Question

For \( \alpha, \beta, \gamma \in \mathbb{R} \), if \[ \lim_{x \to 0} \frac{x^2 \sin(\alpha x) + (\gamma - 1)e^{x^2}}{\sin(2x - \beta x)} = 3, \] then \( \beta + \gamma - \alpha \) is equal to:

• A. 7
• B. 4
• C. 6
• D. –1

Hint:

Use Taylor expansions around \( x = 0 \) to evaluate complex limits.

Answer 1

The Correct Option is A

We are asked to find the value of \( \beta + \gamma - \alpha \) for \( \alpha, \beta, \gamma \in \mathbb{R} \), given the following limit:

\[ \lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1)e^{x^2}}{\sin 2x - \beta x} = 3 \]

Concept Used:

To solve this limit problem, we will use the Maclaurin series (Taylor series expansion around \(x=0\)) for the functions involved. The key series expansions are:

• \( \sin \theta = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \dots \)
• \( e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \dots \)

The limit is in the indeterminate form \( \frac{0}{0} \). For the limit to result in a finite, non-zero constant, the lowest power of \( x \) in the numerator's series expansion must be the same as the lowest power of \( x \) in the denominator's expansion.

Step-by-Step Solution:

Step 1: Determine the value of \( \gamma \).

As \( x \to 0 \), the denominator \( D(x) = \sin 2x - \beta x \to \sin(0) - \beta(0) = 0 \). For the limit to be finite, the numerator \( N(x) \) must also approach 0.

\[ \lim_{x \to 0} \left( x^2 \sin \alpha x + (\gamma - 1)e^{x^2} \right) = 0^2 \sin(0) + (\gamma - 1)e^0 = 0 + (\gamma - 1)(1) = \gamma - 1 \]

Setting the numerator to 0:

\[ \gamma - 1 = 0 \implies \gamma = 1 \]

Step 2: Expand the numerator and denominator using Maclaurin series.

With \( \gamma = 1 \), the numerator becomes:

\[ N(x) = x^2 \sin(\alpha x) \]

Using the expansion for \( \sin(\alpha x) \):

\[ N(x) = x^2 \left( \alpha x - \frac{(\alpha x)^3}{3!} + \dots \right) = \alpha x^3 - \frac{\alpha^3 x^5}{6} + \dots \]

The denominator is:

\[ D(x) = \sin(2x) - \beta x \]

Using the expansion for \( \sin(2x) \):

\[ D(x) = \left( 2x - \frac{(2x)^3}{3!} + \dots \right) - \beta x = (2 - \beta)x - \frac{8x^3}{6} + \dots = (2 - \beta)x - \frac{4}{3}x^3 + \dots \]

Step 3: Determine the value of \( \beta \).

The limit expression is now:

\[ \lim_{x \to 0} \frac{\alpha x^3 - \frac{\alpha^3 x^5}{6} + \dots}{(2 - \beta)x - \frac{4}{3}x^3 + \dots} = 3 \]

The lowest power of \( x \) in the numerator is \( x^3 \). For the limit to be a finite, non-zero value, the lowest power of \( x \) in the denominator must also be \( x^3 \). This means the coefficient of the lower power term (\(x\)) in the denominator must be zero.

\[ 2 - \beta = 0 \implies \beta = 2 \]

Step 4: Determine the value of \( \alpha \).

After setting \( \beta = 2 \), the limit becomes:

\[ \lim_{x \to 0} \frac{\alpha x^3 - \frac{\alpha^3 x^5}{6} + \dots}{-\frac{4}{3}x^3 + \dots} = 3 \]

We can evaluate this limit by taking the ratio of the coefficients of the lowest power terms (\(x^3\)):

\[ \frac{\alpha}{-\frac{4}{3}} = 3 \] \[ \alpha = 3 \times \left(-\frac{4}{3}\right) = -4 \]

Final Computation & Result:

We have found the values of the parameters:

• \( \alpha = -4 \)
• \( \beta = 2 \)
• \( \gamma = 1 \)

Now, we compute the required expression \( \beta + \gamma - \alpha \):

\[ \beta + \gamma - \alpha = 2 + 1 - (-4) = 3 + 4 = 7 \]

Thus, the value of \( \beta + \gamma - \alpha \) is 7.

Answer 2

We apply Taylor series expansions: \[ \lim_{x \to 0} \frac{x^2 (\alpha x) + (\gamma - 1)(1 + x^2 + \cdots) - 3}{\sin 2x - \beta x} \Rightarrow \lim_{x \to 0} \frac{\alpha x^3 + (\gamma - 1) x^2 - 2}{2x - \beta x} \] To get a finite limit of 3, match coefficients: \[ \gamma = 1, \quad \beta = 2, \quad \alpha = -4 \Rightarrow \beta + \gamma - \alpha = 7 \]