I get 1/8R. This probably the most obvious way to do it.
if u look at the circle, you can write L and H in terms of R and x . we dont have to worry what happens to R cos its constant, and x is the only variable.
so we have
[math]\frac{L}{2}=Rcosx[/math]
[math]L=2Rcosx[/math]
and
[math]K=Rsinx[/math]
therefore
[math]H=R-Rsinx[/math]
[math]H=R(1-sinx)[/math]
so
[math]\frac{H}{L^2}=\frac{R(1-sinx)}{4R^{2}\cos^{2}x}[/math]
[math]=\frac{(1-sinx)}{4R(1-\sin^{2}x)}[/math]
[math]=\frac{(1-sinx)}{4R(1-sinx)(1+sinx)}[/math]
[math]=\frac{1}{4R(1+sinx)}[/math]
Letting L tend to 0 is the same as letting x tend to Pi/2
therefore
limit of [math]=\frac{1}{4R(1+sinx)}[/math] as x tends to Pi/2 , by ALG
[math]=\frac{1}{4R(1+1)}[/math]
[math]=\frac{1}{8R}[/math]