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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 $2\pi n <t< \pi/3+2\pi n$ or $2\pi/3 + 2\pi n <t< \pi+2\pi n$. What is the corresponding range of $x$ in each case? – user31373 Sep 27 '12 at 1:18
up vote 1 down vote accepted

inf is -1/pi, sup does not exist

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