Here, $$\mathop {\lim }\limits_{x \to 0} {{{x^2}\sin (\beta x)} \over {\alpha x - \sin x}} = 1$$

$$\mathop {\lim }\limits_{x \to 0} {{{x^2}\left( {\beta x - {{{{(\beta x)}^3}} \over {3!}} + {{{{(\beta x)}^5}} \over {5!}} - ....} \right)} \over {\alpha x - \left( {x - {{{x^3}} \over {3!}} + {{{x^5}} \over {5!}} - ....} \right)}} = 1$$

$$ \Rightarrow \mathop {\lim }\limits_{x \to 0} {{{x^3}\left( {\beta - {{{\beta ^3}{x^2}} \over {3!}} + {{{\beta ^5}{x^4}} \over {5!}} - ....} \right)} \over {(\alpha - 1)x + {{{x^3}} \over {3!}} + {{{x^5}} \over {5!}} - ...}} = 1$$

Limit exists only, when $$\alpha$$ $$-$$ 1 = 0

$$\Rightarrow$$ $$\alpha$$ = 1 ...... (i)

$$\therefore$$ $$ \Rightarrow \mathop {\lim }\limits_{x \to 0} {{{x^3}\left( {\beta - {{{\beta ^3}{x^2}} \over {3!}} + {{{\beta ^5}{x^4}} \over {5!}} - ....} \right)} \over {{x^3}\left( {{1 \over {3!}} - {{{x^2}} \over {5!}} - ...} \right)}} = 1$$

$$\Rightarrow$$ 6$$\beta$$ = 1 ....... (ii)

From Eqs. (i) and (ii), we get

6($$\alpha$$ + $$\beta$$) = 6$$\alpha$$ + 6$$\beta$$

= 6 + 1

= 7