To evaluate the limit:
\[\lim_{x \to 0} \csc{x} \left( \sqrt{2 \cos^2{x} + 3 \cos{x}} - \sqrt{\cos^2{x} + \sin{x} + 4} \right)\]We begin by examining each component of the expression as \( x \to 0 \).
Thus, the limit evaluates to:
\[\lim_{x \to 0} = -\frac{1}{2\sqrt{5}}\]The correct answer is \(- \frac{1}{2\sqrt{5}}\).
We are asked to evaluate the limit:
\( \lim_{x \to 0} \csc{x} \left( \sqrt{2 \cos^2{x} + 3 \cos{x}} - \sqrt{\cos^2{x} + \sin{x} + 4} \right) \)
The expression involves two square roots. To simplify the difference of square roots, we use the identity:
\( \sqrt{A} - \sqrt{B} = \frac{A - B}{\sqrt{A} + \sqrt{B}}. \)
Let \( A = 2 \cos^2{x} + 3 \cos{x} \) and \( B = \cos^2{x} + \sin{x} + 4 \). Thus, we can rewrite the original limit as:
\( \lim_{x \to 0} \csc{x} \left( \frac{A - B}{\sqrt{A} + \sqrt{B}} \right) \)
Now, calculate \( A - B \):
\( A - B = (2 \cos^2{x} + 3 \cos{x}) - (\cos^2{x} + \sin{x} + 4) \)
Simplifying this expression: \[ A - B = 2 \cos^2{x} + 3 \cos{x} - \cos^2{x} - \sin{x} - 4 \] \[ A - B = \cos^2{x} + 3 \cos{x} - \sin{x} - 4 \]
Now, as \( x \to 0 \), we use the small angle approximations:
Substituting these approximations in \( A - B \): \[ A - B \approx 1 + 3(1) - x - 4 = 0 - x. \]
Now, we approximate the denominator \( \sqrt{A} + \sqrt{B} \) at \( x \to 0 \). Using the small angle approximation again: \[ \sqrt{2 \cos^2{x} + 3 \cos{x}} \approx \sqrt{2 + 3} = \sqrt{5}, \] \[ \sqrt{\cos^2{x} + \sin{x} + 4} \approx \sqrt{1 + 0 + 4} = \sqrt{5}. \] Therefore, \( \sqrt{A} + \sqrt{B} \approx 2\sqrt{5} \).
Now, substitute all the approximations into the original expression: \[ \lim_{x \to 0} \csc{x} \left( \frac{0 - x}{2 \sqrt{5}} \right). \] Since \( \csc{x} = \frac{1}{\sin{x}} \approx \frac{1}{x} \) as \( x \to 0 \), we get: \[ \lim_{x \to 0} \frac{1}{x} \times \frac{-x}{2 \sqrt{5}} = -\frac{1}{2\sqrt{5}}. \]
The value of the limit is: \( - \frac{1}{2\sqrt{5}} \).
Let \( M \) and \( m \) respectively be the maximum and the minimum values of \( f(x) = \begin{vmatrix} 1 + \sin^2x & \cos^2x & 4\sin4x \\ \sin^2x & 1 + \cos^2x & 4\sin4x \\ \sin^2x & \cos^2x & 1 + 4\sin4x \end{vmatrix}, \quad x \in \mathbb{R} \) for \( x \in \mathbb{R} \). Then \( M^4 - m^4 \) is equal to:
If $ \theta \in [-2\pi,\ 2\pi] $, then the number of solutions of $$ 2\sqrt{2} \cos^2\theta + (2 - \sqrt{6}) \cos\theta - \sqrt{3} = 0 $$ is:
A thin transparent film with refractive index 1.4 is held on a circular ring of radius 1.8 cm. The fluid in the film evaporates such that transmission through the film at wavelength 560 nm goes to a minimum every 12 seconds. Assuming that the film is flat on its two sides, the rate of evaporation is:
The major product (A) formed in the following reaction sequence is
