Question

`lim_(x -> 0) ((sin(alpha + beta) x + sin(alpha - beta)x + sin 2alpha x))/(cos 2betax - cos 2alphax) * x`

Answer 1

Given, `lim_(x -> 0) ([sin (alpha + beta)x + sin(alpha - beta)x + sin 2alpha * x])/(cos 2betax - cos 2alphax) * x`

= `lim_(x -> 0) ([2 sin alpha x * cos betax + sin 2alpha * x]*x)/(2sin (alpha + beta)x * sin(alpha - beta)x)`   ......`[(because sin C + sin D = 2 sin  (C + D)/2 * cos  (C - D)/2),(cos C - cos D = - 2 sin  (C + D)/2 * sin  (C - D)/2)]`

= `lim_(x -> 0) ([2 sin alphax * cos betax + 2 sin alphax * cos alphax] * x)/(2 sin(alpha + beta)x * sin(alpha - beta)x)`

= `lim_(x -> 0) (2 sin alphax (cos betax + cos alphax)*x)/(2 sin(alpha + beta)x * sin(alpha - beta)x)`

= `lim_(x -> 0) (sin alphax[2 cos((alpha + beta)/2)x * cos((alpha - beta)/2)x]*x)/(sin(alpha + beta)x * sin(alpha - beta)x)`

= `lim_(x -> 0) (sin alphax [2 cos ((alpha + beta)/2)x * cos((alpha - beta)/2)x]*x)/(2 sin((alpha + beta)/2)x * cos((alpha + beta)/2)x) * 2 sin ((alpha - beta)/2)x * cos((alpha - beta)/2)x`   .......`[(because  cos C + cos D = 2 cos  (C + d)/2 cos  (C - D)/2),("and"  sin 2x = 2 sin x cos x)]`

= `lim_(x -> 0) (sin alphax * x)/(2sin((alpha + beta)/2)x sin((alpha - beta)/2) * x)`

= `lim_(x -> 0) 1/2 ((sin alphax)/(alphax) * (alphax) * x)/([(sin ((alpha + beta)/2) x)/(((alpha + beta)/2) * x) xx ((alpha + beta)/2) * x][(sin((alpha - beta)/2)*x)/(((alpha - beta)/2)* x) xx ((alpha - beta))/2 * x])`

= `1/2 * (alphax^2)/(((alpha + beta)/2)x * ((alpha - beta)/2)x)`

= `1/2[alpha/(((alpha + beta)/2)((alpha - beta)/2))]`

= `1/2 * (4alpha)/(alpha^2 - beta^2)`

= `(2alpha)/(alpha^2 - beta^2)`