We see that \(O\left ( \infty \right ) =0\). From the table this means only case 1 and 2 are possible. (since case 2 has no conditions on \(O\left ( \infty \right ) \) and only case 1 allows zero order for \(O\left ( \infty \right ) \)). We see there is a pole at \(x=0\) of order 2. This is allowed by both case 1 and case 2. Hence case 1,2 are possible and \(L=\left [ 1,2\right ] \)