# Does $\displaystyle \lim_{x \rightarrow 0} \frac{\sin\left(x \sin \left( \frac 1x \right) \right)}{x \sin \left( \frac 1x \right)}$ exist?

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?
• 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... Commented Jan 15, 2023 at 18:21
• More specifically, for all $y_0\in[0,\pi)\cup(\pi,2\pi)$ you have $\lim_{n\to\infty}\frac1{y_0+n\pi}\sin(y_0+n\pi)=0$... Commented Jan 15, 2023 at 18:37
• Your limit exists. See the general definition en.wikipedia.org/wiki/Limit_of_a_function#More_general_subsets Commented Jan 15, 2023 at 18:46
• 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). Commented Jan 15, 2023 at 19:13
• – user
Commented Jan 15, 2023 at 19:38

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 $$f$$.
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.