1.2.5 Example 5

\[ r=\frac {-5x+27}{36\left ( x-1\right ) ^{2}}\]

\(O\left ( \infty \right ) =2-1=1\). And \(r\) has pole at \(x=1\) of order 2. We see that \(O\left ( \infty \right ) \) is not satisfied for case 1 and case 3 (case 1 requires even or greater than 2 for \(O\left ( \infty \right ) \) and case 3 requires \(O\left ( \infty \right ) =2\).). So our only hope is case 2. Case 2 has no \(O\left ( \infty \right ) \) conditions. But it needs to have at least one pole of order 2 or a pole which is odd order and greater than 2. This is satisfied here, since pole is order 2. Hence only case 2 is possible. Hence \(L=[2]\). I do not understand why paper says all three cases are possible for this. This seems to be an error in the paper (1).