To solve the given limit:

$\lim_{x \to 0} \left( \cos(2x) \cdot \frac{1}{\sin^2(3x)} \right)$

Step 1: Simplify the expression

First, notice that the expression inside the limit can be written as:

$\lim_{x \to 0} \frac{\cos(2x)}{\sin^2(3x)}$

Step 2: Evaluate each component at the limit

We know that as $x \to 0$:

• $\cos(2x)$ approaches $\cos(0) = 1$.
• $\sin(3x)$ approaches $\sin(0) = 0$.

However, this creates an indeterminate form $\frac{1}{0^2}$, which tends to infinity. Therefore, we need to use the small-angle approximation.

Step 3: Apply small-angle approximations

For small values of $x$, the following approximations hold:

• $\cos(2x) \approx 1 - \frac{(2x)^2}{2} = 1 - 2x^2$
• $\sin(3x) \approx 3x$

Thus, we can rewrite the limit as:

$\lim_{x \to 0} \frac{1 - 2x^2}{(3x)^2}$

Step 4: Simplify the expression further

Simplifying the expression inside the limit:

$\frac{1 - 2x^2}{9x^2}$

This can be separated as:

$\lim_{x \to 0} \left(\frac{1}{9x^2} - \frac{2x^2}{9x^2}\right) = \lim_{x \to 0} \left(\frac{1}{9x^2} - \frac{2}{9}\right)$

As $x \to 0$, the first term $\frac{1}{9x^2}$ tends to infinity, while the second term $\frac{2}{9}$ is a constant.

Step 5: Conclusion

Since $\frac{1}{9x^2}$ dominates as $x$ approaches 0, the entire limit tends towards infinity:

$\lim_{x \to 0} \frac{\cos(2x)}{\sin^2(3x)} = \infty$

Thus, the limit is infinite.

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Would you like further details or have any questions? Here are some related questions:

1. How do you apply the small-angle approximation to other trigonometric functions?
2. What are the conditions for using L'Hôpital's Rule in limits?
3. How can limits approaching infinity be interpreted in real-world problems?
4. What are some common indeterminate forms, and how are they resolved?
5. How would the result change if we evaluated the limit of a different function, like $\frac{\sin(2x)}{\sin^2(3x)}$?

Tip: Always check if a limit yields an indeterminate form; it can guide you toward the appropriate method of resolution.