If $$\lim_{x \to 0} \left[\frac{\alpha x e^x - \beta \log_e(1+x) + \gamma x^2 e^{-x}}{x \sin^2 x}\right] = 10$$, $$\alpha, \beta, \gamma \in R$$, then the value of $$\alpha + \beta + \gamma$$ is ___.

We need $$\lim_{x \to 0} \frac{\alpha x e^x - \beta \ln(1+x) + \gamma x^2 e^{-x}}{x \sin^2 x} = 10$$.

Since $$\sin^2 x \approx x^2$$ near $$x = 0$$, the denominator behaves like $$x^3$$ for small $$x$$. For the limit to be finite and nonzero, the numerator must behave like $$x^3$$ near $$x = 0$$, meaning the coefficients of $$x^0$$, $$x^1$$, and $$x^2$$ in the numerator must vanish.

Expanding using Taylor series: $$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots$$ $$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$$ $$e^{-x} = 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \cdots$$

So: $$\alpha x e^x = \alpha\left(x + x^2 + \frac{x^3}{2} + \cdots\right)$$ $$\beta \ln(1+x) = \beta\left(x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots\right)$$ $$\gamma x^2 e^{-x} = \gamma\left(x^2 - x^3 + \cdots\right)$$

The numerator expanded is: $$(\alpha - \beta)x + \left(\alpha + \frac{\beta}{2} + \gamma\right)x^2 + \left(\frac{\alpha}{2} - \frac{\beta}{3} - \gamma\right)x^3 + \cdots$$

For the limit to exist (and equal 10), the coefficients of $$x^1$$ and $$x^2$$ must be zero: $$\alpha - \beta = 0 \implies \alpha = \beta \quad \cdots (i)$$ $$\alpha + \frac{\beta}{2} + \gamma = 0 \quad \cdots (ii)$$

Using $$\alpha = \beta$$ in (ii): $$\alpha + \frac{\alpha}{2} + \gamma = 0 \implies \gamma = -\frac{3\alpha}{2}$$.

The limit then equals: $$\lim_{x \to 0} \frac{\left(\frac{\alpha}{2} - \frac{\beta}{3} - \gamma\right)x^3}{x \cdot x^2} = \frac{\alpha}{2} - \frac{\alpha}{3} + \frac{3\alpha}{2} = \frac{\alpha}{2} - \frac{\alpha}{3} + \frac{3\alpha}{2} = 2\alpha - \frac{\alpha}{3} = \frac{5\alpha}{3}$$

Setting this equal to 10: $$\frac{5\alpha}{3} = 10 \implies \alpha = 6$$. Then $$\beta = 6$$ and $$\gamma = -9$$.

Therefore $$\alpha + \beta + \gamma = 6 + 6 - 9 = 3$$.