What according to you will be the answer for: $$\lim_{x \to 0} \frac{\sin(1/x)}{\sin(1/x)}$$

My answer was $1$. The book says not defined and the argument is that both numerator and denominator are in $[-1,1]$, so we cannot determine what it exactly is.

My argument is, whatever value it takes, shouldn't it be the same in both numerator and denominator? Moreover, another example may be:

$$\lim_{x \to 0} \frac{1/x}{1/x}$$

I say $1$ again, because it is the same. But, it is not defined if we individually consider numerator and denominator! Still we have cancelled it in this case. Why can't I do it in case of sine function?

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    $\begingroup$ It is $1$. I find it hard to believe the book would say the limit does not exist. My guess is that you are mistranslating/misinterpreting something. $\endgroup$ – Git Gud Jun 1 '14 at 12:39
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    $\begingroup$ Well, perhaps the book meant that in any neighborhood of zero there are infinite values of $\;x\;$ for which $\;\sin\frac1x=0\;$ so the limit is not even well defined as the function is not well defined (you get $\;\frac00\;$ ...)...? And thus the book's right. $\endgroup$ – DonAntonio Jun 1 '14 at 12:40
  • $\begingroup$ @DonAntonio: You mean "is $0$", not "isn't defined"... $\endgroup$ – user21820 Jun 1 '14 at 12:41
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    $\begingroup$ Exactly @user21820, thanks. $\endgroup$ – DonAntonio Jun 1 '14 at 12:41
  • $\begingroup$ Remember! $a / a = 1$ is only true if $a \neq 0$. $\endgroup$ – Paul Draper Jun 2 '14 at 11:36

The limit does not exist, but not because "both numerator and denominator are in [-1,1], so we can not determine what it exactly is". The denominator is $0$ whenever $\frac{1}{x}$ is a multiple of $π$, so in any open interval containing $0$ there is a point at which the expression is not defined, and hence the limit does not exist. This has nothing to do with the oscillatory behaviour of the numerator and denominator.

In your other example, the limit is indeed $1$ because it is indeed defined and equal to $1$ in any open interval of $0$ less $0$ itself.

  • $\begingroup$ Actually, this depends on the definition of limit. In my Analysis textbook (Elon Lages Lima), for $lim_{x\to 0} f( x )$ to be defined, you only need $0$ to be a limit point in the domain of $f$ (and, of course, you need the limit to exist). I this case, the limit exists, and (obviously) is $1$. $\endgroup$ – fonini Jun 2 '14 at 3:28
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    $\begingroup$ @fonini: Yes it's certainly possible to have all kinds of definitions for limits, so it's definitely good to always be clear exactly what the definition in use means when applied to each example. I had simply assumed the 'standard definition' since that is the only reasonable explanation for the book's statement. $\endgroup$ – user21820 Jun 2 '14 at 7:39
  • $\begingroup$ As of today, WolframAlpha gets this limit right! I submitted a help desk report on 1 June 2014 and they emailed me back today that they'd fixed it. $\endgroup$ – wchargin May 12 '15 at 4:44
  • $\begingroup$ @WChargin: Do you know what they did to fix it? As I expected, they hard-coded that case in, so it fails on wolframalpha.com/input/…. Please don't submit any more help desk reports, because students need to be taught the hard way that wolfram alpha is often wrong. $\endgroup$ – user21820 May 12 '15 at 6:43
  • $\begingroup$ @user21820 Oh, that's disappointing. It doesn't even work on $\csc(1/x)/\csc(1/x)$! I know nothing about how they "fixed" it. All they said was: "The issue you reported has been fixed on the site. See <URL>…" I guess they won't be getting any more tickets from me… $\endgroup$ – wchargin May 12 '15 at 14:16

When considering $\lim_{x\to 0} f(x)$, it does not matter whether or not $f(0)$ is defined at all. However, we still should have a look at the domain where your function $f(x)=\frac{\sin(1/x)}{\sin(1/x)}$ is defined. The problem is that we may sometimes divide by $0$, namely first of all $\frac1x$ is not defined for $x=0$, but also $\sin(1/x)=0$ whenever $\frac1x$ is an integer multiple of $\pi$. Hence the maximal domain of $f$ is $$D=\mathbb R\setminus\bigl(\{0\}\cup\{\,\tfrac1{k\pi}\mid k\in\mathbb Z\setminus\{0\}\,\}\bigr).$$ We note that for arbitrary $\epsilon>0$, the set $D\cap(-\epsilon,\epsilon)$ has more gaps than just at $x=0$ and I can only imagine that this is what your book objects to. On the other hand, it makes sense to define $\lim_{x\to a} f(x)$ for all $a\in\overline D$ and that would include tha case $a=0$. Only an "unlucky" wording of the definition of limit may prevent this, so you should check back what the exact definition of $\lim_{x\to a}f(x)$ is in your book (including, what conditions are imposed about where $f$ is defined).

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    $\begingroup$ I don't think "unlucky" is the word I'd use here. This seems to be a rather basic exercise in elementary calculus, so asking for advanced wording for limits instead of the usual, basic ones is, perhaps, a little too much at this stage. $\endgroup$ – DonAntonio Jun 1 '14 at 12:51
  • $\begingroup$ I had not considered this point, thank you! So in a nutshell, when sin(1/x) is exactly zero because, 1/x may be n$pi$, it will be exactly zero divided by zero, which is not defined. $\endgroup$ – Saurabh Raje Jun 1 '14 at 13:22
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    $\begingroup$ @DonAntonio Yeah, if you have a better word, even with scary-quotes, feel free to improve ... $\endgroup$ – Hagen von Eitzen Jun 2 '14 at 9:12

I don't know if it should be a comment or an answer:

In most books, the definition of limit is: given $ \epsilon > 0$, there exists $\delta > 0$ such that $|f(x) - l| < \epsilon$ for all $0<|x| < \delta$. Using this definition, there is no limit, as explained by many other answers.

However, some books use: given $ \epsilon > 0$, there exists $\delta > 0$ such that $|f(x) - l| < \epsilon$ for all $x$ in the domain of $f$ satisfying $0<|x| < \delta$. With this second definition, then your function is constant on its domain of definition, and the limit exists.

I guess your book is using the first definition.


Go back to basics: if you would like to show the limit is 1 as x approaches 0, then given $ \epsilon > 0$ can you find $\delta > 0$ such that $|f(x) - 1| < \epsilon$ for ALL $0<|x| < \delta$ ?

The answer is NO because there are an (infinite) number of points in $(-\delta, 0) \cup (0,+\delta)$ where $f(x)$ is undefined - i.e. all those points as noted in previous answers where $1/x$ is a multiple of $\pi$.


My view of the problem: you function $f$ is defined on $D=(0,1]\setminus\{x\in [0,1]:\frac{1}{x\pi}\in\Bbb N\}$. Easy to see that $f\equiv 1$ on $D$. Therefore we can, by continuity, say that $f=1$ on $\bar D=[0,1]$.

So, in other words, the limit does not exist as it is, because in each neighbourhood of $0$ there's a point where the function is not defined explicitly. However, in all of those points we can define our function by continuity and then pass to the limit $x\to 0$.


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