6 нояб. 2011 г. · This implies that neither sin(1/x) nor cos(x)/x is uniformly continuous on (0,1). ... Is f(x)=xsin(x−2) uniformly continuous on (0,1)? Is f(x)=sin ...

16 мар. 2017 г. · h(x)=xsin(1x) can be extended to a continuous function on [0,1] by setting h(0)=0. As [0,1] is compact it follows that h(x) is uniformly ...

8 нояб. 2020 г. · A continuous function on (0,1) is uniformly continuous if and only if it can be extended continuously to [0,1]. See for example math.

21 нояб. 2014 г. · As x changes from an infinitely small nonzero "number" to another infinitely small nonzero "number", sin(1/x) goes all the way from 1 to −1.

9 нояб. 2018 г. · However, we know that xsin1x→0 as x→0 and this implies continuous extendibility to [0,1) and, hence uniform continuity on (0,1). The argument ...

24 нояб. 2015 г. · Prove f(x)=xsin(1/x) is uniformly continuous on (0,1). I have tried to use the definition to prove this. |x−y|<δ⟹|f(x)−f(y)|<ϵ.

9 нояб. 2013 г. · I think it is also uniformly continuous on (0,1). If (xn) and (yn) are sequences that converge to 0, then since sin is bounded, xnsin(1/xn) and ...

27 сент. 2021 г. · Despite |y−x|<δ for all δ>0. As ε was arbitrary, this concludes the proof that sin(1/x) is not uniformly continuous in (0,1).

4 февр. 2016 г. · So I am tasked with finding whether 1sin(x)−1x is uniformly continuous on the open interval I=(0,1). To look at the "simple" ways to prove it ...

3 мая 2020 г. · Let f:(0,1]→R be continuous on the domain. I want the condition of f(x) where g(x)=f(x)sin(1/x) is uniformly continuous on (0,1]. I expect that ...