The problem you've uploaded is:

$\lim_{x \to 0} \frac{\sin x}{\cos 2x - 1}$

To solve this limit, let's proceed step by step:

Step 1: Simplify the denominator

Notice that the denominator contains $\cos 2x - 1$, and we can use the trigonometric identity for cosine:

$\cos 2x = 1 - 2\sin^2 x$

Substituting this identity into the denominator:

$\cos 2x - 1 = (1 - 2\sin^2 x) - 1 = -2\sin^2 x$

So, the limit becomes:

$\lim_{x \to 0} \frac{\sin x}{-2\sin^2 x}$

Step 2: Simplify the expression

Now, we can cancel out one factor of $\sin x$ from both the numerator and denominator:

$\lim_{x \to 0} \frac{1}{-2\sin x}$

Step 3: Apply the limit

As $x \to 0$, $\sin x$ approaches 0. Therefore, the expression becomes:

$\lim_{x \to 0} \frac{1}{-2 \times 0} = -\infty$

Thus, the limit does not exist in the usual finite sense because the expression tends to negative infinity.

Final Answer:

$\lim_{x \to 0} \frac{\sin x}{\cos 2x - 1} = -\infty$

Would you like a further breakdown of any part of this solution, or have any questions?

Here are 5 related questions you might want to explore:

1. What happens if we apply L'Hopital's Rule to this limit?
2. How can we approach limits that tend to infinity?
3. What are other important trigonometric identities used in limit problems?
4. Can we handle this limit using series expansion instead of identities?
5. How does the behavior of the sine function affect the limit as $x$ approaches 0?

Tip: Always check if a trigonometric identity can simplify the problem before applying other methods like L'Hopital's Rule!