https://file.aiquickdraw.com/mathbot/file/1726197822912vavms6me.jpg

The given image shows the mathematical expression:

\[
\lim_{{x \to 0}} \frac{x^2}{\sin(x) \cdot \tan(2x)}
\]

To solve this limit, let’s analyze it step by step.

### Step 1: Use approximations for small values of \( x \)

As \( x \to 0 \), we can use the standard approximations for sine and tangent:
- \( \sin(x) \approx x \)
- \( \tan(2x) \approx 2x \)

This simplifies the expression to:

\[
\frac{x^2}{x \cdot 2x} = \frac{x^2}{2x^2} = \frac{1}{2}
\]

### Step 2: Conclusion
Thus, the limit evaluates to:

\[
\lim_{{x \to 0}} \frac{x^2}{\sin(x) \cdot \tan(2x)} = \frac{1}{2}
\]

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

### Follow-up Questions:
1. What is L'Hopital's Rule and how could it apply to this limit?
2. How are small-angle approximations useful for limits involving trigonometric functions?
3. Can you explain why \( \sin(x) \approx x \) as \( x \to 0 \)?
4. How does the tangent approximation work for very small angles?
5. How would this problem change if the limit were \( x \to \infty \)?

**Tip:** When dealing with limits involving trigonometric functions, small-angle approximations like \( \sin(x) \approx x \) and \( \tan(x) \approx x \) are highly useful.

Limit analysis for the expression \(\lim_{{x \to 0}} \frac{x^2}{\sin(x) \cdot \tan(2x)}\).

Solve the limit \(\lim_{{x \to 0}} \frac{x^2}{\sin(x) \cdot \tan(2x)}\) using small-angle approximations for trigonometric functions. This step-by-step solution demonstrates the use of \(\sin(x) \approx x\) and \(\tan(2x) \approx 2x\) as \(x \to 0\). Ideal for calculus students in Grades 11-12 or undergraduates.

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