I was playing around with the function $f : \mathbb R \rightarrow \mathbb R$, defined as follows. $$f(x) = \frac{\sin\left(x \sin \left( \frac 1x \right) \right)}{x \sin \left( \frac 1x \right)}$$ This function is undefined at $x = \frac{1}{n\pi}$ for all $n \in \mathbb N$. Namely, this suggests

$$\forall \delta > 0 : \exists x : 0 < |x| < \delta \wedge f(x) \text{ is undefined.}$$

I'm curious about $\lim_{x \rightarrow 0} f(x)$. If this value exists, say set $\lim_{x \rightarrow 0} f(x) = L$, we naturally must have that

$$\forall \varepsilon > 0 : \exists \delta > 0 : 0 < |x| < \delta \Rightarrow |f(x) - L| < \varepsilon.$$

But, since $f$ is not defined for all $x$ in the set $0 < |x| < \delta$, the conclusion $|f(x) - L| < \varepsilon$ cannot always be guaranteed to hold. So, I'm inclined to say that the limit $L$ does not exist.

That being said, the graph looks like this and based on the graph, $f$ appears to satisfy this inequality: $$\forall x : \frac{\sin x}{x} \leq f(x) < 1$$

Since $\displaystyle \lim_{x\rightarrow 0} \frac{\sin x}{x} = 1$, this suggests in a squeeze-theorem-esque way that $L$ "wants" to take the value $1$, but of course we cannot actually apply the squeeze theorem here for the same reason that we couldn't apply the $\varepsilon$-$\delta$ definition directly.

So, my questions are:

  1. Is my analysis correct that $\displaystyle \lim_{x \rightarrow 0} \frac{\sin\left(x \sin \left( \frac 1x \right) \right)}{x \sin \left( \frac 1x \right)}$ does not exist?
  2. Is there any looser definition of a limit in common use (say, ${\lim}^\star$) that would set $\displaystyle {\lim_{x \rightarrow 0}}^\star \frac{\sin\left(x \sin \left( \frac 1x \right) \right)}{x \sin \left( \frac 1x \right)} = 1$ in a way that captures the spirit of what I described above?
  • $\begingroup$ Well, perhaps the term $x\sin \frac 1x$ could be "unwound" a bit by substitution of $y=\frac 1x$, then $\frac 1y\sin y$ would be a more "stable" quantity and perhaps the limit could be determined... $\endgroup$
    – abiessu
    Commented Jan 15, 2023 at 18:21
  • $\begingroup$ More specifically, for all $y_0\in[0,\pi)\cup(\pi,2\pi)$ you have $\lim_{n\to\infty}\frac1{y_0+n\pi}\sin(y_0+n\pi)=0$... $\endgroup$
    – abiessu
    Commented Jan 15, 2023 at 18:37
  • 1
    $\begingroup$ Your limit exists. See the general definition en.wikipedia.org/wiki/Limit_of_a_function#More_general_subsets $\endgroup$ Commented Jan 15, 2023 at 18:46
  • $\begingroup$ Since $\lim_{t\rightarrow0}\frac{\sin t}{t}=1$, there is no harm in extending the value of $\sin t/t$ ad $1$ when $t=0$. Notice that $x\sin(1/x)\xrightarrow{x\rightarrow0}0$ and so, your limit exists (and equals 1). $\endgroup$
    – Mittens
    Commented Jan 15, 2023 at 19:13
  • $\begingroup$ Related 1 and Related 2 $\endgroup$
    – user
    Commented Jan 15, 2023 at 19:38

2 Answers 2


First of all the limit exists: if you substitute $y=x\sin(\frac{1}{x})$ then you have $y \to 0$ for $x \to 0$ (sin is a bounded function so $\lim_{x \to 0} x\sin\frac{1}{x} = 0$). So $\lim_{x \to 0}\frac{\sin(x\sin(\frac{1}{x})}{x\sin(\frac{1}{x})}=\lim_{y \to 0} \frac{\sin(y)}{y}=1$. Secondly of course if $f$ in $x$ is not defined also $f(x)-L$ is not defined, in the definition of a limit you have that all $x$ have to be in the domain of $f$.

  • $\begingroup$ What's the justification that a substitution like that is valid? $\endgroup$ Commented Jan 15, 2023 at 20:09
  • $\begingroup$ It's a theorem math.stackexchange.com/a/167948/1139301 $\endgroup$ Commented Jan 15, 2023 at 20:27
  • 1
    $\begingroup$ Oh nice! I had entirely forgotten about limit substitutions lol. Originally I was thinking of using sandwich theorem on y=-x^2+1 and y=1 but was afraid it wouldn’t be as rigorous as needed. This works much better imo $\endgroup$ Commented Jan 15, 2023 at 20:52

While $f$ is undefined at $\frac{1}{n\pi}$ the limits $\lim_{x\rightarrow \frac{1}{n\pi}}f(x)$ seem to exist according to the visualization (you would have to check that) so you could consider the continuous extension of $f$ and take $\lim_{x\rightarrow 0}$ of that.

  • $\begingroup$ Thank you, yes, that's basically the closest thing I had in mind. $\endgroup$ Commented Jan 15, 2023 at 20:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .