Let `alpha, beta in R " be such that " lim_( x to 0) (x^(2)sin (beta x))/(ax - sin x) = 1."Then,"6(alpha+beta)"equals"`

Answer

`(7)`

Explanation

Here, ` underset( x to 0) lim (x^(2) sin (beta x))/(ax - sin x ) = 1`
` rArr underset( x to 0) lim (x^(2)(beta x-((betax)^(3))/(3!) +((betax)^(5))/(5!) - ...))/(ax - (x - x^(3)/(3!) + x^(5)/(5!) - ...))= 1`
`rArr underset(x to 0) lim (x^(3)(beta-(beta^(3) x^(2))/(3!) + (beta^(5)x^(4))/(5!) - ...))/((alpha-1)x+x^(3)/(3!) +x^(5)/(5!)-...)=1`
Limit exists only, when `alpha - 1 = 0`
`rArr" " alpha = 1` ....(i)
`:. underset( x to 0) lim (x^(3) (beta-(beta^(3)x^(2))/(3!)+(beta^(5)x^(4))/(5!)-...))/(x^(3)(1/(3!)-x^(2)/(5!)-...))= 1 `
` rArr" " 6beta=1` ...(ii)
From Eqs. (i) and (ii) , we get
` 6(alpha + beta) = 6alpha + 6 brta`
` = 6 + 1 = 7`