JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Limits

For the limit to exist and equal 3, the denominator must approach zero, so the numerator must also approach zero.

As $x\to 0$: The numerator approaches $\gamma -1$, so $\gamma =1$.

For the denominator: $\sin 2x-\beta x\approx 2x-\beta x=(2-\beta )x$ as $x\to 0$. For this to approach zero, we need $\beta =2$.

Expanding using Taylor series:

• $\sin \alpha x\approx \alpha x$
• $e^{x^2}\approx 1+x^2$
• $\sin 2x\approx 2x-\frac{8x^3}{6}$

The limit becomes:


This gives $\alpha =-4$.

Therefore, $\beta +\gamma -\alpha =2+1-(-4)=7$.