JEE Main 2025 Jan 24 Shift 1, Mathematics Q7: Methods to Evaluate Limits

The expression is of the form $\frac{\sqrt{A}-\sqrt{B}}{\sin x}$, with both $\sqrt{A}$ and $\sqrt{B}$ tending to $\sqrt{5}$ as $x\to 0$. Rationalise:

$L=\underset{x\to 0}{\lim }\frac{A-B}{\sin x(\sqrt{A}+\sqrt{B})}$, where $A=2{\cos }^{2}x+3\cos x$ and $B={\cos }^{2}x+\sin x+4$.

Compute the numerator:

$A-B={\cos }^{2}x+3\cos x-\sin x-4$

Use ${\cos }^{2}x=1-{\sin }^{2}x$:

$A-B=-{\sin }^{2}x+3\cos x-\sin x-3=-{\sin }^{2}x-\sin x+3(\cos x-1)$

For small $x$: $\sin x\approx x$, ${\sin }^{2}x\approx x^2$, $\cos x-1\approx -\frac{x^2}{2}$.

$A-B\approx -x^2-x+3\left(-\frac{x^2}{2}\right)=-x-\frac{5x^2}{2}$

Divide by $\sin x\approx x$: $\frac{A-B}{\sin x}\to -1$ as $x\to 0$.

Denominator factor $\sqrt{A}+\sqrt{B}\to 2\sqrt{5}$.

$L=\frac{-1}{2\sqrt{5}}$.

$\therefore$ Option (4).