JEE Mathematics 2025 Question with Solution

Given: Evaluate $\lim_{x \to 0} \csc{x} \left( \sqrt{2 \cos^2{x} + 3 \cos{x}} - \sqrt{\cos^2{x} + \sin{x} + 4} \right)$

Find: The value of the limit and the correct option.

Use the identity

$$\sqrt{A} - \sqrt{B} = \frac{A-B}{\sqrt{A}+\sqrt{B}}$$

where

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

Then

$$\lim_{x \to 0} \csc{x} \left( \sqrt{A} - \sqrt{B} \right) = \lim_{x \to 0} \csc{x} \left( \frac{A-B}{\sqrt{A}+\sqrt{B}} \right)$$

Now,

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

so

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

As $x \to 0$, use the small angle approximations $\cos{x} \approx 1$ and $\sin{x} \approx x$. Then

$$A-B \approx 1 + 3(1) - x - 4 = -x$$

Also,

$$\sqrt{A} + \sqrt{B} \approx \sqrt{5} + \sqrt{5} = 2\sqrt{5}$$

Hence the expression becomes

$$\lim_{x \to 0} \csc{x} \left( \frac{-x}{2\sqrt{5}} \right)$$

Since

$$\csc{x} = \frac{1}{\sin{x}} \approx \frac{1}{x}$$

we get

$$\lim_{x \to 0} \frac{1}{x} \cdot \frac{-x}{2\sqrt{5}} = -\frac{1}{2\sqrt{5}}$$

Therefore, the value of the limit is $-\frac{1}{2\sqrt{5}}$. The correct option is D.

Given: $\lim_{x \to 0} \csc{x} \left( \sqrt{2 \cos^2{x} + 3 \cos{x}} - \sqrt{\cos^2{x} + \sin{x} + 4} \right)$

Find: The limit using direct first-order approximation.

As $x \to 0$,

$$\cos{x} \approx 1, \qquad \sin{x} \approx x$$

So,

$$\sqrt{2\cos^2{x} + 3\cos{x}} \approx \sqrt{5}$$

and

$$\sqrt{\cos^2{x} + \sin{x} + 4} \approx \sqrt{5+x}$$

Now use

$$\sqrt{5+x} \approx \sqrt{5}\left(1 + \frac{x}{10}\right)$$

Therefore,

$$\sqrt{5} - \sqrt{5+x} \approx \sqrt{5} - \sqrt{5}\left(1 + \frac{x}{10}\right) = -\frac{\sqrt{5}\,x}{10}$$

Also, $\csc{x} \approx \frac{1}{x}$. Hence

$$\lim_{x \to 0} \frac{1}{x} \left(-\frac{\sqrt{5}\,x}{10}\right) = -\frac{\sqrt{5}}{10} = -\frac{1}{2\sqrt{5}}$$

Therefore, the value of the limit is $-\frac{1}{2\sqrt{5}}$. The correct option is D.