Limits

To solve the problem, we need to evaluate the limit:

$ \sin 3x \sin x x \to 0 \sin 3x 3\sin x 0 0 0 \frac{0}{0} \sin 3x \sin x 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 3\sin x = 3x - \frac{3x^3}{6} + \frac{3x^5}{120} - \cdots = 3x - \frac{3x^3}{6} + \frac{3x^5}{120} \sin 3x - 3\sin x = (3x - \frac{27x^3}{6} + \frac{243x^5}{120}) - (3x - \frac{3x^3}{6} + \frac{3x^5}{120}) = 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 x^3 \lim_{x \to 0} \frac{-4x^3 + 2x^5}{x^3} = \lim_{x \to 0} (-4 + 2x^2) x \to 0 2x^2 \to 0 \lim_{x \to 0} (-4 + 0) = -4 {-4} $.