We are asked to evaluate the limit:

$\lim_{x \to 0} \sin(2x) \cdot \tan(-10x)$

Let's break it down step by step.

Step 1: Simplify the Trigonometric Functions

1. The sine function is: $\sin(2x)$ As $x$ approaches 0, $2x$ also approaches 0. We know from the small angle approximation that: $\sin(2x) \approx 2x \text{ as } x \to 0$

2. The tangent function is: $\tan(-10x)$ As $x$ approaches 0, $-10x$ also approaches 0. We also know from the small angle approximation that: $\tan(-10x) \approx -10x \text{ as } x \to 0$

Step 2: Substitute the Approximations into the Limit

Using these approximations: $\lim_{x \to 0} \sin(2x) \cdot \tan(-10x) \approx \lim_{x \to 0} (2x) \cdot (-10x)$

Step 3: Simplify the Expression

$\lim_{x \to 0} (2x) \cdot (-10x) = \lim_{x \to 0} (-20x^2)$

Step 4: Evaluate the Limit

Now, we can evaluate the limit: $\lim_{x \to 0} -20x^2 = -20 \cdot 0^2 = 0$

Thus, the limit is:

$\boxed{0}$

Would you like further details or have any questions?

Related Questions:

1. How do we derive the small angle approximations for sine and tangent?
2. What happens if we use higher-order approximations for the sine and tangent functions?
3. Can this limit be solved using L'Hôpital's rule?
4. What is the geometric interpretation of this limit?
5. How does this problem change if we consider limits as $x$ approaches infinity?

Tip: When dealing with trigonometric limits as $x$ approaches 0, small angle approximations can greatly simplify the problem.