Blasius Boundary Layer
Note that for my third boundary condition, I used 10,000,000 as a placeholder for "infinity"...

ClearAll["Global`*"]
eqs={ 2f'''[η]+f[η] f''[η] ==0, f[0]==0,f'[0]==0,f'[10000000.0]==1};
f[η_]=f[η]/.NDSolve[eqs,f[η],{η,0,20}][[1]];g0[η_]=D[f[η],η];
 c=η/.FindRoot[g0[η]-0.99,{η,5}];η[g_]=InverseFunction[g0][g];
(* Done with Blasius already! He got his PhD for doing (and thinking of doing...) those above 3 lines *)
(* Now let's also add in Von Karman's approximation...*)
us0[η_]=(-((η^2)/30)+(Sqrt[(2/15)])  η);ηvk[g_]=InverseFunction[us0][g]//Quiet;
Blasius = Plot[η[g],{g,0,1},PlotRange->{{0,1.2},{0,(Sqrt[30])}},AspectRatio-> 1,PlotStyle->Directive[RGBColor[1,0,0],AbsoluteThickness[2]]];
vonKarman = Plot[ηvk[g],{g,0,1},PlotRange->{{0,1.2},{0,(Sqrt[30])}},AspectRatio-> 1,PlotStyle->Directive[RGBColor[0,0,1],AbsoluteThickness[2]]];
Print["The coefficient for δ at 0.99 \!\(\*SubscriptBox[\(U\), \(0\(\\ \)\)]\) is C = ",c, " so (δ/x)=",c," ((Subscript[Re, x])^(-0.5)) (Blasius)"]
Legended[Show[vonKarman,Blasius,AxesStyle->Black,FrameLabel->{"(SuperStar[u])","η"},Frame->True,ImageSize->300],Placed[LineLegend[{Blue,Red},{"von Karman","Blasius"},LabelStyle->Directive[{Bold}],LegendFunction->Framed],{{0.05,0.6},{0,0}}]]