Limit – Trigonometric Limit – JEE Main 24 January 2025 Shift 1

Solution:

Let $L = \lim_{x\rightarrow0}\frac{\sqrt{2 \cos^{2}x+3 \cos x}-\sqrt{\cos^{2}x+\sin x+4}}{\sin x}$

Rationalize the numerator:
$$ L = \lim_{x\rightarrow0}\frac{(2 \cos^{2}x+3 \cos x) – (\cos^{2}x+\sin x+4)}{\sin x (\sqrt{2 \cos^{2}x+3 \cos x}+\sqrt{2 \cos^{2}x+3 \cos x + \dots})} $$
(Note: The denominator term approaches $\sqrt{2+3}+\sqrt{1+0+4} = \sqrt{5}+\sqrt{5} = 2\sqrt{5}$)

Simplify the numerator:
$$ \text{Num} = \cos^2 x + 3\cos x – \sin x – 4 $$
$$ \text{Num} = (\cos^2 x + 3\cos x – 4) – \sin x $$
$$ \text{Num} = (\cos x + 4)(\cos x – 1) – \sin x $$

Substitute back into the limit:
$$ L = \frac{1}{2\sqrt{5}} \lim_{x\rightarrow0} \frac{(\cos x + 4)(\cos x – 1) – \sin x}{\sin x} $$
$$ L = \frac{1}{2\sqrt{5}} \lim_{x\rightarrow0} \left[ (\cos x + 4)\frac{\cos x – 1}{\sin x} – 1 \right] $$

Using half-angle formulas: $\cos x – 1 = -2\sin^2(x/2)$ and $\sin x = 2\sin(x/2)\cos(x/2)$:
$$ \frac{\cos x – 1}{\sin x} = \frac{-2\sin^2(x/2)}{2\sin(x/2)\cos(x/2)} = -\tan(x/2) $$
As $x \to 0$, $-\tan(x/2) \to 0$.

So the term inside the bracket becomes:
$$ [ (1+4)(0) – 1 ] = -1 $$

Final Result:
$$ L = \frac{1}{2\sqrt{5}} (-1) = -\frac{1}{2\sqrt{5}} $$

Ans. (4)