If _ x → 0 α e ^ x + β e ^ - x + γ sinx x ^ 2 x = (2)/(3) , where...

Given,

$\lim_{x \to 0}\dfrac{\alpha{e}^{x} + \beta{e}^{- x} + \gamma sinx}{x\sin^{2}x} = \dfrac{2}{3}$

$⇒ \lim_{x \to 0}\dfrac{\alpha{e}^{x} + \beta{e}^{- x} + \gamma sinx}{{x}^{3}\dfrac{\sin^{2}x}{{x}^{2}}} = \dfrac{2}{3}$

$⇒ \lim_{x \to 0}\dfrac{\alpha{e}^{x} + \beta{e}^{- x} + \gamma sinx}{{x}^{3}} = \dfrac{2}{3}$

Now using expansion of given function in terms of $x$ we get,

$= \lim_{x \to 0}\dfrac{\alpha\left(1 + x + \dfrac{{x}^{2}}{2 !} + \dfrac{{x}^{3}}{3 !} + \ldots\right) + \beta\left(1 - x + \dfrac{{x}^{2}}{2 !} - \dfrac{{x}^{3}}{3 !} + \ldots\right) + \gamma\left(x - \dfrac{{x}^{3}}{3 !} + \ldots\right)}{{x}^{3}}$

Now for limit to exist constant terms should be zero

$⇒ a + \beta = 0 . . . . \left(1\right)$

Also coefficient of $x$ should be zero

$⇒ \alpha - \beta + \gamma = 0 . . . . . . . \left(2\right)$

Also coefficient of ${x}^{2}$ should be zero

$⇒ \dfrac{\alpha}{2} + \dfrac{\beta}{2} = 0 . . . . . . . \left(3\right)$

So limit becomes, $\lim_{x \to 0}\dfrac{{x}^{3}\left(\dfrac{\alpha}{3 !} - \dfrac{\beta}{3 !} - \dfrac{\gamma}{3 !}\right) + {x}^{4}\left(\dfrac{\alpha}{3 !} - \dfrac{\beta}{3 !} - \dfrac{\gamma}{3 !}\right)}{{x}^{3}} = \dfrac{2}{3}$

Now putting the value of limit we get,

$\dfrac{\alpha}{6} - \dfrac{\beta}{6} - \dfrac{\gamma}{6} = \dfrac{2}{3} . . . . . . . . \left(4\right)$

Now on solving equation $1, 2, 3 \text{ and } 4$ we get,

$⇒ \alpha = 1, \beta = - 1, \gamma = - 2$

Now putting these in all option we get, $\alpha{\beta}^{2} + \beta{\gamma}^{2} + \gamma{\alpha}^{2} + 3 = 0$