$\lim_{x\to\infty} x^2\operatorname{cos}\left(\frac{3x+2}{x^2}-1\right)$ without using L’Hôpital. Can I calculate the following limit without applying L'Hôpital?

$$\lim_{x\to\infty} x^2\operatorname{cos}\left(\frac{3x+2}{x^2}-1\right)$$

With L'Hôpital it gives me as a result $\frac{-9}{2}$.

We have to find the limit of :
$$\lim\limits_{x \to \infty}x^2(cos(\frac{ax+b}{x^2})-1)=?$$
If we put the substitution $x=\frac{1}{y}$ and use the formula $1-2sin(x)^2=cos(2x)$ we get :
$$\lim\limits_{y \to 0}\frac{-2sin(\frac{(ay+by^2)}{2})^2}{y^2}$$
Wich equivalent to :
$$\lim\limits_{y \to 0}\frac{(ay+by^2)^2}{4}\frac{-2sin(\frac{(ay+by^2)}{2})^2}{y^2\frac{(ay+by^2)^2}{4}}$$
Or 
$$\lim\limits_{y \to 0}\frac{-sin(\frac{(ay+by^2)}{2})^2}{\frac{(ay+by^2)^2}{4}}=-1$$
Because : $\lim\limits_{z \to 0}\frac{sin(z)}{z}=1$
So we have :
$$\lim\limits_{y \to 0}\frac{(ay+by^2)^2}{4}\frac{-2sin(\frac{(ay+by^2)}{2})^2}{y^2\frac{(ay+by^2)^2}{4}}=\lim\limits_{y \to 0}\frac{(ay+by^2)^2}{4}\frac{-2}{y^2}=\frac{-a^2}{2}$$
