1.2.3 Example 3

\begin{align*} r & =\frac {1}{x}-\frac {3}{16x^{2}}\\ & =\frac {16x-3}{16x^{2}}\end{align*}

There is pole at \(x=0\) of order \(2\). And \(O\left ( \infty \right ) =2-1=1\). Case 1 is not satisfied, since \(O\left ( \infty \right ) \) is not greater than 2. Also case 3 can not hold, since case 3 requires \(O\left ( \infty \right ) \) be at least order 2 and here it is 1. Only possibility left is case 2. There is one pole of order 2. Since case 2 have no conditions on \(O\left ( \infty \right ) \) to satisfy, then case 2 has been met. So this is case 2 only. \(L=\left [ 2\right ] \)