Recently, I showed my Calculus students how to show that

$$ \lim_{x \to \infty} \frac{\sin(x)}{x} = 0, $$

by using the squeeze theorem.

An interesting question that I was asked several times was, "how come we couldn't just conclude that it is zero, without using the squeeze theorem, since it is obvious that the limit is zero."

I told them that the function is still oscillating, even though the oscillations are tiny, and so we need to make it rigorous by applying the squeeze theorem. They weren't convinced ...

What else could I say to them? Is there a nice counter-example / explanation?

(No epsilon-delta arguments, please ... )


  • $\begingroup$ It's hard to know what an acceptable level of "rigor" may exist for a class learning the very basics of math. I could restructure their argument to be "sin(x) is bounded, while x is not, so the only possible limit to sin(x)/x is 0" which I think would fly in most applied math courses. $\endgroup$
    – Kaynex
    Commented Nov 4, 2017 at 9:29
  • 6
    $\begingroup$ It's also “obvious” that, since the base is $>1$, $\lim_{x\to\infty}\left(1+\dfrac{1}{x}\right)^x=\infty$. ;-) $\endgroup$
    – egreg
    Commented Nov 4, 2017 at 10:10
  • $\begingroup$ Why is that obvious? $\endgroup$
    – Kaynex
    Commented Nov 4, 2017 at 10:17
  • 1
    $\begingroup$ Here is possible related question 'Obvious' theorems that are actually false. By the way it is hard to give you advice since they didnt present reason why it is obvious, it is equivalent of saying this is true, and I dont need to give reasons why, that is a bad practice not only in mathematics... $\endgroup$
    – Sil
    Commented Nov 4, 2017 at 10:29
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    $\begingroup$ @D.Hutchinson Intuition can lead to the correct result, but can also lead to disaster; this should teach why rigor is necessary and, in the present case, rigor is provided by the squeeze theorem. $\endgroup$
    – egreg
    Commented Nov 4, 2017 at 11:26

2 Answers 2


Apply the Socratic Method of dialectics. I would focus on asking them how do they perceive it as obvious. Unless they were simply being hasty which I don't think was the case, they probably had an intuition of the fact that $x \rightarrow \infty$ "fast enough" to negate the influence the numerator has over the value of the fraction, that is they understand that the numerator is bounded whereas the denominator is not.

If this is the case, then show them that the Squeeze Theorem is simply making use of this "obvious" fact. If they insist on its obsolescence then ask one-the most capable of the deniers will do-and ask him to show how he would prove it. Guide him step by step and help him express his intuition in a rigorous manner-and that would be the Squeeze Theorem.

As for counterexamples if the above is still not sufficient perhaps try with something like $$\lim_{x\rightarrow+\infty}\dfrac{\ln(x)}{x^2}$$

(though convincing them that $\ln(x)\lt x$ could be tricky unless presented as fact)


Well, think about it for a second. The sine of anything can't be less than $-1$ and more than $1$ ($-1\leqslant\sin{x}\leqslant1$). In other words, whatever $\sin{x}$ yields, the number in the numerator always stays within the closed interval $[-1, 1]$. As the $x$ in the denominator gets larger and larger, the fraction overall should get smaller and smaller from both the negative and positive sides because the numerator just oscillates between two fixed numbers while being divided by an ever growing number. The entire thing just has got to go to zero.

  • $\begingroup$ No, my argument has nothing to do with your example of the infinite sum of $\frac{1}{n}$. We're not dealing with a sum here. Here, we've got a series of numbers that approach zero at each step as $x$ gets larger. $\endgroup$ Commented Nov 4, 2017 at 10:05
  • $\begingroup$ And my example is not supposed to be super mathematically sound, but it certainly intuitively makes perfect sense. I thought that's what the original poster was asking for. $\endgroup$ Commented Nov 4, 2017 at 10:11
  • $\begingroup$ This is precisely the squeeze theorem, i.e. the argument that the OP wants to avoid using. Read the question carefully. $\endgroup$
    – Alex M.
    Commented Nov 4, 2017 at 16:12

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