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Image for reference

I was reading about discontinuous functions when I came across this function. It was stated that the function has no limiting value at $0$ since it rapidly oscillated between $[-1, 1]$ as we approached $0$.

My question is, thinking graphically, as we approach close to $0$ the inclination of the graph should be almost $90$ degrees ACW since the rate of oscillation increases exponentially as we move toward $0$ which means the graph can appear to pass through the origin (since it is an odd function; although it is undefined at origin) which should mean the limit is $0$.

Where am I going wrong with this intuition?

Edit: Added a drawing for more clarity on my question enter image description here

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    $\begingroup$ It is not defined at 0 however you can construct a sequence which takes your favourite value arrbitrarily close to zero $\endgroup$ Jun 25, 2021 at 9:42
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    $\begingroup$ Can you reconcile your intuition with the fact that no matter how small you choose $x>0$, there are still infinitely many $0<y<x$ such that $\sin(1/y)=1$ and infinitely many $0<z<x$ such that $\sin(1/z)=-1$? $\endgroup$ Jun 25, 2021 at 9:53
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    $\begingroup$ For more fun, look at the graphs of $y = \sin\left(\frac{1}{\sin (1/x)}\right)$ and $y = \sin\left(\frac{1}{\sin\left(\frac{1}{\sin (1/x)}\right)}\right)$. For the last one you'll probably have to rely mostly on what you know happens, because most of the details can't be shown graphically, unless done in a very schematic way. $\endgroup$ Jun 25, 2021 at 11:21
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    $\begingroup$ @Draculin, a last argument: let's assume that you actually could "zoom in" like you've shown in your picture. The curve is vertical, which means the derivative must tend to either $+\infty$ or $-\infty$. Importantly, that's one or the other, right? The derivative is fairly easy to calculate: $\frac{-\cos(1/x)}{x^2}$. So what is the limit of that derivative as $x$ tends to $0$? Is it $+\infty$, $-\infty$ or is it in fact indeterminate? (hint: it's the last one) $\endgroup$ Jun 25, 2021 at 15:06
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    $\begingroup$ To further extend what @user3733558 said, for each extended real number $r$ there exists a sequence $\{x_n\}$ such that $x_n \rightarrow 0$ (even a one-sided such sequence) such that the limit of the difference quotients evaluated relative to $x=0$ and $x=x_n$ approaches $r$ (in fact, when $r$ is not infinite, the sequence can be chosen so that all the difference quotients are equal to $r).$ In the case that $r$ is finite, simply consider the intersections of $y = \sin(1/x)$ with the line $y = rx$ to come up with the desired sequence $\{x_n\}.$ $\endgroup$ Jun 25, 2021 at 15:32

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The rate of the oscillation does indeed "blow up" around $x=0$. What this means is that there are infinitely many waves in the interval $(-\delta,\delta)$ for any positive $\delta$, and in each of these waves you can find $y$-values of between $-1$ and $1$. Since the $y$-values don't settle upon a single value, the limit does not exist. Try comparing the behaviour of $\sin(1/x)$ around $x=0$ with $x\sin(1/x)$.

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