I want to calculate the following limits
lim
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
Your second limit is correct.
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