$\lim_{x \to 0}\frac{\sin^2(3x)}{x^3-x}$ - Did I solve this Correctly?

Question

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

$\lim_{x \to 0}\frac{\sin(3x)\sin(3x)}{x(x-1)(x+1)}$

$3\lim_{x \to 0}\frac{\sin(3x)}{3x}\cdot \lim_{x\to 0}\frac{\sin(3x)}{(x-1)(x+1)}$

$3\cdot 1 \cdot 0 = 0$

My textbook solved this differently I'm not sure where they got the extra $x$ to make another $\frac{\sin3x}{3x}.$

Answer 1

They got their extra $x$ from multiplying through by $\frac{x}{x}$; \begin{align*} \lim_{x\rightarrow 0}\frac{\sin^{2}(3x)}{x^{3}-x} &= \lim_{x\rightarrow 0}\frac{\sin(3x)}{x}\sin(3x)\frac{1}{x^{2}-1}\\ &= \lim_{x\rightarrow 0}\frac{\sin(3x)}{x}\sin(3x)\frac{x}{x}\frac{1}{x^{2}-1}\\ &= \lim_{x\rightarrow 0}\frac{\sin(3x)}{x}\frac{\sin(3x)}{x}\frac{x}{x^{2}-1}\\ \Rightarrow \lim_{x\rightarrow 0}\frac{\sin^{2}(3x)}{x^{3}-x} &= \lim_{x\rightarrow 0}\frac{\sin(3x)}{x}\frac{\sin(3x)}{x}\frac{1}{x-\frac{1}{x}}\\ \end{align*}

That being said, your method looks fine to me too!

Answer 2

That's fine, as an alternative

$$\frac{\sin^2(3x)}{x^3-x}=\frac{\sin^2(3x)}{(3x)^2}\frac{9x}{x^2-1}\to 1\cdot 0=0$$