3 x 2 − 5 x sin 3 x

To find the limit

$\lim_{x \to 0} \frac{3x^2 - 5x \sin(3x)}{x}$

we can first simplify the expression. Dividing each term in the numerator by $x$ gives us:

$\lim_{x \to 0} \left(3x - 5\sin(3x)\right)$

As $x$ approaches $0$, we can use the Taylor series expansion for $\sin(3x)$:

$\sin(3x) \approx 3x - \frac{(3x)^3}{6} + O(x^5) = 3x - \frac{27x^3}{6} + O(x^5)$

Substituting this back into the limit gives:

$3x - 5\left(3x - \frac{27x^3}{6}\right) = 3x - 15x + \frac{135x^3}{6} = -12x + \frac{135x^3}{6}$

As $x \to 0$, both terms approach $0$. So we evaluate:

$\lim_{x \to 0} (-12x + O(x^3)) = 0$

Thus, the limit is:

$\lim_{x \to 0} \frac{3x^2 - 5x \sin(3x)}{x} = 0$

Would you like further details or explanations on any part of this solution? Here are some related questions you might consider:

1. How does the Taylor series expansion work for functions near zero?
2. What is the significance of limits in calculus?
3. Can you explain L'Hôpital's Rule and when to apply it?
4. What are common mistakes to avoid when calculating limits?
5. How do you approach limits involving trigonometric functions?

Tip: Always check if direct substitution leads to an indeterminate form, as it may require further techniques like factoring or L'Hôpital's Rule.