If lim _{x rightarrow 0} frac{alpha x e^{x}-b | vedclass

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(C) Given $\lim _{x \rightarrow 0} \frac{\alpha x e^{x}-\beta \log _{e}(1+x)+\gamma x^{2} e^{-x}}{x \sin ^{2} x}=10$.Since $\sin x \approx x$ as $x \r

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$3$

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(C) Given $\lim _{x \rightarrow 0} \frac{\alpha x e^{x}-\beta \log _{e}(1+x)+\gamma x^{2} e^{-x}}{x \sin ^{2} x}=10$.Since $\sin x \approx x$ as $x \rightarrow 0$,the expression becomes $\lim _{x \rightarrow 0} \frac{\alpha x e^{x}-\beta \log _{e}(1+x)+\gamma x^{2} e^{-x}}{x^{3}}=10$.Using Taylor series expansions:$e^{x} = 1+x+\frac{x^{2}}{2}+\dots$$\log _{e}(1+x) = x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\dots$$e^{-x} = 1-x+\frac{x^{2}}{2}-\dots$Substituting these:$\lim _{x}$ ${\rightarrow 0} \frac{\alpha x(1+x+\frac{x^{2}}{2}) - \beta(x-\frac{x^{2}}{2}+\frac{x^{3}}{3}) + \gamma x^{2}(1-x)}{x^{3}} = 10$$\lim _{x}$ ${\rightarrow 0} \frac{x(\alpha-\beta) + x^{2}(\alpha+\frac{\beta}{2}+\gamma) + x^{3}(\frac{\alpha}{2}-\frac{\beta}{3}-\gamma)}{x^{3}} = 10$For the limit to exist and equal $10$,coefficients of $x$ and $x^{2}$ must be $0$:$1) \alpha-\beta = 0 \Rightarrow \alpha = \beta$$2) \alpha+\frac{\beta}{2}+\gamma = 0$ $\Rightarrow \alpha+\frac{\alpha}{2}+\gamma = 0$ $\Rightarrow \gamma = -\frac{3\alpha}{2}$$3) \frac{\alpha}{2}-\frac{\beta}{3}-\gamma = 10$Substituting $\beta$ and $\gamma$ in $(3)$:$\frac{\alpha}{2}-\frac{\alpha}{3}-(-\frac{3\alpha}{2}) = 10$$\frac{3\alpha-2\alpha+9\alpha}{6} = 10$ $\Rightarrow \frac{10\alpha}{6} = 10$ $\Rightarrow \alpha = 6$.Thus,$\alpha = 6, \beta = 6, \gamma = -9$.$\alpha+\beta+\gamma = 6+6-9 = 3$.

If $\lim _{x \rightarrow 0} \frac{\alpha x e^{x}-\beta \log _{e}(1+x)+\gamma x^{2} e^{-x}}{x \sin ^{2} x}=10$,where $\alpha, \beta, \gamma \in R$,then the value of $\alpha+\beta+\gamma$ is:

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