Let me explain. Suppose we want to calculate $\lim\limits_{n\to\infty} n^2-n$. Since this limit is indeterminate, one way to do it is to write it as $\lim\limits_{n\to\infty} n^2(1-1/n)$. Since $n^2$ goes to infinity and $1-1/n$ goes to $1$, the limit is $\infty$. If this was part of a bigger expression, we would leave it as it is and then at the end look at the limits of all the individual factors. This is the way I've learned it and the way I've always done it.

However, I've noticed that some students do the following:

$$\lim\limits_{n\to\infty} n^2-n = \lim\limits_{n\to\infty} n^2(1-1/n) = \lim\limits_{n\to\infty} n^2 = \infty$$

It is the second equality I'm concerned with. It's not, strictly speaking, wrong: after all, all the limits here are equal to each other. And yet I've been telling them not to do it. My way is to do whatever you want with your expression, and then take all limits in a single step. Now that I think about it, however, I can't find a reason not to let the $1/n$ go to $0$ before taking the rest of the limit: all the equalities are correct, and it simplifies the expression.

Is there any reason why the students should be discouraged from doing this? Or am I just enforcing a rule for no reason?

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    $\begingroup$ $$ 1= \lim_{n \to \infty} 1 = \lim_{n \to \infty} n \frac{1}{n} = \lim_{n \to \infty} n(0) =\lim_{n \to \infty} 0 = 0 $$ $\endgroup$ Jun 3, 2015 at 23:56
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    $\begingroup$ @StevenGubkin, you should put your comment as an answer. $\endgroup$
    – JRN
    Jun 4, 2015 at 1:16
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    $\begingroup$ I always loved $\lim\limits_{n\to\infty} (1+1/n)^n$. if you eliminate $1/n$ your answer is certainly wrong. $\endgroup$
    – oerkelens
    Jun 4, 2015 at 11:24

3 Answers 3


The important thing is whether students' reasoning is logically valid — and in particular, that they only use the conclusion of a theorem after they've checked that all its hypotheses hold — not whether they follow any particular arbitrary rules or procedures. In this case, the relevant theorem is the following:

Theorem. Let $f$ and $g$ be real-valued functions defined on a subset of the real line, and let $c$ be either a real number or $\pm \infty$. If $\lim_{t \to c} g(t)$ exists and is equal to a nonzero real number, then $\lim_{t \to c} f(t) g(t)$ exists if and only if $\lim_{t \to c} f(t)$ exists, in which case $$\lim_{t \to c} f(t) g(t) = [\lim_{t \to c} f(t)] \cdot [\lim_{t \to c} g(t)].$$

In the example you gave, the problem isn't that the students fail to adhere to the (needless) rule of taking all limits in a single step. The problem is that they omit an important part of the reasoning: since $\lim_{n \to \infty} (1 - 1/n)$ exists and is equal to $1$, $$ \lim_{n \to \infty} n^2 (1 - 1/n) = (\lim_{n \to \infty} n^2) \cdot [\lim_{n \to \infty} (1 - 1/n)] = (\lim_{n \to \infty} n^2) \cdot 1 = \lim_{n \to \infty} n^2. $$ In Steven Gubkin's example, neither $n$ nor $1/n$ has a limit that exists and is a nonzero real number, so the theorem doesn't apply. In both cases, the key is to use precise, logically valid reasoning.

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    $\begingroup$ The reality is that almost no practitioners of calculus will ever remember the exact formal statements of a half-dozen theorems such as this one. Scientists and engineers who have developed competence in this kind of thing use informal modes of reasoning based on their correct insights into how the expression in question works. $\endgroup$
    – user507
    Nov 3, 2018 at 23:21
  • $\begingroup$ @DanielHast: but $\lim_{n\to \infty} n^2$ doesn't exist. So why are your students allowed to apply the theorem you quote? $\endgroup$ Nov 4, 2018 at 11:01
  • $\begingroup$ @MichaelBächtold: $c$ is allowed to be $\pm \infty$. $\endgroup$
    – user507
    Nov 7, 2018 at 16:12
  • $\begingroup$ @BenCrowell: the problem is that $\lim_{n\to \infty}n^2=\infty$, so I don't see how one is allowed to apply the theorem. If the theorem would apply, we should also be allowed to use it in Steven Gubkin's example. $\endgroup$ Nov 8, 2018 at 8:24
  • $\begingroup$ $\lim_{n \to \infty} (1 - 1/n)$ exists and is equal to a nonzero real number, and $\lim_{n \to \infty} n^2$ exists (in the extended real line), so the theorem applies with $g(n) = (1 - 1/n)$ and $f(n) = n^2$. $\endgroup$ Nov 10, 2018 at 3:48

To extend the answer by Daniel Hast: One theorem one might want to use is:

If $(a_n)_{n\in\mathbb N}$ and $(b_n)_{n\in\mathbb N}$ are convergent sequences then $$\begin{align} \lim_{n\to\infty} (a_n \pm b_n) &= \lim_{n\to\infty} a_n \pm \lim_{n\to\infty} b_n \\ \lim_{n\to\infty} (a_n \cdot b_n) &= \lim_{n\to\infty} a_n \cdot \lim_{n\to\infty} b_n \\ \lim_{n\to\infty} \frac{a_n}{b_n} &= \frac{\lim_{n\to\infty} a_n}{\lim_{n\to\infty} b_n} \end{align}$$ For the last equation one also needs $\lim_{n\to\infty} b_n \neq 0$.

Now, one can do something like $$\lim_{n\to\infty} \tfrac 1n + \sqrt[n]{4}=\lim_{n\to\infty} \tfrac 1n + \lim_{n\to\infty} \sqrt[n]{4}=0+1=1 \qquad(1)$$

The problem here is, that one applies the above theorem before showing, that the subsequences converge. So a better way to do it would be:

$$\lim_{n\to\infty} \tfrac 1n = 0 \land \lim_{n\to\infty} \sqrt[n]{4} = 1 \Rightarrow \lim_{n\to\infty} \tfrac 1n + \sqrt[n]{4}=\lim_{n\to\infty} \tfrac 1n + \lim_{n\to\infty} \sqrt[n]{4}=0+1=1\qquad (2)$$

(2) is the way to actually write down the proof and (1) is the way to find the proof (This need to be taught to students because they sometimes think that a proof and the solution process are the same). (2) also prevents you from failures because $\lim_{n\to\infty} a_n=\infty$ means, that $(a_n)$ diverges. I say to students:

$\infty$ is no real number. So $\lim_{n\to\infty} a_n=\infty$ means, that one cannot apply the above theorem, because $(a_n)$ diverges. But $\lim_{n\to\infty} a_n=\infty$ sometimes behave in computing with limits like it would converges. For example $\lim_{n\to\infty}(a_n+b_n)=\lim_{n\to\infty}(a_n+b_n)=\lim_{n\to\infty}a_n+\lim_{n\to\infty}b_n$ if one has limits of the form $\infty+\infty$ or $\infty+c$ [Of course one need to introduce the relevant theorems here]. But for $\infty-\infty$ one can never apply the above theorem. So if one limit is $\infty$ one needs to be careful for limits of the form $\infty-\infty$, $\infty\times 0$, $\frac{ \infty}{\infty}$.

  • $\begingroup$ @BenjaminDickman Thers's a big diverence: In (1) you apply the theorem without checking the premises of the theorem. You do it just later at the second equal sign. In (2) you first show the convergence of the subsequences and then apply the theorem. Note that $\lim_{n\to\infty} a_n=\infty$ means, that you cannot apply the cited theorem... $\endgroup$ Jun 5, 2015 at 12:52
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    $\begingroup$ Ah, okay... now I know what you meant... ;-) $\endgroup$ Jun 5, 2015 at 16:05

I can not count the times that I used or read expressions as $\epsilon+\epsilon^2=\epsilon$ or $(1+dx)(1+dy)-1=dx+dy$. It is usual infinitesimal reasoning.

Another example, any time that in the evaluation of a limit we use a Taylor series ended as dots we are implicitly doing an intermediate limit evaluation:

$$ \lim_{x \to 0}~ \sqrt{\frac{1+2x}{x^2}}-\frac{1}{x} = \lim_{x \to 0}~{\frac{1}{x}+1-\frac{x}{2}+\dots-\frac{1}{x}}=1 $$

Thus, state rules ( "never replace parts of an expression by their limits when taking a limit", "not to do it. ... take all limits in a single step") is not the way (never it is, rules are ok only for computers) and, when done, it is full of exceptions. Instead it is better help students to practice, learn the dangers, advantages, problems and concepts under some approach. In this way, students experience and acquire the concept of infinitesimal, asymptotic behavior, ... .

From this point of view, a student who writes:

$$ \lim_{n \to \infty}~ n^2-n = \lim_{n \to \infty}~ n^2 $$


$$ \lim_{n \to 0}~ (n^2+n)/n = \lim_{n \to 0}~ n/n = 1 $$

is showing he/she understands the relation between infinity/infinitesimal and powers, the concept of asymptotic behavior, ....

  • $\begingroup$ Right, sorry, the limit was at zero. Nevermind. $\endgroup$
    – Tommi
    Nov 2, 2018 at 12:06
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    $\begingroup$ This is certainly an accurate description of how, in my experience, scientists and engineers reason about this kind of thing. I don't think the relevant distinction is between limits and infinitesimals, but rather between formal rigor and correct informal reasoning by experienced practitioners. Correct reasoning about infinitesimals can be formal (using non-standard analysis) or informal. Correct reasoning about limits can likewise be formal or informal. The thing to keep in mind is that real-world practitioners essentially never use formal reasoning for this kind of thing. $\endgroup$
    – user507
    Nov 3, 2018 at 23:25

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