If $$\lim\limits_{x \rightarrow 0} \frac{\alpha \mathrm{e}^{x}+\beta \mathrm{e}^{-x}+\gamma \sin x}{x \sin ^{2} x}=\frac{2}{3}$$, where $\alpha, \beta, \gamma \in \mathbf{R}$, then which of the following is NOT correct?

<p>$$\mathop {\lim }\limits_{x \to 0} {{\alpha {e^x} + \beta {e^{ - x}} + \gamma \sin x} \over {x{{\sin }^2}x}} = {2 \over 3}$$</p> <p>$\Rightarrow \alpha + \beta = 0$ (to make indeterminant form) ...... (i)</p> <p>Now,</p> <p>$$\mathop {\lim }\limits_{x \to 0} {{\alpha {e^x} - \beta {e^{ - x}} + \gamma \cos x} \over {3{x^2}}} = {2 \over 3}$$ (Using L-H Rule)</p> <p>$\Rightarrow \alpha - \beta + \gamma = 0$ (to make indeterminant form) ...... (ii)</p> <p>Now,</p> <p>$$\mathop {\lim }\limits_{x \to 0} {{\alpha {e^x} + \beta {e^{ - x}} - \gamma \sin x} \over {6x}} = {2 \over 3}$$ (Using L-H Rule)</p> <p>$\Rightarrow {{\alpha - \beta + \gamma } \over 6} = {2 \over 3}$</p> <p>$\Rightarrow \alpha - \beta + \gamma = 4$ ...... (iii)</p> <p>$\Rightarrow \gamma = - 2$</p> <p>and (i) + (ii)</p> <p>$2\alpha = - \gamma$</p> <p>$\Rightarrow \alpha = 1$ and $\beta = - 1$</p> <p>and $$\alpha {\beta ^2} + \beta {\gamma ^2} + \gamma {\alpha ^2} + 3 = 1 - 4 - 2 + 3 = - 2$$</p>