To evaluate the limit:

$\lim_{x \to 0} \frac{3x - \sin(3x)}{3x - \tan(3x)},$

we will proceed as follows:

Step 1: Recall the small-angle approximations

As $x \to 0$, the following approximations are valid:

• $\sin(3x) \approx 3x - \frac{(3x)^3}{6} = 3x - \frac{27x^3}{6}$,
• $\tan(3x) \approx 3x + \frac{(3x)^3}{3} = 3x + \frac{27x^3}{3}$.

Step 2: Simplify the numerator and denominator

Using the approximations:

Numerator: $3x - \sin(3x) \approx 3x - \left(3x - \frac{27x^3}{6}\right) = \frac{27x^3}{6} = \frac{9x^3}{2}.$

Denominator: $3x - \tan(3x) \approx 3x - \left(3x + \frac{27x^3}{3}\right) = -\frac{27x^3}{3} = -9x^3.$

Step 3: Write the simplified fraction

The limit becomes: $\lim_{x \to 0} \frac{\frac{9x^3}{2}}{-9x^3}.$

Simplify the fraction: $\frac{\frac{9x^3}{2}}{-9x^3} = \frac{9}{-18} = -\frac{1}{2}.$

Final Answer:

$\lim_{x \to 0} \frac{3x - \sin(3x)}{3x - \tan(3x)} = -\frac{1}{2}.$

Would you like further clarification on any step?

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Relative Questions:

1. How are small-angle approximations derived?
2. What other trigonometric approximations are useful for limits?
3. How does the Taylor expansion improve accuracy in such problems?
4. Could L'Hôpital's Rule also solve this limit? How?
5. Why is this limit indeterminate initially, and how is it resolved?

Tip: Always verify the validity of approximations by considering higher-order terms when precision matters.