JEE Mains · Maths · STD 11 - 12. limits
For \(\alpha, \beta, \gamma, \in \mathbf{R}\), if \(\lim _{x \rightarrow 0} \frac{x^2 \sin \alpha x+(\gamma-1) e^{x^2}}{\sin 2 x-\beta x}=3\), then \(\beta+\gamma-\alpha\) is equal to:
- A 7
- B 4
- C 6
- D \(-1\)
Answer & Solution
Correct Answer
(A) 7
Step-by-step Solution
Detailed explanation
\begin{aligned} & \lim _{x \rightarrow 10} \frac{x^2(\alpha x)+(\gamma-1)\left(1+\frac{x^2}{1}\right)}{2 x-\frac{8 x^3}{6}-\beta x}=3 \\ & \lim _{x \rightarrow 0} \frac{(\gamma-1)+(\gamma-1) x^2+\alpha x^3}{(2-\beta) x-\frac{4}{3} x^3}=3 \\ & \gamma-1, \beta=2, \frac{-3…
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