Question

\( \lim_{x \to 0} \frac{x \cdot \cot(4x)}{\sin(2x) \cdot \cot^2(2x)} = ? \)

Answer 1

Given Limit:
We are tasked with solving the following limit: \[ \lim_{x \to 0} \frac{x \cdot \cot(4x)}{\sin(2x) \cdot \cot^2(2x)} \]

Step 1: Express Trigonometric Functions in Terms of Sine and Cosine:
Recall that: \[ \cot(x) = \frac{\cos(x)}{\sin(x)} \] So, we can rewrite the limit as: \[ \lim_{x \to 0} \frac{x \cdot \frac{\cos(4x)}{\sin(4x)}}{\sin(2x) \cdot \left( \frac{\cos(2x)}{\sin(2x)} \right)^2} \]

Step 2: Simplify the Expression:
Simplify the denominator and the trigonometric expressions: \[ \lim_{x \to 0} \frac{x \cdot \cos(4x)}{\sin(4x)} \cdot \frac{\sin^2(2x)}{\cos^2(2x) \cdot \sin(2x)} \] \[ = \lim_{x \to 0} \frac{x \cdot \cos(4x) \cdot \sin(2x)}{\sin(4x) \cdot \cos^2(2x)} \]

Step 3: Apply Small Angle Approximations:
As \( x \to 0 \), we can use the approximations \( \sin(x) \approx x \) and \( \cos(x) \approx 1 \). Using these approximations: \[ \sin(4x) \approx 4x, \quad \sin(2x) \approx 2x, \quad \cos(4x) \approx 1, \quad \cos(2x) \approx 1 \] Substituting these into the expression: \[ \lim_{x \to 0} \frac{x \cdot 1 \cdot 2x}{4x \cdot 1^2} = \lim_{x \to 0} \frac{2x^2}{4x} = \lim_{x \to 0} \frac{2x}{4} = \frac{1}{2} \]

Final Answer:
Therefore, the value of the limit is: \[ \lim_{x \to 0} \frac{x \cdot \cot(4x)}{\sin(2x) \cdot \cot^2(2x)} = 1 \]