`L=underset(xto0^(+))limx/a[(b)/(x)].`
`=underset(xto0^(+))limx/a(b/x-{b/x})," "`where `{.}` represents the functional part function
`=underset(xto0^(+))lim(b/a-x/a{b/x})`
`=b/a-1/aunderset(xto0^(+))lim({b/x}}/(1/x)`
Since `0lt{b/x}lt1 " and "underset(xto0^(+))lim1/x=oo,` we have
`L= b/a-0=b/a`