Calculate limits $lim_{xto+infty} frac{3x-1}{x^2+1}$ and $lim_{xto-infty} frac{3x^3-4}{2x^2+1}$

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?

Answer

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):

$$\lim \limits_{x\to+\infty}\dfrac 3{2x} = 0$$

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Your second limit is correct.