If _ x → 0 [ e ^ x - β _ e ( 1 + x ) + γ x ^ 2 e ^ - x x ^ 2 x ] =...

We have,

$\lim_{x \to 0}\left[\dfrac{\alphax{e}^{x} - \beta{\log}_{e} ( 1 + x ) + \gamma{x}^{2}{e}^{- x}}{x\sin^{2}x}\right] = 10$

$⇒ \lim_{x \to 0}\left[\dfrac{\alphax\left(1 + x + \dfrac{{x}^{2}}{2 !} + . .\right) - \beta\left(x - \dfrac{{x}^{2}}{2} + \dfrac{{x}^{3}}{3} - . . .\right) + \gamma{x}^{2}\left(1 - x + \dfrac{{x}^{2}}{2 !} - . .\right)}{{x}^{3}\left(\dfrac{\sin^{2}x}{{x}^{2}}\right)}\right] = 10$

$⇒ \lim_{x \to 0}\left[\dfrac{\alphax\left(1 + x + \dfrac{{x}^{2}}{2}\right) - \beta\left(x - \dfrac{{x}^{2}}{2} + \dfrac{{x}^{3}}{3}\right) + \gamma{x}^{2}\left(1 - x\right)}{{x}^{3}}\right] = 10$

Neglecting higher powers.

$\lim_{x \to 0}\dfrac{x ( \alpha - \beta ) + {x}^{2}\left(\alpha + \dfrac{\beta}{2} + \gamma\right) + {x}^{3}\left(\dfrac{\alpha}{2} - \dfrac{\beta}{3} - \gamma\right)}{{x}^{3}} = 10$

For limit to exist

$\alpha - \beta = 0$

$\alpha + \dfrac{\beta}{2} + \gamma = 0$

$\dfrac{\alpha}{2} - \dfrac{\beta}{3} - \gamma = 10 \ldots \left(i\right)$

$\beta = \alpha, \gamma = - \dfrac{3\alpha}{2}$

Putting in \left(i\right)$, we get</p><p>$\dfrac{\alpha}{2} - \dfrac{\alpha}{3} + \dfrac{3\alpha}{2} = 10$</p><p>$⇒ \dfrac{\alpha}{6} + \dfrac{3\alpha}{2} = 10$</p><p>$⇒ \dfrac{\alpha + 9\alpha}{6} = 10$</p><p>$⇒ \alpha = 6$</p><p>Hence,</p><p>$\alpha = 6, \beta = 6, \gamma = - 9$</p><p>$⇒ \alpha + \beta + \gamma = 3