\lim _{x\to 0}\frac{2\:sin\:x\:cos\:x}{3x}

To solve the limit \(\lim_{x \to 0} \frac{2 \sin x \cos x}{3x}\), we can begin by simplifying the expression.

### Step-by-Step Solution

1. **Rewrite the Trigonometric Expression:**
   The numerator \(2 \sin x \cos x\) can be rewritten using the double-angle identity for sine:
   \[
   2 \sin x \cos x = \sin(2x)
   \]
   So, the limit becomes:
   \[
   \lim_{x \to 0} \frac{\sin(2x)}{3x}
   \]

2. **Use the Sine Limit Identity:**
   We know from the limit identity \(\lim_{u \to 0} \frac{\sin u}{u} = 1\). To use this, let’s make a substitution to match the form of this identity.

   Substitute \(u = 2x\), which implies \(du = 2 dx\), or equivalently, \(x = \frac{u}{2}\).

   Substituting into the limit:
   \[
   \lim_{x \to 0} \frac{\sin(2x)}{3x} = \lim_{u \to 0} \frac{\sin(u)}{3 \cdot \frac{u}{2}}
   \]

3. **Simplify the Expression:**
   Simplify the denominator:
   \[
   \lim_{u \to 0} \frac{\sin(u)}{\frac{3u}{2}} = \lim_{u \to 0} \frac{2\sin(u)}{3u}
   \]

4. **Apply the Limit Identity:**
   Now, apply the limit identity:
   \[
   \lim_{u \to 0} \frac{2\sin(u)}{3u} = \frac{2}{3} \lim_{u \to 0} \frac{\sin(u)}{u} = \frac{2}{3} \times 1 = \frac{2}{3}
   \]

### Conclusion

Therefore, the limit is:
\[
\lim_{x \to 0} \frac{2 \sin x \cos x}{3x} = \frac{2}{3}
\]

Would you like further details or have any questions about this solution? 

Here are some related questions for further exploration:

1. What is the double-angle identity for sine and how is it derived?
2. How can we apply L'Hopital's Rule to solve \(\lim_{x \to 0} \frac{2 \sin x \cos x}{3x}\)?
3. What is the geometric interpretation of the limit \(\lim_{x \to 0} \frac{\sin x}{x} = 1\)?
4. How do we use substitution to solve limits involving trigonometric functions?
5. How would the solution change if the function was \(\lim_{x \to 0} \frac{\sin^2(x)}{3x}\)?

**Tip:** Remember to use trigonometric identities to simplify expressions before applying limit rules.

Learn how to solve the limit \(\lim_{x \to 0} \frac{2 \sin x \cos x}{3x}\) with a detailed step-by-step solution. This problem involves trigonometric functions and the double-angle identity, showcasing advanced high school-level mathematics.

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