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?
 A: 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.
A: 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.
A: I think we can change the limit $\to0$ to even avoid L'Hosiptal's Rule as follows:
Putting $\frac1x=h$
$$\lim_{x\to\infty}\frac{3x-1}{x^2+1}=\lim_{h\to0}\frac{\frac3h-1}{\frac1{h^2}+1}=\lim_{h\to0}\frac{(3-h)h}{1+h^2}=0$$ 
$$\lim_{x\to\infty}\frac{3x^3-4}{2x^2+1}=\lim_{h\to0}\frac{\frac3{h^3}-4}{\frac2{h^2}+1}=\lim_{h\to0}\frac{3-4h^3}{h(2+h^2)}=\frac30$$
A: Here's an alternative, that I think is more elementary.
To start with: the constant terms are going to become increasingly irrelevant as the $x$ terms get really large. To prove this, you could explicitly write the differences from the version without constants, and show the differences disappear:
\begin{align}
\lim_{x\to+\infty} \frac{3x-1}{x^2+1} 
  &= \lim_{x\to+\infty}\frac{3x-1}{3x}\cdot\frac{3x}{x^2}\cdot\frac{x^2}{x^2+1} \\
  &= \lim_{x\to+\infty}\frac{3x-1}{3x}\cdot\lim_{x\to+\infty}\frac{3x}{x^2}\cdot\lim_{x\to+\infty}\frac{x^2}{x^2+1} \\
  &= \lim_{x\to+\infty}\left(1 - \frac{1}{3x}\right)\cdot\lim_{x\to+\infty}\frac{3x}{x^2}\cdot\lim_{x\to+\infty}\left(1 - \frac{1}{x^2+1}\right) \\
  &= \lim_{x\to+\infty}\frac{3x}{x^2}
\end{align}
Now you've got rid of the constant terms, what's left is easy.
Of course, in a pedantic presentation of this result, the second equality is only correct if the three RHS limits exist, so you have to sort of attach that precondition until you successfully compute them, at which point you can say "so that was all right after all".
This works for all polynomials, and is an illustration of the general fact that to see what a polynomial does as its argument goes to infinity, you can safely throw away all but the leading term. With experience, you'll do this without any of the tedious intermediate steps I showed above: "the limit of $(3x-1)/(x^2+1)$ as $x \to \infty$ is zero because quadratic beats linear".
