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?

In the first case, once you no longer have an indeterminate form, L'Hospital is not needed. 
So... After one application of LH, you should obtain $\lim \limits_{x\to+\infty}\dfrac 3{2x}$, which has the "form" $\dfrac 3{\infty} $ and not $\dfrac 3 0$, and in any case, is not an indeterminate form, so L'Hospital is no longer warranted. What we do have is the following (after one meager application of LH):
$$\lim \limits_{x\to+\infty}\dfrac 3{2x} = 0$$ 

Your second limit is correct.
