24 сент. 2013 г. · It is continuous. However, it is not uniformly continuous. Suppose it were; then for every ϵ, there exists a δ for which |x−y|<δ⟹|x2−y2|<ϵ.

26 июн. 2015 г. · The easiest way to actually prove that a function is not uniformly continuous is to show that δ can't exist.

25 июл. 2015 г. · In your proof, you need |x+x0|<2 to prove uniform continuity, which is true on [0,1] but not on R. So the proof cannot be extended to R.

13 янв. 2020 г. · Definition : f(x)=1/x2 is uniformly continuous on (0,1) if, given ϵ>0, there exists a δ>0 such that for all x,y∈(0,1) with |x−y|<δ, we have |1x2 ...

3 февр. 2019 г. · Proving 1x2 is not uniformly continuous on (0,2] Since both x,y∈(0,2] above eq. reduces to |1/x2−1/y2|=|y−x|4<δ4(by the definition of ...

23 апр. 2018 г. · Using this theorem, you need to show that f(0) cannot be defined so that f is continuous. Simply saying that "f(0) is not defined" is not quite ...

16 нояб. 2020 г. · Let xn=nandyn=n+2n. we have limn→+∞|xn−yn|=0. But. limn→+∞|x2n−y2n|=4≠0. thus x↦x2 is not uniformly continuous at R. We proved that with ϵ=3 and ... Не найдено: infinity) | Нужно включить: infinity)

3 нояб. 2017 г. · A function f:A→R fails to be uniformly continuous if and only if there exists a particular ϵ0>0 and two sequences (xn) and (yn) in A satisfying ...

1 мар. 2017 г. · Let me show one of my favourite ways with an example. Let's show that f(x)=x2 is not uniformly continuous on [0,+∞).

22 апр. 2019 г. · Actually 1/x2 is not uniformly continuous on (0,∞), since at every point x you need a different constant δ in the definition of continuity. As x ... Не найдено: infinity) | Нужно включить: infinity)