Instead try to write a program which proves that the limit sin(1/x) x->0 does not exist. (Remember proving that a single function doesn't work isn't enough. You have to prove no such function exists.)
I think it may be possible to phrase a claim for this.
for any almostZero, it is possible to find an integer m1, so that the m1.2.pi + pi/2 is greater than 1/almostZero and therefore for which sin(m1.2.pi+pi/2) =1; 1/(m1.2.pi + pi/2) will be smaller than 1/almostZero. It is also possible to find an integer m2, so that m2.pi+pi is greater than 1/almostZero and therefore for which sin(m2.pi+pi) =0; 1/(m2.pi+pi) will be smaller than 1/almostZero.
The claim: find me an almostZero for it is not possible to find such m1,m2.
This could be a possibility, but I haven't written the program yet ;-)
I suspect that such program could revolve around showing that for any arbitrarily small number -- almostZero -- you can always find two smaller numbers x1,x2 as such that sin(1/x1) is smaller and sin(1/x2) is larger than sin(1/almostZero). The Popper-compliant claim could be: find me an almostZero for which the defeating function will be incapable of producing such x1 and x2. I am not sure about all of this, though. Unless you have investigated something very similar before, it takes time to investigate things like that.
for any almostZero, it is possible to find an integer m1, so that the m1.2.pi + pi/2 is greater than 1/almostZero and therefore for which sin(m1.2.pi+pi/2) =1; 1/(m1.2.pi + pi/2) will be smaller than 1/almostZero. It is also possible to find an integer m2, so that m2.pi+pi is greater than 1/almostZero and therefore for which sin(m2.pi+pi) =0; 1/(m2.pi+pi) will be smaller than 1/almostZero. The claim: find me an almostZero for it is not possible to find such m1,m2.
This could be a possibility, but I haven't written the program yet ;-)