Tuesday, 10 January 2017

analysis - Prove that f(x)=x1/5 is continuous everywhere

Need to prove that f(x)=x1/5 is continuous everywhere, where f:RR:



from definition we need to show that given ϵ>0 δ>0 s.t. |xx0|<δ|x15x150|<ϵ for any point x0R



I have a proof but it's somewhat unjustified:



consider
|x15x150||x15||x150| from the triangle inequality since |x15|<|x| and |x150|<|x0| then |x15x150||x15||x150|<|x||x0|=|xx0|=δ so we can choose δ=ϵ?




Overall I'm not happy with the proof, in the last inequality I don't think I can just state that delta = epsilon and be done, but I have no idea what else to do. I also am not sure about this step |x||x0|=|xx0|and |x15|<|x| and |x150|<|x0| that step also...



if anyone could help me out.. thank you

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