Invullen geeft:
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\( \mbox{R} = \lim_{\, radius \rightarrow 0} \,\, \frac{6}{radius^ 2} \left ( 1 \, - \, \frac{circumference}{2 \pi \, radius} \right ) \)
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\( \mbox{R} = \lim_{\, r \rightarrow 0} \,\, \frac{6}{r^ 2} \left ( 1 \, - \, \frac{\mbox{O}(r)}{2 \pi \, r} \right ) \)
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\( \mbox{R} = \lim_{\, r \rightarrow 0} \,\, \frac{6}{r^ 2} \left ( 1 \, - \, \frac{ 2 \pi \, \mathcal{R} \sin \left (\frac{r}{\mathcal{R}} \right )}{2 \pi \, r} \right ) \)
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\( \mbox{R} = \lim_{\, r \rightarrow 0} \,\, \frac{6}{r^ 2} \left ( 1 \, - \, \frac{\sin \left (\frac{r}{\mathcal{R}} \right )}{ \frac{r}{\mathcal{R}}} \right ) \)
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\( \mbox{R} = \lim_{\, \varphi \rightarrow 0} \,\, \frac{6}{( \varphi \mathcal{R})^ 2} \left ( 1 \, - \, \frac{\sin(\varphi)}{ \varphi} \right ) \)
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\( \mbox{R} = \frac{6}{\mathcal{R}^2} \cdot \lim_{\, \varphi \rightarrow 0} \,\, \frac{1}{ \varphi^ 2} \left ( 1 \, - \, \frac{\sin(\varphi)}{ \varphi} \right ) \)
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Ik weet niet of dit de bedoeling is....