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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?

Ali Caglayan
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dato datuashvili
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    L'Hopital is overkill. For the first problem, divide numerator and denominator by $x^2$. This gives you the expression $\textstyle{3\over x}-{1\over x^2}\over {1+{1\over x^2}}$. Taking the limit of this as $x$ tends to infinity is straightforward. Do something similar for the second problem (or for any limit at infinity of a rational expression). – David Mitra Sep 14 '13 at 13:26
  • @DavidMitra Or just use the simple rule of thumb that only the highest power of the numerator and denominator of a rational function matter when taking a limit at infinity. – AJMansfield Sep 14 '13 at 16:44

4 Answers4

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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.

amWhy
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  • first one was my typo,i have mistakenly typed – dato datuashvili Sep 14 '13 at 13:31
  • Don't you just hate typos! :^) – amWhy Sep 14 '13 at 13:33
  • Actually, L'Hôpital is valid when the denominator has infinite limit (it doesn't matter what the numerator does). What the OP did for the first problem is correct. – David Mitra Sep 14 '13 at 13:33
  • Thanks, @David. Not necessary to apply twice. And I've just seen too many cases where L'Hôpital is repeatedly applied when it is not valid to do so, so I usually suggest applying only when limit evaluates as an indeterminate form. – amWhy Sep 14 '13 at 13:34
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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.

amWhy
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Shuchang
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I think we can change the limit $\to0$ to even avoid L'Hosiptal's Rule as follows:

Putting $\frac1x=h$

$$\lim_{x\to\infty}\frac{3x-1}{x^2+1}=\lim_{h\to0}\frac{\frac3h-1}{\frac1{h^2}+1}=\lim_{h\to0}\frac{(3-h)h}{1+h^2}=0$$

$$\lim_{x\to\infty}\frac{3x^3-4}{2x^2+1}=\lim_{h\to0}\frac{\frac3{h^3}-4}{\frac2{h^2}+1}=\lim_{h\to0}\frac{3-4h^3}{h(2+h^2)}=\frac30$$

lab bhattacharjee
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    Nice approach, but you are making some critical errors, for example $\lim_{x\to a}f(x)\cdot g(x) = \lim_{x\to a}f(x) \cdot \lim_{x\to a}g(x)$ iff all of those limits exist. In order to be able to use any of the limit laws, you have to be able to (without them) show that the limits exists. The fact that you are skipping some steps, and not showing the intermediate limits, doesn't help this. – AJMansfield Sep 14 '13 at 13:51
  • To summarize, this answer is not correct. – AJMansfield Sep 14 '13 at 13:56
  • @AJMansfield, did you mean this? – lab bhattacharjee Sep 14 '13 at 14:04
  • I think you make some mistakes in reducing the substitution, you need to actually substitute $\frac{1}{h}$ in for each instance of $x$, and solve it. – AJMansfield Sep 14 '13 at 14:06
  • Oh, I see, I just plugged the substitution into wolfram|alpha; you did that part correctly. – AJMansfield Sep 14 '13 at 14:09
  • @AJMansfield, added one step for better clarity – lab bhattacharjee Sep 14 '13 at 14:09
  • This is good, I will reinstate my upvote. – AJMansfield Sep 14 '13 at 14:10
  • @AJMansfield: this may or may not have been what you meant, but you don't need to assume a priori that the left-hand-side limit exists. The rule goes, if the limit of two things exists, then the limit of their product exists and is equal to the product of the limits. – Ben Millwood Sep 14 '13 at 16:01
  • @BenMillwood I'm sorry for the confusion. I didn't intend to say the limit of the product has to exist a priori. – AJMansfield Sep 14 '13 at 16:38
  • @AJMansfield: I suspected as much but thought it worth clarifying anyway :) – Ben Millwood Sep 14 '13 at 16:46
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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".

Ben Millwood
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