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?

For the first limit, you cannot use twice L'Hôpital law since $3/2x$ is not $0/0$ or $\infty/\infty$. Actually, it's zero.
The second is right.
