Friday, 6 June 2014

calculus - Proving uniform continuity of trigonometric function arctan

In my assignment I have to prove that the following function is uniformly continuous:



f(x)=arctanxsin(1x) in the open interval (0,).



I thought I'd prove the one side limits x0+ 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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