Key Concepts and Formulas

• L'Hôpital's Rule: If a limit of a function of the form $\frac{f(x)}{g(x)}$ results in an indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ as $x \to c$, then $\lim\limits_{x \to c} \frac{f(x)}{g(x)} = \lim\limits_{x \to c} \frac{f'(x)}{g'(x)}$, provided the latter limit exists.
• Taylor Series Expansions (near x=0):
  • $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$
  • $e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \dots$
  • $\sin x = x - \frac{x^3}{3!} + \dots$
• Limit Properties: The limit of a sum/difference is the sum/difference of the limits, and the limit of a product is the product of the limits.

Step-by-Step Solution

Step 1: Analyze the Indeterminate Form The given limit is: $\lim\limits_{x \rightarrow 0} \frac{\alpha \mathrm{e}^{x}+\beta \mathrm{e}^{-x}+\gamma \sin x}{x \sin ^{2} x}=\frac{2}{3}$ As $x \to 0$, the denominator $x \sin^2 x \to 0 \cdot 0^2 = 0$. For the limit to be a finite non-zero value ($\frac{2}{3}$), the numerator must also approach 0. So, at $x=0$: $\alpha \mathrm{e}^{0}+\beta \mathrm{e}^{0}+\gamma \sin 0 = \alpha(1) + \beta(1) + \gamma(0) = \alpha + \beta$ Thus, for the indeterminate form $\frac{0}{0}$, we must have: $\alpha + \beta = 0 \quad (\text{Equation 1})$

Step 2: Apply L'Hôpital's Rule for the First Time Since the numerator is 0 at $x=0$, we can apply L'Hôpital's Rule. Let $f(x) = \alpha \mathrm{e}^{x}+\beta \mathrm{e}^{-x}+\gamma \sin x$ and $g(x) = x \sin^2 x$. Then $f'(x) = \alpha \mathrm{e}^{x} - \beta \mathrm{e}^{-x} + \gamma \cos x$. And $g'(x) = (1) \sin^2 x + x (2 \sin x \cos x) = \sin^2 x + 2x \sin x \cos x$. The limit becomes: $\lim\limits_{x \rightarrow 0} \frac{\alpha \mathrm{e}^{x} - \beta \mathrm{e}^{-x} + \gamma \cos x}{\sin^2 x + 2x \sin x \cos x}$ As $x \to 0$, the denominator $\sin^2 x + 2x \sin x \cos x \to 0^2 + 2(0)(0)(1) = 0$. For the limit to be finite, the numerator must also be 0 at $x=0$: $\alpha \mathrm{e}^{0} - \beta \mathrm{e}^{0} + \gamma \cos 0 = \alpha - \beta + \gamma$ So, we must have: $\alpha - \beta + \gamma = 0 \quad (\text{Equation 2})$

Step 3: Apply L'Hôpital's Rule for the Second Time We apply L'Hôpital's Rule again to the expression from Step 2. Let $f_1(x) = \alpha \mathrm{e}^{x} - \beta \mathrm{e}^{-x} + \gamma \cos x$ and $g_1(x) = \sin^2 x + 2x \sin x \cos x$. Then $f_1'(x) = \alpha \mathrm{e}^{x} + \beta \mathrm{e}^{-x} - \gamma \sin x$. And $g_1'(x) = 2 \sin x \cos x + (2 \sin x \cos x + 2x \cos^2 x - 2x \sin^2 x)$ $g_1'(x) = 4 \sin x \cos x + 2x (\cos^2 x - \sin^2 x) = 2 \sin(2x) + 2x \cos(2x)$. The limit becomes: $\lim\limits_{x \rightarrow 0} \frac{\alpha \mathrm{e}^{x} + \beta \mathrm{e}^{-x} - \gamma \sin x}{2 \sin(2x) + 2x \cos(2x)}$ As $x \to 0$, the denominator $2 \sin(0) + 2(0) \cos(0) = 0 + 0 = 0$. For the limit to be finite, the numerator must also be 0 at $x=0$: $\alpha \mathrm{e}^{0} + \beta \mathrm{e}^{0} - \gamma \sin 0 = \alpha + \beta$ This is consistent with Equation 1, $\alpha + \beta = 0$.

Step 4: Apply L'Hôpital's Rule for the Third Time We apply L'Hôpital's Rule again. Let $f_2(x) = \alpha \mathrm{e}^{x} + \beta \mathrm{e}^{-x} - \gamma \sin x$ and $g_2(x) = 2 \sin(2x) + 2x \cos(2x)$. Then $f_2'(x) = \alpha \mathrm{e}^{x} - \beta \mathrm{e}^{-x} - \gamma \cos x$. And $g_2'(x) = 2 \cos(2x) \cdot 2 + (2 \cos(2x) + 2x (-\sin(2x) \cdot 2))$ $g_2'(x) = 4 \cos(2x) + 2 \cos(2x) - 4x \sin(2x) = 6 \cos(2x) - 4x \sin(2x)$. The limit becomes: $\lim\limits_{x \rightarrow 0} \frac{\alpha \mathrm{e}^{x} - \beta \mathrm{e}^{-x} - \gamma \cos x}{6 \cos(2x) - 4x \sin(2x)}$ Now, we evaluate this limit by substituting $x=0$: Numerator: $\alpha \mathrm{e}^{0} - \beta \mathrm{e}^{0} - \gamma \cos 0 = \alpha - \beta - \gamma$. Denominator: $6 \cos(0) - 4(0) \sin(0) = 6(1) - 0 = 6$. So, the limit is: $\frac{\alpha - \beta - \gamma}{6}$ We are given that this limit is $\frac{2}{3}$. $\frac{\alpha - \beta - \gamma}{6} = \frac{2}{3}$ $\alpha - \beta - \gamma = 6 \cdot \frac{2}{3} = 4 \quad (\text{Equation 3})$

Step 5: Solve the System of Linear Equations We have the following system of equations:

1. $\alpha + \beta = 0$
2. $\alpha - \beta + \gamma = 0$
3. $\alpha - \beta - \gamma = 4$

From Equation 1, $\beta = -\alpha$. Substitute $\beta = -\alpha$ into Equation 2: $\alpha - (-\alpha) + \gamma = 0 \implies 2\alpha + \gamma = 0 \implies \gamma = -2\alpha$.

Substitute $\beta = -\alpha$ and $\gamma = -2\alpha$ into Equation 3: $\alpha - (-\alpha) - (-2\alpha) = 4$ $\alpha + \alpha + 2\alpha = 4$ $4\alpha = 4$ $\alpha = 1$ Now, find $\beta$ and $\gamma$: $\beta = -\alpha = -1$. $\gamma = -2\alpha = -2(1) = -2$.

So, the values are $\alpha = 1$, $\beta = -1$, and $\gamma = -2$.

Step 6: Verify the Values and Check the Options Let's check if these values satisfy the original limit condition. Numerator: $1 \cdot e^x + (-1) \cdot e^{-x} + (-2) \sin x = e^x - e^{-x} - 2 \sin x$. Using Taylor series: $e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \dots$ $e^{-x} = 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \dots$ $e^x - e^{-x} = (1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \dots) - (1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \dots) = 2x + \frac{2x^3}{6} + \dots = 2x + \frac{x^3}{3} + \dots$ $-2 \sin x = -2(x - \frac{x^3}{6} + \dots) = -2x + \frac{2x^3}{6} + \dots = -2x + \frac{x^3}{3} + \dots$ Numerator $\approx (2x + \frac{x^3}{3}) + (-2x + \frac{x^3}{3}) = \frac{2x^3}{3}$.

Denominator: $x \sin^2 x$. Using Taylor series for $\sin x \approx x$: $x \sin^2 x \approx x (x)^2 = x^3$. The limit is $\lim\limits_{x \to 0} \frac{\frac{2x^3}{3}}{x^3} = \frac{2}{3}$. This confirms our values are correct.

Now, let's check the given options:

(A) $\alpha^{2}+\beta^{2}+\gamma^{2}=6$ $(1)^2 + (-1)^2 + (-2)^2 = 1 + 1 + 4 = 6$ This statement is CORRECT.

(B) $\alpha \beta+\beta \gamma+\gamma \alpha+1=0$ $(1)(-1) + (-1)(-2) + (-2)(1) + 1 = -1 + 2 - 2 + 1 = 0$ This statement is CORRECT.

(C) $\alpha\beta^{2}+\beta \gamma^{2}+\gamma \alpha^{2}+3=0$ $(1)(-1)^2 + (-1)(-2)^2 + (-2)(1)^2 + 3 = (1)(1) + (-1)(4) + (-2)(1) + 3$ $= 1 - 4 - 2 + 3 = -2$ The expression evaluates to -2, not 0. So, the statement is: $-2 + 3 = 1 \neq 0$ This statement is INCORRECT.

(D) $\alpha^{2}-\beta^{2}+\gamma^{2}=4$ $(1)^2 - (-1)^2 + (-2)^2 = 1 - 1 + 4 = 4$ This statement is CORRECT.

The question asks for the statement that is NOT correct.

Common Mistakes & Tips

• Incorrectly applying L'Hôpital's Rule: Ensure that the form is indeed indeterminate ($\frac{0}{0}$ or $\frac{\infty}{\infty}$) before applying the rule.
• Algebraic Errors: Solving the system of linear equations requires careful calculation. Double-check your arithmetic.
• Taylor Series Approximation: While L'Hôpital's rule is more rigorous for finding the values of $\alpha, \beta, \gamma$, understanding Taylor series expansions can help verify the limit and understand the behavior of the functions near $x=0$.
• Denominator Differentiation: The derivative of $x \sin^2 x$ can be tricky. Remember to use the product rule and the chain rule correctly.

Summary

The problem involves evaluating a limit that results in an indeterminate form. We used L'Hôpital's Rule multiple times to simplify the expression and obtain a finite limit. The condition that the limit is $\frac{2}{3}$ led to a system of linear equations for $\alpha, \beta,$ and $\gamma$. Solving this system yielded $\alpha=1, \beta=-1, \gamma=-2$. We then substituted these values into the given options to determine which statement was not correct. Option (C) yielded a value of 1, not 0, making it the incorrect statement.

The final answer is \boxed{A}.