Hi guys. Final tomorrow and i had some last minute questions for proving/disproving a function is uniformly cont. Basically i want to know if the following proofs are acceptable Consider f(x)=1/x for x element (0, 2) = I Proof 1: f(x) does not converge uniformly on I. In order for a function to be uniformly continuous on an open interval, say (a,b) it must be able to be extended to a continuous function on [a,b]. Since lim f(x) = Infinity , the extension function cannot be continuous, therefore f(x) is x-->0 not uniformly continuous on I.// Proof 2: Thm: if f is uniformly continuous on S, and sn is a Cauchy sequence in S, then f(sn) is also a Cauchy sequence. consider the sequence sn= 1/n which is clearly Cauchy on I if f is uniformly continuous on I, then f(sn) will also be Cauchy. But f(sn) = n which is not a Cauchy sequence. thus, f is not uniformly continuous on I.// Are those 2 proofs acceptable? Any feed back would greatly be appreciated Also i wanna know if my epsilon delta proof is correct. Prove that X^2 is uniformly continuous on [0,3] (i know i can use that one theorem, that states if its continuous on a closed interval, its uniformly continuous but i wanna practice epsilon delta) Scratch work for each epsilon> 0 there exists delta>0 such that |x-y|< delta ==> |x^2 - y^2| < epsilon |x^2 - y^2| = |(x-y)(x+y)|< |(x-y)*9|<epsilon thus choose delta = epsilon/9 Forward Proof for each epsilon>0 there exists a delta>0 s.t. |x-y|<delta= epsilon/9 implies |f(x)-f(y)|= |(x-y)(x+y)|< |(x-y)*9| < |(epsilon/9)*9| = epsilon // Is this done correctly? Am i forgetting to do something. Any help is appreciated. Sorry for such a long post... Please reply over!