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What is the limit:

$lim_{x to 0}sin(frac{1}{x})$?

I plotted this function on https://www.desmos.com/

And this was the result:

I searched the web for this function and it was straight forward stated as "The limit doesn't exist" without any theoretical proof. Could anyone provide a theoretical proof or something a bit more concrete?

Suppose that $ell=lim_{Xto 0}sin(1/x)$. Using the definition of the limit with $varepsilon=1/2$, we get the existence of a positive $delta$ such that if $0lt leftlvert xrightrvertltdelta$, then $leftlvert sin(1/x)-ellrightrvertlt 1/2$.
In particular, if $0lt s,tlt delta$,
$$
leftlvert sinleft(frac1sright)-sinleft(frac1tright)rightrvert
leqslantleftlvert sinleft(frac1sright)-ellrightrvert+leftlvert ell-sinleft(frac1tright)rightrvertlt 1 $$
hence we reach a contradiction by choosing $$s=frac1{2pi n+pi/2}mbox{ and }t=frac1{2pi n-pi/2}$$
where $ngeqslant 1$ is such that $2pi n+pi/2>delta^{-1}$.

Recall that when a limit

$$lim_{xto x_0} f(x)=L$$

exists it is unique and it is the same for all the subsequences, that is

$$forall x_n to x_0 implies f_n=f(x_n) to L$$

Therefore to prove that a limit doesn't exist it suffices to show that at least two subsequences exist with different limit.

In this case let consider

$$x_n=frac2{pi n}to 0^+$$

then

$$sinleft(frac{1}{x_n}right)=sinleft(nfrac pi 2right)$$

What can we conclude form here? (try for example with $n=4k$ and $n=4k+1$)

The reason is essentially because the function "oscillates infinitely back and forth and does not settle on a single point". If this does not satisfy you, we may prove this formally with the following theorem

$lim_{xto c}f(x)=L$ if and only if, for every sequence $(x_n)inmathbb R$ tending to $c$, it is true that $(f(x_n))$ tends to $L$.

This is sometimes known as the sequential criterion for the convergence of a function. But in our case, this condition means that the limit does not exist. This is because we can just pick two sequences: the first so that the sequence $(f(x_n))$ corresponds to the points at the "top" of the graph; i.e., $f(x_n)=1$ for all $n$. You can find an explicit formula if you wish. Similarly pick our second sequence so that $f(y_n)=-1$ for all $n$. If we make sure both sequences are, say, positive, and decreasing to $0$, then we now have two sequences converging to $0$, but giving a different limit: $1$ and $-1$ respectively. If the limit existed, then by the theorem, $1=-1$, which is obviously absurd. We are done.

(BTW, if even this isn't rigorous enough for you, you can prove $(x_n),(y_n)to 0$ by the monotone convergence theorem, and that $(f(x_n))to1$, $(f(y_n))to-1$ by using the formal definition for the limit.)

I will use the property that for all $Ninmathbb{R}$ there exists $x,y>N$ such that $sin(x)=1$ and $sin(y)=-1$. If you also want a proof of this, just tell me.

Assume for the contrary that $lim_{xto0}sin(frac1x)$ exists, so it equals some $Linmathbb{R}$. Then, by definition of the limit, for $varepsilon:=frac12>0$ we get some $delta>0$ such that for all $x$ with $|x|<delta$ we have $|sin(frac1x)-L|<varepsilon=frac12$. Now use the beforementioned property that for $N:=frac1{delta}$ there exist $x',y'>N$ such that $sin(x')=1$ and $sin(y')=-1$. Define $x:=frac1{x'}$ and $y:=frac1{y'}$. Then $|x|,|y|<delta$ and $sin(frac1x)=1$ and $sin(frac1y)=-1$. However, we then find by the triangle inequality $2=|(-1-L)-(1-L)|leq|sin(frac1y)-L|+|sin(frac1x)-L|<2varepsilon=1$. A contradiction.

Let’s look at half the limit: approaching from the right, say. Note that $1/xtoinfty$ as $xto 0^+$, so we may rewrite the limit as

$$lim_{xto 0^+} sin(1/x) = lim_{xtoinfty} sin(x).$$

Can you see why $sin(x)$ doesn’t have a limit in this case?