Key Concepts and Formulas

• Limits of Indeterminate Forms: When evaluating a limit results in an indeterminate form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$, we can use L'Hôpital's Rule or Taylor series expansions.
• Taylor Series Expansions (around $x=0$):
  • $\sin u = u - \frac{u^3}{3!} + \frac{u^5}{5!} - \dots$
  • $e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \dots$
• L'Hôpital's Rule: If $\lim_{x \to c} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$, provided the latter limit exists.

Step-by-Step Solution

Step 1: Analyze the Limit and Identify the Indeterminate Form. The given limit is $\lim _{x \rightarrow 0} \frac{x^2 \sin \alpha x+(\gamma-1) \mathrm{e}^{x^2}}{\sin 2 x-\beta x}=3$. As $x \rightarrow 0$, the denominator $\sin 2x - \beta x \rightarrow \sin(0) - \beta(0) = 0$. For the limit to be a finite non-zero value (3 in this case), the numerator must also approach 0. Numerator as $x \rightarrow 0$: $x^2 \sin \alpha x + (\gamma-1) \mathrm{e}^{x^2} \rightarrow 0^2 \sin(0) + (\gamma-1) \mathrm{e}^0 = 0 + (\gamma-1)(1) = \gamma-1$. For the limit to be finite, we must have $\gamma-1 = 0$. Thus, $\gamma = 1$. This gives us the indeterminate form $\frac{0}{0}$.

Step 2: Simplify the Expression with $\gamma=1$ and Use Taylor Series Expansions. With $\gamma=1$, the limit becomes $\lim _{x \rightarrow 0} \frac{x^2 \sin \alpha x}{\sin 2 x-\beta x}=3$. Now, we use the Taylor series expansions for $\sin u$ around $u=0$. For the numerator: $\sin \alpha x = \alpha x - \frac{(\alpha x)^3}{3!} + \dots$ So, $x^2 \sin \alpha x = x^2 \left(\alpha x - \frac{\alpha^3 x^3}{3!} + \dots\right) = \alpha x^3 - \frac{\alpha^3 x^5}{3!} + \dots$ For the denominator: $\sin 2x = 2x - \frac{(2x)^3}{3!} + \dots$ So, $\sin 2x - \beta x = \left(2x - \frac{(2x)^3}{3!} + \dots\right) - \beta x = (2-\beta)x - \frac{8x^3}{3!} + \dots$

Substituting these into the limit expression: $\lim _{x \rightarrow 0} \frac{\alpha x^3 - \frac{\alpha^3 x^5}{3!} + \dots}{(2-\beta)x - \frac{8x^3}{3!} + \dots} = 3$

Step 3: Determine the Value of $\beta$. For the limit to be finite and non-zero, the lowest power of $x$ in the denominator must be cancelled by a term in the numerator, or the lowest power of $x$ in the denominator must be of a higher order than the lowest power of $x$ in the numerator if the limit is to be 0, or of a lower order if the limit is to be infinite. In our case, the lowest power of $x$ in the numerator is $x^3$ (assuming $\alpha \neq 0$). The lowest power of $x$ in the denominator is $x$ (from the term $(2-\beta)x$). If $2-\beta \neq 0$, then the denominator's lowest power is $x^1$, and the numerator's lowest power is $x^3$. The limit would be $\lim_{x \to 0} \frac{\alpha x^3}{(2-\beta)x} = \lim_{x \to 0} \frac{\alpha x^2}{2-\beta} = 0$, which contradicts the given limit of 3. Therefore, the coefficient of the lowest power of $x$ in the denominator must be zero. So, $2-\beta = 0$, which implies $\beta = 2$.

Step 4: Re-evaluate the Limit with $\beta=2$ and Determine $\alpha$. With $\beta=2$, the denominator becomes: $\sin 2x - 2x = \left(2x - \frac{(2x)^3}{3!} + \dots\right) - 2x = -\frac{8x^3}{6} + \dots = -\frac{4x^3}{3} + \dots$ The limit expression now is: $\lim _{x \rightarrow 0} \frac{\alpha x^3 - \frac{\alpha^3 x^5}{3!} + \dots}{-\frac{4x^3}{3} + \dots} = 3$

We can factor out $x^3$ from both the numerator and the denominator: $\lim _{x \rightarrow 0} \frac{x^3 \left(\alpha - \frac{\alpha^3 x^2}{3!} + \dots\right)}{x^3 \left(-\frac{4}{3} + \dots\right)} = 3$

Cancel $x^3$ and evaluate the limit: $\frac{\alpha - 0}{-\frac{4}{3}} = 3$ $\frac{\alpha}{-\frac{4}{3}} = 3$ $\alpha = 3 \times \left(-\frac{4}{3}\right)$ $\alpha = -4$.

Step 5: Calculate $\beta+\gamma-\alpha$. We have found $\gamma=1$, $\beta=2$, and $\alpha=-4$. Now, we compute $\beta+\gamma-\alpha$: $\beta+\gamma-\alpha = 2 + 1 - (-4)$ $\beta+\gamma-\alpha = 3 + 4$ $\beta+\gamma-\alpha = 7$.

Common Mistakes & Tips

• Incorrect Taylor Expansion: Ensure you use the correct Taylor series for $\sin u$ and $e^u$ and expand them to a sufficient order to resolve the indeterminate form. For this problem, expanding $\sin u$ up to the $u^3$ term and $e^u$ up to the $u^2$ term was sufficient for the numerator's initial analysis, but a more complete expansion of $\sin(\alpha x)$ and $\sin(2x)$ was needed for the denominator analysis.
• Forgetting the Numerator's Initial Condition: The initial step of setting the numerator to zero at $x=0$ (i.e., $\gamma-1=0$) is crucial. Skipping this might lead to an incorrect expression for the limit.
• Algebraic Errors: Be meticulous with algebraic manipulations, especially when dealing with fractions and negative signs, as these can easily lead to incorrect final answers.

Summary

The problem requires evaluating a limit that results in an indeterminate form. We first established that for the limit to be finite, the numerator must be zero at $x=0$, which yielded $\gamma=1$. Then, using Taylor series expansions for the trigonometric functions, we simplified the limit expression. By ensuring the denominator's lowest power of $x$ matched the numerator's lowest power of $x$ to avoid a zero limit, we found $\beta=2$. Finally, by comparing the coefficients of the $x^3$ terms in the numerator and denominator, we determined $\alpha=-4$. Calculating $\beta+\gamma-\alpha$ gave the final answer of 7.

Final Answer

The final answer is $\boxed{7}$.