In my assignment I have to prove that the following function is uniformly continuous:
f(x)=arctanx⋅sin(1x) in the open interval (0,∞).
I thought I'd prove the one side limits x→0+ and x→∞−.
lim
For the second limit I thought using the squeeze theorem. Please let me know if I made a mistake:
0\le \arctan x \cdot \sin \left(\frac{1} {x} \right)\le \arctan x .
Therefore, from the squeeze theorem:
\lim_{x\to 0^+} \arctan x \cdot\sin\left(\frac{1} {x}\right) =0
Since \arctan and \sin are continuous always when it is defined, we can say that the function is uniformly continuous, since the one side limits exists.
I have only one concern here: zero is not defined here, in the open interval from the beginning. Is it a problem to use the squeeze theorem here? If it is, thought I'd use \frac {x} {x+1} instead of 0.
Thanks,
Alan
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