The fact that, when solving limits of sequences ($n \in \mathbb{N}$), it strikes me as very bizarre that, even though I follow all the (elementary) rules for solving limits (of sequences), I can get different results for the same example! Consider the following:

$$ \lim_{n \to \infty} \left( \frac{3n + 1}{2n - 1} \right) = \frac{\lim_{n \to \infty}(3n + 1)}{\lim_{n \to \infty} (2n - 1)} = \frac{\lim_{n \to \infty}(3n) + \lim_{n \to \infty}(1)}{\lim_{n \to \infty} (2n) - \lim_{n \to \infty}(1)} = \frac{\infty + 1}{\infty - 1} \Longrightarrow \text{This sequence does not have a limit.} $$

I followed all of the rules for solving limits, yet this conclusion is wrong. The correct approach is the following.

$$ \lim_{n \to \infty} \left( \frac{3n + 1}{2n - 1} \right) = \lim_{n \to \infty} \left( \frac{n(3 + \frac{1}{n})}{n(2 - \frac{1}{n})} \right) = \lim_{n \to \infty} \left( \frac{3 + \frac{1}{n}}{2 - \frac{1}{n}} \right) =\frac{\lim_{n \to \infty}(3 + \frac{1}{n})}{\lim_{n \to \infty} (2 - \frac{1}{n})} = \frac{\lim_{n \to \infty}(3) + \lim_{n \to \infty}(\frac{1}{n})}{\lim_{n \to \infty} (2) - \lim_{n \to \infty}(\frac{1}{n})} = \frac{3 + 0}{2 - 0} = \frac{3}{2}. $$

My question is: why do I get the correct answer only if I arrange the sequence a certain way?

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    $\begingroup$ The error is in the first part: you are assuming that if you get $\frac{\infty+1}{\infty-1}$, then the limit doesn't exist. This is wrong. You cannot conclude anything about the existence of the limit if you get a so-called "indeterminate form," such as $\frac{\infty}{\infty}$ or $\frac{0}{0}$. $\endgroup$ Commented Sep 10, 2019 at 19:56
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    $\begingroup$ Take the limit of x/x as x approaches zero. Obviously the answer is 1, but taking the limit of the top and bottom separately gives 0/0. This is an indeterminate form, not a limit that doesn't exist. $\endgroup$
    – Gabe
    Commented Sep 10, 2019 at 20:03
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    $\begingroup$ Alternatively you can first apply multiplication and addition laws to see that $$ \lim_{n\to\infty}\frac{3n+1}{2n−1}=\frac32\lim_{n\to\infty}\frac{6n+2}{6n−3} =\frac32+\frac32\lim_{n\to\infty}\frac{5}{6n−3} $$ $\endgroup$ Commented Sep 10, 2019 at 20:23
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    $\begingroup$ @CarlWitthoft That is simultaneously overkill and circular logic (how would you be able to differentiate a linear function if you don't already know how to take the limit of a fraction where the numerator and the denominator are linear functions?) $\endgroup$
    – Arthur
    Commented Sep 11, 2019 at 13:09
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    $\begingroup$ @Arthur What I meant is that he is not trying to formalize the whole theory of Calculus (in which using L'Hopital to define derivatives would lead to a circular argument) and, therefore, the suggestion of using L'Hopital is of practical value. $\endgroup$
    – rafa11111
    Commented Sep 11, 2019 at 18:29

2 Answers 2


You did not in fact follow all the rules (and this is a good example of why understanding why the rules are true is important). The relevant rule is:

If both $\lim_{n\rightarrow a}f(n)$ and $\lim_{n\rightarrow a}g(n)$ exist and are finite (and $\lim_{n\rightarrow a}g(n)\not=0$), then $$\lim_{n\rightarrow a}{f(n)\over g(n)}={\lim_{n\rightarrow a}f(n)\over \lim_{n\rightarrow a}g(n)}.$$

There are various other rules of similar flavor. But the point is that the "rule" you've tried to apply is not one of them. (And it's a good exercise at this point to go through your textbook and look at what the rules do in fact say, and note that none of them actually get you what you want.)

Indeed, the example you give is a good example of the limitations of these rules: while we can often manipulate limits "algebraically," we can't do this in all cases, and some hypotheses are needed. It's also a good example of why proofs are important, since there are plenty of plausible-sounding statements which are in fact false.

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    $\begingroup$ Hmmm, thank you very much, I didn't know about this caveat! We only scratched the surface of what we can do with limits, as I only studied them in the context of sequences and series. Jesus bless and keep you always. Amen. $\endgroup$
    – God bless
    Commented Sep 10, 2019 at 20:05
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    $\begingroup$ @GregorPerčič I remember when I studied analysis my professor went a great length talking about "pseudo-rules" that a lot of students use and are wrong. These "pseudo-rules" as he called them where, like the above example, correct rules but without checking preconditions, or simply wrong but "inferred" by the students from the others (like: since in some cases the limit of a product is the product of the limits, then it must also be true for the sum/difference). You really need to be careful with limit rules. BTW: generally you should be taught their proofs also... $\endgroup$ Commented Sep 11, 2019 at 12:05
  • $\begingroup$ You don't need "and are finite" (which words are not in OP's source). An infinite limit does not exist. It's just that infinite limits occur so frequently and are generally useful to talk about so we have words and notation to describe this particular kind of nonexistent limit. $\endgroup$ Commented Sep 11, 2019 at 15:55
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    $\begingroup$ This comment might not apply to the OP, but I've seen many students to whom it applies, so it seems worth recording here: There's a tendency to think that math consists of formulas, and so learning math means learning formulas. Students with this view of math are very likely to learn formulas, like the one used by the OP, without learning their limitations, simply because the limitations are expressed in words (the limits of $f$ and $g$ exist) rather than formulas. $\endgroup$ Commented Sep 11, 2019 at 18:59
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    $\begingroup$ @EricTowers I'm actually with TonyK on this one: although I recognize that plenty of texts (most?) use the opposite convention, I strongly believe that the "right" approach is that infinite limits do exist (or more snappily, that the extended real line is a better context for calculus). But ultimately this really is just a matter of taste; I phrased my answer so that it applies regardless of what convention is used. $\endgroup$ Commented Sep 11, 2019 at 19:04

Your reference page here starts with this introduction (emphasis mine):

Listed here are a couple of basic limits and the standard limit laws which, when used in conjunction, can find most limits. They are listed for standard, two-sided limits, but they work for all forms of limits. However, note that if a limit is infinite, then the limit does not exist.

So everywhere below that warning, if there is a statement that a limit must exist, then it's implicit that the limit must not be $\infty$.

  • $\begingroup$ For this reason I wonder if we should stop using $=\infty$ for infinite limits. The equals sign implies a value for the limit, and tempts the student into arithmetic with $\infty$ (as this question demonstrates). $\endgroup$ Commented Sep 11, 2019 at 17:19

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