Computing $\lim_{x\to 0}\frac{1-cos(3x)}{5x^2}$ not using derivative

Question

How can we compute the following limit not using derivative?

$\lim_{x\to 0}\frac{1-\cos(3x)}{5x^2}$

I know $\lim_{x\to 0}\frac{\cos(x)-1}{x}=0$. But $\lim_{x\to 0}\frac{1-\cos(3x)}{5x^2}=\lim_{x\to 0}\frac{\cos(3x)-1}{3x}\frac{(-3)}{5x^2}=0.-\infty$

Answer 1

$$\lim_{x\to 0}\dfrac{1-\cos3x}{5x^2}\cdot \dfrac{1+\cos3x}{1+\cos3x}$$

$$ = \lim_{x\to 0}\dfrac{\sin^23x}{5x^2}\cdot \lim_{x\to 0}\dfrac{1}{1+\cos3x}$$

$$ = \lim_{x\to 0}\dfrac{\sin^23x}{(3x)^2}\cdot \lim_{x\to 0}\dfrac{9}{5+5\cos3x}$$

$$ = 1\cdot \dfrac{9}{5+5}=\dfrac{9}{10}$$

Answer 2

Hint: $$\sin^2\left(\frac 32 x \right)=\frac{1-\cos 3x}{2}$$ and $$\lim_{x\to 0}\dfrac{\sin^2 ax}{(ax)^2}=1$$