How do I find the $\lim_{x\to 0}\frac{7x^3-4x^2}{\sin(3x^2)}$ without using L’Hopital’s? $$\Large{\lim_{x\to 0}\frac{7x^3-4x^2}{\sin(3x^2)}}$$
I know how to find this Limit only by using L’Hopital’s. Here's How:
$$={\lim_{x\to 0}\frac{7x^3-4x^2}{\sin(3x^2)}}$$
$$={\lim_{x\to 0}\frac{21x^2-8x}{6x\cos(3x^2)}}$$
$$={\lim_{x\to 0}\frac{x(21x-8)}{6x\cos(3x^2)}}$$
$$={\lim_{x\to 0}\frac{(21x-8)}{6\cos(3x^2)}}$$
$$=\frac{21\cdot 0 - 8}{6 \cos 0}$$
$$=\frac{-4}{3}$$
My Question is, is there a way to do this without using the L’Hopital’s? If so, How?
Also Note: We have not been taught the series expansion of $\sin x$.

So we have $$\lim_{x\to 0} \frac{7x^3 - 4x^2}{\sin(3x^2)}.$$
First, we have to factor the numerator so $x^3 - 4x^2 = x^2(7x - 4)$.
Therefore, we have $$\lim_{x\to 0} \frac{x^2(7x - 4)}{\sin(3x^2)}.$$
After this, we multiply the numerator and the denominator by $3x^2$.
Note $\frac{3x^2}{3x^2}=1$ so we aren't doing anything arbitrarily and it is still equivalent.  We know that $$\lim_{x\to 0} \frac{\sin(x)}{x}=1 \text{ and } \lim_{x\to 0}\frac{x}{\sin(x)}=1$$ (you have to use a calculator or L'Hopital's theorem to check this). By doing this multiplication, we get $$\lim_{x\to 0} \frac{3x^2 x^2  (7x - 4)}{3x^2 \sin (3x^2)}.$$ And since this is similar to $x / \sin(x)$ because $x / \sin(x)$ is essentially $1 \cdot 1 / (1 \cdot \sin(1 \cdot x))$, the limit as $x$ approaches $0$ of $(x / \sin(x)) = (3x^2 / 3x^2 \sin(3x^2)$ must be true due to proportionality or similarity. After this, we can simplify the limit as $x$ approaches $0$ of $(3x^2 \cdot x^2 \cdot (7x - 4))/(3x^2\sin(3x^2))$ to the limit as $x$ approaches $0$ of $(x^2(7x - 4)/x^2)$. After this, we simplify the $x^2$ in the numerator and the denominator thus leaving the limit as $x$ approaches $0$ of $(7x -4)$ which is $-4/3$.  I hope this makes sense though! 
