Let's solve the problem step by step.

We are tasked to evaluate the limit:

$\lim_{x \to 0} \left( (x - \sin 4x) \cdot \cot 2x \right)$

Step 1: Expand $\sin 4x$ using its Taylor series

The Taylor expansion of $\sin 4x$ around $x = 0$ is: $\sin 4x \approx 4x - \frac{(4x)^3}{6} + \mathcal{O}(x^5)$

Thus: $x - \sin 4x \approx x - \left( 4x - \frac{(4x)^3}{6} \right) = -3x + \frac{64x^3}{6}$

Step 2: Expand $\cot 2x$

The cotangent function is defined as: $\cot 2x = \frac{\cos 2x}{\sin 2x}$

For small $x$, we use the approximations: $\sin 2x \approx 2x, \quad \cos 2x \approx 1$

Thus: $\cot 2x \approx \frac{1}{2x}$

Step 3: Multiply the approximations

Now substitute the approximations into the given limit: $(x - \sin 4x) \cdot \cot 2x \approx \left( -3x + \frac{64x^3}{6} \right) \cdot \frac{1}{2x}$

Simplify: $\left( -3x + \frac{64x^3}{6} \right) \cdot \frac{1}{2x} = \frac{-3x}{2x} + \frac{\frac{64x^3}{6}}{2x}$

Simplify each term: \frac{-3x}{2x} = -\frac{3}{2}, \quad \frac{\frac{64x^3}{6}}{2x} = \frac{64x^2}{12x} = \frac{64x}{12} \to 0 \quad \text{(as x \to 0)}.

Thus, the dominant term is: $\lim_{x \to 0} \left( (x - \sin 4x) \cdot \cot 2x \right) = -\frac{3}{2}.$

Final Answer:

The correct option is 3) $-\frac{3}{2}$.

Would you like a detailed explanation of any step?
Here are 5 related questions to expand on this topic:

1. How is the Taylor series expansion derived for trigonometric functions?
2. Why does $\cot 2x$ simplify to $\frac{1}{2x}$ for small $x$?
3. How does the behavior of limits differ for higher-order terms as $x \to 0$?
4. What are some real-world applications of trigonometric limits in physics or engineering?
5. Can L’Hôpital’s Rule be applied to this problem as an alternative method?

Tip: For limits involving trigonometric functions, Taylor expansions or small-angle approximations often provide the fastest solutions!