For $\alpha,\beta,\gamma\in\mathbb{R}$, if $\displaystyle\lim_{x\to0}\frac{x^2\sin(\alpha x)+(\gamma-1)e^{x^2}}{\sin(2x)-\beta x}=3$, then $\beta+\gamma-\alpha$ is equal to:

Step 1: Analyze the Denominator

First, let's analyze the denominator of the given limit. For the limit to be finite and non-zero, the denominator must approach zero as $x \to 0$. We use the Taylor series expansion for $\sin(2x)$ around $x=0$, which is $\sin(u) = u - \frac{u^3}{3!} + \dots$. Substituting $u=2x$, we get the expansion for $\sin(2x)$. Then we combine it with $-\beta x$.

Step 2: Determine the value of $\beta$

If $2-\beta \neq 0$, the denominator would be of order $x$, while the numerator (as we will see) would be of order $x^2$ or higher, making the limit $0$ or $\infty$, not $3$. Therefore, for the limit to be finite and non-zero, the coefficient of $x$ in the denominator must be zero. This implies $2-\beta = 0$, so $\beta = 2$.

Step 3: Analyze the Numerator

Now, let's analyze the numerator. We use the Taylor series expansions for $\sin(\alpha x)$ and $e^{x^2}$ around $x=0$. The expansion for $\sin(u)$ is $u - \frac{u^3}{3!} + \dots$, so $\sin(\alpha x) = \alpha x - \frac{(\alpha x)^3}{3!} + \dots$. The expansion for $e^u$ is $1 + u + \frac{u^2}{2!} + \dots$, so $e^{x^2} = 1 + x^2 + \frac{(x^2)^2}{2!} + \dots$. We substitute these into the numerator and collect terms.

Step 4: Determine the value of $\gamma$

For the limit to be finite and non-zero, the lowest power of $x$ in the numerator must be at least $x^3$ (since the denominator is of order $x^3$ after setting $\beta=2$). This means the constant term and the $x^2$ term in the numerator must be zero. Thus, $\gamma-1 = 0$, which implies $\gamma = 1$.

Step 5: Evaluate the Limit and Determine $\alpha$

With $\beta=2$ and $\gamma=1$, the limit simplifies to $\displaystyle\lim_{x\to0}\frac{\alpha x^3 + O(x^4)}{-\frac{4x^3}{3} + O(x^5)}$. We can cancel $x^3$ from the numerator and denominator, and then evaluate the limit by taking the ratio of the coefficients of $x^3$. We set this equal to $3$ and solve for $\alpha$.

Step 6: Calculate $\beta+\gamma-\alpha$

Finally, we substitute the values of $\alpha$, $\beta$, and $\gamma$ that we found into the expression $\beta+\gamma-\alpha$ to get the final answer.

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