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You have provided the following mathematical limit problem:

\[
\lim_{{x \to 0}} \frac{\sin x}{\cos(2x) - 1}
\]

Let's solve this step-by-step.

### Step 1: Evaluate the limit directly
When we directly substitute \(x = 0\):

\[
\frac{\sin(0)}{\cos(2(0)) - 1} = \frac{0}{\cos(0) - 1} = \frac{0}{1 - 1} = \frac{0}{0}
\]

This results in an indeterminate form of \( \frac{0}{0} \). Hence, we will need to apply a different method, such as **L'Hôpital's Rule** or **series expansion**.

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form \( \frac{0}{0} \), we can apply **L'Hôpital's Rule**, which states:

\[
\lim_{{x \to c}} \frac{f(x)}{g(x)} = \lim_{{x \to c}} \frac{f'(x)}{g'(x)}
\]

provided that both \( \lim_{{x \to c}} f(x) = 0 \) and \( \lim_{{x \to c}} g(x) = 0 \).

Let's compute the derivatives of the numerator and the denominator.

- Derivative of the numerator \( \sin(x) \) is \( \cos(x) \).
- Derivative of the denominator \( \cos(2x) - 1 \) is:
  \[
  \frac{d}{dx} [\cos(2x) - 1] = -2 \sin(2x)
  \]

Now, we can rewrite the limit as:

\[
\lim_{{x \to 0}} \frac{\cos(x)}{-2 \sin(2x)}
\]

### Step 3: Simplify the expression
Substitute \( x = 0 \) into the new expression:

\[
\frac{\cos(0)}{-2 \sin(2(0))} = \frac{1}{-2 \cdot 0} = \frac{1}{0}
\]

Since division by zero is undefined, this indicates that the limit does **not exist** and the value approaches infinity.

Thus, the limit is:
\[
\boxed{-\infty}
\]

Would you like a further breakdown or any clarifications?

### Follow-up questions:
1. What is L'Hôpital's Rule and when is it applied?
2. Can limits involving trigonometric functions often result in indeterminate forms?
3. How do Taylor series expansions work for trigonometric limits?
4. How does multiplying or dividing terms affect limit evaluation?
5. What are other methods to resolve indeterminate forms besides L'Hôpital's Rule?

**Tip:** When using L'Hôpital's Rule, always check the conditions for applying it by confirming the indeterminate form before differentiation.

Learn how to solve the limit lim x -> 0 (sin x) / (cos 2x - 1) using L'Hôpital's Rule. The problem involves key trigonometric functions and applies derivative rules for limit evaluation. Suitable for advanced high school or undergraduate students.

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