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: \( - \frac{1}{2\sqrt{5}} \).
If \( \alpha>\beta>\gamma>0 \), then the expression \[ \cot^{-1} \beta + \left( \frac{1 + \beta^2}{\alpha - \beta} \right) + \cot^{-1} \gamma + \left( \frac{1 + \gamma^2}{\beta - \gamma} \right) + \cot^{-1} \alpha + \left( \frac{1 + \alpha^2}{\gamma - \alpha} \right) \] is equal to:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,