Continuity and Differentiability
Limits Involving Trigonometric Functions
JEE Advanced 2016 Paper 1
Grade 12
Question:
Let $\alpha, \beta \in \mathbb{R}$ be such that $\lim_{x \to 0} \frac{x^2 \sin(\beta x)}{\alpha x - \sin x} = 1$. Find $6(\alpha + \beta)$.
Step-by-Step Solution
Key Concept: Expand: numerator $\approx \beta x^3$, denominator $\approx (\alpha - 1)x + x^3/6 - \dots$ . For the limit to be finite (and equal to 1) at $x \to 0$, the leading term of denominator must cancel with numerator: need $\alpha = 1$. Then denominator $\approx x^3/6$, giving $\beta x^3/(x^3/6) = 6\beta = 1 \Rightarrow \beta = 1/6$. So $6(\alpha + \beta) = 6(1 + 1/6) = 7$.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: 7