Correct Answer - C
`underset(xrarr0)(lim)(f(1-cosx))/(g(x)sin^(2)x)`
`" "=underset(xrarr0)(lim)(f(2sin^(2).(x)/(2)))/(g(x)(2sin^(2).(x)/(2)))xx((2sin^(2).(x)/(2))^(2))/(4(sin^(2)(x)/(2))(cos^(2).(x)/(2)))`
`" "=underset(xrarr0)(lim)(a)/(g(x))xxtan^(2).(x)/(2)`
`" "a underset(xrarr0)(lim)(((x)/(2))^(2))/(g(x))xx(tan^(2).(x)/(2))/(((x)/(2))^(2))`
`" "=a underset(xrarr0)(lim)(x^(2))/(4g(x))`
`therefore" "underset(xrarr0)(lim)(x^(2))/(g(x))=(4b)/(a)`
Now, `underset(xrarr0)(lim)(g(2sin^(2))x)/(x^(4))`
`" "underset(xrarr0)(lim)(g(sin^(2)x))/((2sin^(2)x)^(2))xx((2sin^(2)x)^(2))/(x^(4))`
`" "=(a)/(4b)xx4=(a)/(b)`