17 нояб. 2013 г. · Since [0,1] is a compact set, accoring to the Heine-Cantor theorem f(x)=√x is uniformly continuous on [0,1]. Moving along to [1,+∞) ...

21 апр. 2017 г. · Hence, √x is uniformly continuous on [1,∞). √x is continuous on [0,1] , so √x is uniformly continuous on [0,1].

27 окт. 2017 г. · f is differentiable on (0,∞) and has a bounded derivative on, say, [1,∞). Therefore f is Lipschitz, hence uniformly continuous, on [1,∞).

3 дек. 2012 г. · 1/√x is not uniformly continuous on the interval (0,1). No. Otherwise it can be extended continuously to the end points 0 & 1. But the given ...

14 апр. 2018 г. · The function is continuous on [0,1] so therefore it is uniformly continuous on that interval. From 1 to infinity, the derivative is bounded.

11 нояб. 2014 г. · Show that f(x)=1√x is uniformly continuous on the domain (1,∞) but not on the domain (0,1).

11 июн. 2014 г. · How to show √x is uniformly continuous at [0,1] and [1,∞) ... From the definition if we choose δ=ϵ2 |√x−√y|2≤|√x−√y||√x+√y|=|x−y|<ϵ2⟹|√x−√y|<ϵ.

17 нояб. 2016 г. · Accordingly, you can prove that √x is uniformly continuous on [0,1] using any method (Heine-Cantor would be the easiest), and then you can show ...

17 нояб. 2020 г. · The function √ x is uniformly continuous on [0,∞). It is uniformly continuous on [0,1] because it is continuous, and because [0,1] is compact.

12 нояб. 2015 г. · √2 is a constant, not a function. If you mean f(x)=√2 for all values of x, it would be uniformly continuous.