Correct Answer - `(7)`
`underset(xto0)lim(x^(2){betax-((betax)^(3))/(3!)+...})/(alphax-(x-(x^(3))/(3!)+...))=1`
`implies" "underset(xto0)lim(x^(3)(beta-(beta^(3)x^(2))/(3!)+...))/((alpha-1)x+(x^(3))/(3!)+(x^(5))/(5!)+...)=1`
`implies" "underset(xto0)lim(x^(2)(beta-(beta^(3)x^(2))/(3!)+...))/((alpha-1)+(x^(2))/(3!)+(x^(4))/(5!)+...)=1`
`implies" "underset(xto0)lim(beta-(beta^(3))/(3!)x^(2)...)/((1)/(3!)-(x^(2))/(5!)+...)=1`
`:." "beta=(1)/(3!)=(1)/(6)`
`:." "6(alpha+beta)=6(1+(1)/(6))=7`