**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.