]$infty$;
endtikzpicture
enddocument 

With pgfplots it would be much easier.

documentclass[tikz]standalone
usetikzlibrarypositioning,intersections
usepackagepgfplots
pgfplotssetcompat=1.16
begindocument
begintikzpicture
 beginaxis[axis lines=middle,xtick=empty,ytick=empty,ymin=-2,ymax=5,
 xmin=-2,xmax=45]
 addplot[domain=0.1:45,samples=101,smooth,name path global=curve,
 very thick] ((1*exp(-x/10))/(1-1*exp(-x/10)))+((0.00001*exp(-x/10))/((1-1*exp(-x/10))*(1-1*exp(-x/10))));
 path[name path global=top] (-2,4.99) -- (45,4.99);
 endaxis
 draw[-stealth,very thick,name intersections=of=top and curve] 
 ([yshift=-0.1pt]intersection-1) -- ++(0,0.5) node[midway,right]$infty$;
endtikzpicture
enddocument 

Interestingly the plots are not the same, most likely because of the better precision of pgfplots. Mathematica seems to be on the side of pgfplots. So I increased the samples in the TikZ picture to match this (but I feel pgfplots is more elegant anyway).