Question

For $ \alpha, \beta, \gamma \in \mathbb{R} $, if $$ \lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1)e^{x^2} - 3}{\sin 2x - \beta x} = 3, $$ 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 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: \[ \gamma = 1, \quad \beta = 2, \quad \alpha = -4 \Rightarrow \beta + \gamma - \alpha = 7 \]