# Analysis of limit of $\sin(1/x)$

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

• It is not defined at 0 however you can construct a sequence which takes your favourite value arrbitrarily close to zero Jun 25, 2021 at 9:42
• 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 $\sin(1/z)=-1$? Jun 25, 2021 at 9:53
• 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. Jun 25, 2021 at 11:21
• @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) Jun 25, 2021 at 15:06
• 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 $\{x_n\}.$ Jun 25, 2021 at 15:32

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 $$x\sin(1/x)$$.