Cara Elisa,
poiché \[KB=AB-AK=2r-AD\cos 2x=2r-2r{{\cos }^{2}}2x=2r{{\sin }^{2}}2x\] \[HB=CB\sin x=2r{{\sin }^{2}}x\] si ha:
\[\underset{D\to B}{\mathop{\lim }}\,\frac{KB}{HB}=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\sin }^{2}}2x}{{{\sin }^{2}}x}=4\underset{x\to 0}{\mathop{\lim }}\,{{\left( \frac{x}{\sin x}\frac{\sin 2x}{2x} \right)}^{2}}=4\quad .\]
Massimo Bergamini
