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My original approach was to let $f = \sin (1/x)$ and do regular calculus. However, I found that it wasn't so simple. I graphed the function on Mathematica and it was even worse!

It seems like the function tends to infniity near the origin (or negative infinity?) IN either case, I don't think it is bounded above nor does a supermum exist because of the singularity.

Any insights?

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    Let $t=1/x$ and begin with the observation: 0<\sin t<1/2 if and only if either 2\pi n or 2\pi/3 + 2\pi n . What is the corresponding range of $x$ in each case?2012-09-27

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inf is -1/pi, sup does not exist