evaluate the limit as x approaches 0 of (sin3x)/x

To evaluate the limit \(\lim_{x \to 0} \frac{\sin(3x)}{x}\), we can use the standard trigonometric limit:

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

We will manipulate the expression to use this standard limit.

Rewrite the expression by introducing a factor of 3 in the denominator:

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

Now, notice that the limit inside the product can be recognized as a form of the standard trigonometric limit:

\[
\lim_{x \to 0} \frac{\sin(3x)}{3x} = 1.
\]

Thus, we can evaluate the original limit as follows:

\[
\lim_{x \to 0} \frac{\sin(3x)}{x} = \left( \lim_{x \to 0} \frac{\sin(3x)}{3x} \right) \cdot 3 = 1 \cdot 3 = 3.
\]

Therefore, the limit is:

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

Would you like further details or have any questions?

Here are 5 related questions that expand on this topic:
1. How would you evaluate \(\lim_{x \to 0} \frac{\sin(5x)}{x}\)?
2. What is the limit \(\lim_{x \to 0} \frac{\sin(x)}{\sin(2x)}\)?
3. How can L'Hopital's rule be applied to solve limits involving trigonometric functions?
4. Why is the standard limit \(\lim_{x \to 0} \frac{\sin(x)}{x} = 1\) important in calculus?
5. How would you find \(\lim_{x \to 0} \frac{\tan(2x)}{x}\)?

**Tip:** Whenever evaluating limits involving trigonometric functions, see if you can rewrite them to use standard limits like \(\lim_{x \to 0} \frac{\sin(x)}{x} = 1\).

Learn how to evaluate the limit \(\lim_{x \to 0} \frac{\sin(3x)}{x}\) using trigonometric identities and the standard limit formula. This problem is suitable for advanced high school students and covers trigonometric functions and limits in calculus.

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