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.




limx=π20=0.



For the second limit I thought using the squeeze theorem. Please let me know if I made a mistake:



0arctanxsin(1x)arctanx.



Therefore, from the squeeze theorem:



limx0+arctanxsin(1x)=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 xx+1 instead of 0.



Thanks,



Alan

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