The rational expression is the following
[tex]\frac{\frac{x-1}{6xy}}{\frac{5}{2y}}[/tex]
By applying the sandwich rule:
we have
[tex]\frac{\frac{x-1}{6xy}}{\frac{5}{2y}}=\frac{(x-1)(2y)}{(6xy)(5)}[/tex]
which gives
[tex]\frac{\frac{x-1}{6xy}}{\frac{5}{2y}}=\frac{(x-1)(2y)}{(6xy)(5)}=\frac{(x-1)(2y)}{(2\cdot3xy)(5)}=\frac{(x-1)(2y)}{(3x\cdot2y)(5)}=\frac{x-1}{3x\cdot5}=\frac{x-1}{15x}[/tex]
then, the answer is option 2.
