The limit expression you provided is:

$\lim_{x \to 0} \left( \cos(2x) \right)^{\frac{\cot(x)}{\sin(x)}}$

To solve this, we'll break it down step by step.

Step 1: Simplifying the base

The base of the expression is $\cos(2x)$. Since $\cos(2x) \to \cos(0) = 1$ as $x \to 0$, we have:

$\cos(2x) \to 1 \text{ as } x \to 0.$

Step 2: Analyzing the exponent

The exponent is $\frac{\cot(x)}{\sin(x)}$, where $\cot(x) = \frac{\cos(x)}{\sin(x)}$. So, the exponent becomes:

$\frac{\cot(x)}{\sin(x)} = \frac{\cos(x)}{\sin^2(x)}.$

As $x \to 0$, we know that $\cos(0) = 1$ and $\sin(0) = 0$. To deal with this limit, we apply the standard small-angle approximations for trigonometric functions near zero:

• $\sin(x) \approx x$,
• $\cos(x) \approx 1$.

Thus, near zero:

$\frac{\cos(x)}{\sin^2(x)} \approx \frac{1}{x^2}.$

Step 3: Revisiting the limit

Now, we can rewrite the expression as:

$\lim_{x \to 0} \left( \cos(2x) \right)^{\frac{1}{x^2}}.$

Since $\cos(2x) \to 1$ as $x \to 0$, and we're raising 1 to a power that becomes very large, this simplifies to:

$1^{\infty}.$

This is an indeterminate form, so we need to apply L'Hopital's Rule or logarithmic methods to resolve it.

Step 4: Applying logarithms

Take the natural logarithm of the expression:

$y = \lim_{x \to 0} \left( \cos(2x) \right)^{\frac{1}{x^2}},$ $\ln y = \lim_{x \to 0} \frac{1}{x^2} \ln (\cos(2x)).$

Now, use the approximation $\cos(2x) \approx 1 - 2x^2$ for small $x$:

$\ln (\cos(2x)) \approx \ln(1 - 2x^2) \approx -2x^2.$

Thus, the limit becomes:

$\ln y = \lim_{x \to 0} \frac{1}{x^2} (-2x^2) = -2.$

Finally, exponentiate both sides to get $y$:

$y = e^{-2}.$

Final Answer:

$\lim_{x \to 0} \left( \cos(2x) \right)^{\frac{\cot(x)}{\sin(x)}} = e^{-2}.$

Would you like further clarification on any step, or additional details?

Here are 5 related questions you could explore:

1. How does L'Hopital's Rule work for indeterminate forms?
2. Why is the small-angle approximation useful for solving limits?
3. How do we handle limits involving trigonometric functions raised to exponents?
4. Can logarithmic differentiation be used to simplify other types of limits?
5. What are other common indeterminate forms, and how are they resolved?

Tip: Always remember to apply the small-angle approximations for trigonometric functions when dealing with limits as $x \to 0$, as they simplify many complex expressions.