<s>[INST] Calculate limits $ \lim_{x\to+\infty} \frac{3x-1}{x^2+1}$ and $\lim_{x\to-\infty} \frac{3x^3-4}{2x^2+1}$ I want to calculate the following limits
$$\begin{matrix}
\lim_{x\to+\infty} \frac{3x-1}{x^2+1} & \text{(1)} \\
\lim_{x\to-\infty} \frac{3x^3-4}{2x^2+1} & \text{(2)}
\end{matrix}$$
In both cases we have indeterminate forms. Using L'Hôpital's rule on $\text{(1)}$  gives
$$\lim_{x\to+\infty} \frac{3x-1}{x^2+1} = \lim_{x\to+\infty}\frac{3}{2x} = 0$$
Using L'Hôpital's rule on $\text{(2)}$ gives
$$\lim_{x\to-\infty} \frac{3x^3-4}{2x^2+1} = \lim_{x\to-\infty}\frac{9x^2}{4x} = \lim_{x\to-\infty}\frac{18x}{4} = -\infty$$
Is this correct?
 [/INST] I think we can change the limit $\to0$ to even avoid L'Hosiptal's Rule as follows:
Putting $\frac1x=h$
$$\lim_{x\to\infty}\frac{3x-1}{x^2+1}=\lim_{h\to0}\frac{\frac3h-1}{\frac1{h^2}+1}=\lim_{h\to0}\frac{(3-h)h}{1+h^2}=0$$ 
$$\lim_{x\to\infty}\frac{3x^3-4}{2x^2+1}=\lim_{h\to0}\frac{\frac3{h^3}-4}{\frac2{h^2}+1}=\lim_{h\to0}\frac{3-4h^3}{h(2+h^2)}=\frac30$$
</s>