JEE Main 2022 (Online) 26th July Evening ShiftQuestion 1
Let $\beta=\lim_{x\to0}\frac{\alpha x-(e^{3x}-1)}{\alpha x(e^{3x}-1)}$ for some $\alpha\in\mathbb{R}$. Then the value of $\alpha+\beta$ is :
Limits & Continuity
Solution
Concept. The denominator is of order $x^2$; for a finite limit the $x$ term of the numerator must vanish. Formulas used. (1) $e^{3x}-1=3x+\frac{9x^2}2+\dots$Step 1 - Orders. By (1), numerator $=(\alpha-3)x-\frac92x^2+\dots$, denominator $\sim3\alpha x^2$. Step 2 - Finite. Need $\alpha=3$. Step 3 - $\beta$. Numerator $\sim-\frac92x^2$, denominator $\sim9x^2$: $\beta=-\frac12$. Step 4. $\alpha+\beta=\frac52$. Answer: C