JEE Main 2025 Apr 2 Shift 1, Mathematics Q13: Methods to Evaluate Limits

Use the series expansions near $x=0$:

$\sin (\alpha x)=\alpha x-\frac{{\alpha }^3{x}^3}{6}+O(x^5)$, so $x^2\sin (\alpha x)=\alpha x^3+O(x^5)$.

$e^{x^2}=1+x^2+\frac{x^4}{2}+O(x^6)$, so $(\gamma -1)e^{x^2}=(\gamma -1)+(\gamma -1)x^2+O(x^4)$.

$\sin 2x=2x-\frac{(2x{)}^3}{6}+O(x^5)=2x-\frac{4x^3}{3}+O(x^5)$.

So the numerator is $(\gamma -1)+(\gamma -1)x^2+\alpha x^3+O(x^4)$ and the denominator is $(2-\beta )x-\frac{4x^3}{3}+O(x^5)$.

For a finite limit at $x=0$, the constant term in the numerator must vanish:

$$\gamma -1=0\Rightarrow \gamma =1$$

The numerator now starts with $x^2$, and for the limit to be finite we also need the leading $x$ term in the denominator to vanish:

$$2-\beta =0\Rightarrow \beta =2$$

Now the leading orders are $(\gamma -1)x^2+\alpha x^3=\alpha x^3$ over $-\frac{4}{3}{x}^3$, so:

$$\underset{x\to 0}{\lim }\frac{\alpha x^3}{-\frac{4}{3}{x}^3}=-\frac{3\alpha }{4}=3\Rightarrow \alpha =-4$$

Finally:

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

$\therefore$ Option (1) is correct.