For $\alpha, \beta, \gamma \in \mathbf{R}$, if $\lim _\limits{x \rightarrow 0} \frac{x^2 \sin \alpha x+(\gamma-1) \mathrm{e}^{x^2}}{\sin 2 x-\beta x}=3$, then $\beta+\gamma-\alpha$ is equal to :

<p>$$\begin{aligned} & \text { At } x \rightarrow 0 \\ & \sin 2 x-\beta x \rightarrow 0 \\ & \Rightarrow \quad \frac{0}{0} \text { form } \\ & \Rightarrow \quad(\gamma-1) e^0+0 \sin (\alpha x) \rightarrow 0 \\ & \Rightarrow \quad(\gamma-1)=0 \\ & \Rightarrow \quad \gamma=1 \\ & \Rightarrow \quad \lim _{x \rightarrow 0} \frac{x^2 \sin (\alpha x)}{(\sin 2 x-\beta x)}=3 \end{aligned}$$</p> <p>$\Rightarrow \lim _{x \rightarrow 0} \frac{x^2\left[\alpha x-\frac{(\alpha x)^3}{3!}+\frac{(\alpha x)^5}{5!}-\cdots\right]}{\left[(2 x)-\frac{(2 x)^3}{3!}+\frac{(2 x)^5}{5!}\right]-\beta x}$</p> <p>$\Rightarrow \lim _{x \rightarrow 0} \frac{\alpha x^3-\frac{\alpha^3 x^5}{3!}+\frac{\alpha^5 x^7}{5!}-\cdots}{x(2-\beta)-\frac{8 x^3}{6}+\frac{2^5 \cdot x^5}{5!}-\cdots}=3$</p> <p>$$\begin{aligned} \Rightarrow & 2-\beta=0 \text { and } \frac{\alpha}{\frac{-8}{6}}=3 \\ \Rightarrow & \beta=2 \\ & \alpha=3\left(-\frac{8}{6}\right)=-4 \\ \Rightarrow & \gamma=1, \beta=2, \alpha=-4 \\ \Rightarrow & \beta+\gamma-\alpha=7 \end{aligned}$$</p>