https://file.aiquickdraw.com/mathbot/file/1725686664916662rtgzx.jpg

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From the image, the formula appears to resemble:

\[
\lim_{{x \to 0}} \frac{{\sin(3x)}}{x} = 3
\]

This is a well-known limit in calculus. Let me explain the reasoning behind this result.

### Explanation:
We are evaluating the limit:

\[
\lim_{{x \to 0}} \frac{{\sin(3x)}}{x}
\]

Using the substitution \( u = 3x \), we rewrite the limit as:

\[
\lim_{{x \to 0}} \frac{{\sin(3x)}}{x} = \lim_{{u \to 0}} \frac{{\sin(u)}}{\frac{u}{3}} = 3 \lim_{{u \to 0}} \frac{{\sin(u)}}{u}
\]

It is a standard result that:

\[
\lim_{{u \to 0}} \frac{{\sin(u)}}{u} = 1
\]

Therefore, the limit simplifies to:

\[
3 \times 1 = 3
\]

Thus, the solution is:

\[
\lim_{{x \to 0}} \frac{{\sin(3x)}}{x} = 3
\]

Would you like more details, or do you have any questions about this solution?

### Related Questions:
1. What are the common limit properties of trigonometric functions like sine and cosine?
2. How does L'Hôpital's Rule apply to trigonometric limits?
3. What other substitution methods work for solving trigonometric limits?
4. How can this type of limit be generalized for other angles, such as \( \sin(nx) \)?
5. Can this limit be solved using series expansions of sine?

**Tip**: For evaluating limits involving trigonometric functions, always check if you can apply standard limits like \( \lim_{{x \to 0}} \frac{{\sin(x)}}{x} = 1 \), which simplifies many problems.

What is the limit of \( \frac{\sin(3x)}{x} \) as \( x \to 0 \)?

Understand how to solve the limit of \( \frac{\sin(3x)}{x} \) as \( x \to 0 \) using standard limit theorems and trigonometric substitutions. This problem demonstrates core concepts in calculus and trigonometry, suitable for students in Grades 11-12.

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