Limits ( algebriac help) with variable substitution and Trig

Question

$$\lim_{x\to0}\frac{\sin^2(3x)}{x^2\cos x}=\lim_{u\to0}\frac{\sin^2(u)}{\left(\dfrac u3\right)^2}=9\left(\lim_{u\to0}\frac{\sin u}u\right)^2.$$

A bit of confusion on the little steps that are missing. 1. Why does $\sin(x)$ go away. It is because $\cos(0)=1$? 2. How does the change of variable work, completely lost?

Answer 1

Hint:$$\quad{\lim_{x\to0}\frac{\sin^2(3x)}{x^2\cos x}=\\\lim_{x\to0}\frac{\sin(3x)}{x}.\lim_{x\to0}\frac{\sin(3x)}{x}.\lim_{x\to0}\frac{1}{\cos x}=\\ \lim_{x\to0}\color{red} {3}\frac{\sin(3x)}{\color{red} {3}x}.\lim_{x\to0}\color{red} {3}\frac{\sin(3x)}{\color{red} {3}x}.\lim_{x\to0}\frac{1}{\cos x}=\\ \color{red} {9}\underbrace{\lim_{x\to0}\frac{\sin(3x)}{\color{red} {3}x}}_{1}.\underbrace{\lim_{x\to0}\frac{\sin(3x)}{\color{red} {3}x}}_{1}.\lim_{x\to0}\frac{1}{\cos x}=\\9.1.1.\lim_{x\to0}\frac{1}{\cos x}=9.1.1.1}$$

Answer 2

$$\lim_{x\to0}\frac{\sin^2(3x)}{x^2\cos x}=3^2\left(\lim_{x\to0}\dfrac{\sin3x}{3x}\right)^2\cdot\dfrac1{\lim_{x\to0}\cos x}=?$$