Limit as r tends towards the radial center of a spherical quantum well

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Suppose $$R\left ( r \right )=\frac{u\left ( r \right )}{r}$$

Where $$u\left ( r \right )=A\sin\left ( kr \right )+B\cos\left ( kr \right )$$

The boundary condition is $$u\left ( a \right )=0 \wedge |u\left ( 0 \right )|< \infty$$

Indeed, putting the equation in order, we have

$$R\left ( r \right )=\frac{u\left ( r \right )}{r}=\frac{A\sin\left ( kr \right )}{r}+\frac{B\cos\left ( kr \right )}{r}$$

For the boundary condition to hold, we require that B=0 to prevent a blow up of the reciprocal 1 on 0-fair enough

But happens to $$\frac{A\sin\left ( kr \right )}{r}$$ as r tends to zero?

Clearly, the result is indeterminate. How does one get around explaining away the fact that it is alright that is not an issue-physical or mathematical.