Limits – LH rule – JEE Main 2 Apr 2025 Shift 1

Solution:

Given limit:
$$ \lim_{x\rightarrow0}\frac{x^{2}\sin(\alpha x)+(\gamma-1)e^{x^{2}}}{\sin(2x)-\beta x}=3 $$

For the limit to exist, the constant term in the numerator must be zero (since the denominator $\to 0$ as $x \to 0$).
$$ \lim_{x\rightarrow 0} (x^2 \sin(\alpha x) + (\gamma-1)e^{x^2}) = 0 $$
$$ 0 + (\gamma – 1) = 0 \Rightarrow \gamma = 1 $$

Substitute $\gamma=1$ into the limit. Also, for small $x$, $\sin(\alpha x) \approx \alpha x$.
$$ \lim_{x\rightarrow0}\frac{x^{2}(\alpha x)}{\sin(2x)-\beta x} = \lim_{x\rightarrow0}\frac{\alpha x^3}{\sin(2x)-\beta x} $$

This is a $\left(\frac{0}{0}\right)$ form. Applying L’Hopital’s Rule (differentiating numerator and denominator):
$$ \lim_{x\rightarrow0}\frac{3\alpha x^2}{2\cos(2x)-\beta} = 3 $$

For the limit to be finite, the denominator must be zero at $x=0$ (since numerator is 0).
$$ 2\cos(0) – \beta = 0 \Rightarrow 2 – \beta = 0 \Rightarrow \beta = 2 $$

Now the limit becomes:
$$ \lim_{x\rightarrow0}\frac{3\alpha x^2}{2\cos(2x)-2} $$

Applying L’Hopital’s Rule again:
$$ \lim_{x\rightarrow0}\frac{6\alpha x}{-4\sin(2x)} $$

Using the standard limit $\lim_{x\to0} \frac{x}{\sin(kx)} = \frac{1}{k}$:
$$ \frac{6\alpha}{-4} \cdot \frac{1}{2} = 3 $$
$$ \frac{6\alpha}{-8} = 3 \Rightarrow \frac{3\alpha}{-4} = 3 $$
$$ 3\alpha = -12 \Rightarrow \alpha = -4 $$

Finally, calculate the value:
$$ \beta + \gamma – \alpha = 2 + 1 – (-4) = 7 $$

Ans. (1)