<s>[INST] $\lim_{x\to\infty} x^2\operatorname{cos}\left(\frac{3x+2}{x^2}-1\right)$ 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 $\frac{-9}{2}$.
 [/INST] I think he want to type and solve this $$\lim_{x\to\infty} x^2(cos(\frac{3x+2}{x^2})-1)=\frac{-9}{2}$$ this can be solve by multiplying conjugate 
$$\quad{\lim_{x\to\infty} x^2(cos(\frac{3x+2}{x^2})-1)=\\
\lim_{x\to\infty} x^2(cos(\frac{3x+2}{x^2})-1)\times \frac{cos(\frac{3x+2}{x^2})+1}{cos(\frac{3x+2}{x^2})+1}=\\
\lim_{x\to\infty} x^2(-\sin^2(\frac{3x+2}{x^2}))\times \frac{1}{cos(\frac{3x+2}{x^2})+1}=\\}$$ 
as $x\to \infty \implies \sin(\frac{3x+2}{x^2})\to 0 \implies \sin(\frac{3x+2}{x^2})\sim \frac{3x+2}{x^2}\\$so 
$$\quad{\lim_{x\to\infty} x^2(-\sin^2(\frac{3x+2}{x^2}))\times \frac{1}{cos(\frac{3x+2}{x^2})+1}=\\
\lim_{x\to\infty} -x^2(\frac{3x+2}{x^2})^2\times \frac{1}{cos(\frac{3x+2}{x^2})+1}=\\
\lim_{x\to\infty} -(\frac{(3x+2)^2}{x^2})\times \frac{1}{cos(\frac{3x+2}{x^2})+1}=\\\lim_{x\to\infty} -(\frac{9x^2+12x+4}{x^2})\times \frac{1}{cos(\frac{3x+2}{x^2})+1}=\\
\lim_{x\to\infty} -(\frac92)\times \frac{1}{cos(\frac{3x+2}{x^2})+1}=\\
 -(\frac92)\times \frac{1}{cos(0)+1}=\\\frac{-9}{2}}$$
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