Here is the key idea.
However close to 0 we start, it will happen that as x rarr0,
we get infinitely many values of (2pi)/x that are coterminal with pi/2, so they have sin((2pi)/x)=1. (Specifically, for every x=4/(1+4n with n an integer.)
We also get infinitely many values of (2pi)/x that are coterminal with (3pi)/2, so they have sin((2pi)/x)=-1. (Specifically, for every x=4/(3+4n with n an integer.)
sin((2pi)/x) cannot be getting closer and closer to some number L, because we will always have some x's for which the distance is abs(L-1) and others for which the distance is abs(L-(-1)) = abs(L+1). These distances cannot both go to 0.
Using the definition
Recall the definition:
lim_(xrarra)f(x)=L if and only if
For every epsilon > 0, there is a delta > 0 for which
for every x with 0 < abs(x-a) < delta, we have abs(f(x)-L) < epsilon.
To use the definition to show that a limit does not exist requires showing that there is an epsilon> 0 for which, for every delta > 0 there is an x with 0 < abs(x-a) < delta, but abs(f(x)-L) >= epsilon.
Given L (a proposed limit), choose epsilon = 1.
Claim: for every delta > 0 there is an x with 0 < abs(x-0) < delta, but abs(sin((2pi)/x)-L) >= epsilon
Proof of claim: For every delta > 0, there is an integer n for which abs(4/(1+4n)) < delta and abs(4/(3+4n)) < delta.
Let x_1 = 4/(1+4n) and x_2 = 4/(3+4n) for such an n.
Then abs(f(x_1)-L) = abs(1-L) = abs (L-1).
and abs(f(x_2)-L) = abs(-1-L) = abs (L+1).
Claim: we cannot have both abs (L+1) < 1. and abs (L+1) < 1.
Suppose abs (L+1) < 1. Then -1 < L+1 < 1.
But his implies that -3 < L - 1 <-1, so abs(L-1) >=1
This completes the proof.
