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 ... Prove that the function $f(x) = \sin(1/x)$ is not uniformly ... Is $\sin(1/x)$ uniformly continuous on $(0,1) Другие результаты с сайта math.stackexchange.com

Продолжительность: 5:32
Опубликовано: 11 февр. 2021 г.

Продолжительность: 10:32
Опубликовано: 3 авг. 2024 г.

26 июл. 2024 г. · Answer: 1/x and sin(1/x) are not uniformly continuous on (0,1) as they do not send Cauchy sequences to Cauchy sequences. Lets prove it here.

10 июл. 2016 г. · Moreover, for the same reason, f(x)=xαsin(1x2) f ( x ) = x α sin ⁡ ( 1 x 2 ) is uniformly continuous for any α>0 α > 0 on (0,1].

Let f(x) = sin(1/x) for 0 < x ≤ 1. Prove that f is not uniformly continuous on. (0,1]. 2. Let f(x)=1/(1 + |x|) for real x. Prove that f is uniformly continuous ...

14 янв. 2019 г. · Is h(x) = xsin(1/x) uniformly continuous on (0,1)?. 1.2 Good Solution h(x) = xsin(1/x) is uniformly continuous on (0,1). Proof: Let > 0.

To prove that f(x) = sin(1/x) is not uniformly continuous in (0,1), we need to show that there exists an ε > 0 such that for any δ > 0, there are points x, ...

By Nonuniform Continuity Criterion 5.4.2, g(x) := sin(1/x) is not uniformly continuous on B. 4. Show that the function ...

3 нояб. 2010 г. · Hello, I want to show that f ( x ) = x ⋅ sin ⁡ ( 1 x ) in the interval (0, infinity) is uniformly continuous using the following definition: