Limits and Derivatives | JEE Main 2025

JEE MAIN 2025

02-04-2025 SHIFT-1

Question

For $\alpha ,\beta ,\gamma \in R$ , if $\mathop {\lim }\limits_{x \to 0} \frac{{{x^2}\sin \alpha x + (\gamma - 1){e^{{x^2}}}}}{{\sin 2x - \beta x}} = 3$, then $\beta + \gamma - \alpha $ is equal to :

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✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

Question Tags

JEE Main

Mathematics

Easy

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