To resolve the problem, we must compute the limit:
$$ \lim_{x \to 0} \frac{\sin 3x - 3\sin x}{x^3} $$
1. Strategy Identification:
The expression involves $\sin 3x$ and $\sin x$. As $x$ approaches $0$, both the numerator and denominator approach $0$, yielding the indeterminate form $\frac{0}{0}$. This suggests using either series expansion or L'Hôpital's Rule. We will employ the Maclaurin series expansion for simplification.
2. Maclaurin Series Application:
Expand $\sin 3x$ and $\sin x$ using their Taylor series around $x = 0$:
$\sin 3x = 3x - \frac{(3x)^3}{6} + \frac{(3x)^5}{120} - \cdots = 3x - \frac{27x^3}{6} + \frac{243x^5}{120} - \cdots$
$\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots$
Consequently,
$3\sin x = 3\left(x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots\right) = 3x - \frac{3x^3}{6} + \frac{3x^5}{120} - \cdots$
3. Numerator Simplification:
Calculate the difference in the numerator:
$\sin 3x - 3\sin x = \left(3x - \frac{27x^3}{6} + \frac{243x^5}{120}\right) - \left(3x - \frac{3x^3}{6} + \frac{3x^5}{120}\right)$
$= (3x - 3x) + \left(-\frac{27x^3}{6} - (-\frac{3x^3}{6})\right) + \left(\frac{243x^5}{120} - \frac{3x^5}{120}\right)$
$= -\frac{24x^3}{6} + \frac{240x^5}{120}$
$= -4x^3 + 2x^5$
4. Division by $x^3$:
Substitute back into the limit expression:
$\lim_{x \to 0} \frac{-4x^3 + 2x^5}{x^3} = \lim_{x \to 0} \frac{x^3(-4 + 2x^2)}{x^3} = \lim_{x \to 0} (-4 + 2x^2)$
5. Limit Calculation:
As $x$ approaches $0$, $2x^2$ also approaches $0$. Therefore, the limit evaluates to:
$\lim_{x \to 0} (-4 + 0) = -4$
Final Result:
The limit's value is $ {-4} $.