I have been assigned this question but the solution given uses a different method so i don't know if my solution is correct ∃ ϵ ¯ &gt; 0 s.t. | x ¯ − y ¯ | &lt; δ ∃ x ¯ , y ¯ ∈ D s.t. | f ( x ¯ ) − f ( y ¯ ) | ≥ ϵ ∀ δ &gt; 0 ↭ ∀ n ∈ N ∀ δ &gt; 0 ∃ n ∈ N s.t. 1 n &lt; δ | x ¯ − y ¯ | &lt; 1 n &lt; δ x ¯ = x ¯ n y ¯ = y ¯ n | x ¯ n − y ¯ n | &lt; 1 n y ¯ n = x ¯ n − 1 2 n | x ¯ n − y ¯ n | = | x ¯ n − x ¯ n + 1 2 n | = 1 2 n &lt; 1 n y ¯ n = x ¯ n − 1 2 n | x ¯ n − y ¯ n | = 1 2 n &lt; 1 n choose ϵ ¯ = 1 | f ( x ¯ ) − f ( y ¯ ) | ≥ ϵ ¯ = 1 | x ¯ n 3 − ( x ¯ n 3 − 3 x ¯ n 2 2 n + 3 x ¯ n 4 n 2 − 1 8 n 3 ) | = | 3 x ¯ n 2 2 n − 3 x ¯ n 4 n 2 + 1 8 n 3 | choose x ¯ n = 2 n 2 = | 3 ( 2 n 2 ) 2 2 n − 3 ( 2 n 2 ) 4 n 2 + 1 8 n 3 | = | 6 n 3 − 3 2 + 1 8 n 3 | ≥ 1 thus f ( x ) = x 3 , ( x ∈ R ) is not uniformly continuous, as required

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I have been assigned this question but the solution given uses a different method so i don&#039;t know if my solution is correct  ∃            ϵ      ¯        &amp;gt;  0    s.t.        |              x      ¯        −            y      ¯            |    &amp;lt;  δ  ∃            x      ¯        ,            y      ¯        ∈  D    s.t.        |    f  (            x      ¯        )  −  f  (            y      ¯        )      |    ≥  ϵ  ∀  δ  &amp;gt;  0  ↭  ∀  n  ∈      N    ∀  δ  &amp;gt;  0    ∃  n  ∈      N      s.t.        1    n    &amp;lt;  δ      |              x      ¯        −            y      ¯            |    &amp;lt;      1    n    &amp;lt;  δ            x      ¯        =                    x        ¯              n                y      ¯        =                    y        ¯              n        |                      x        ¯              n    −                    y        ¯              n        |    &amp;lt;      1    n                        y        ¯              n    =                    x        ¯              n    −      1          2      n            |                      x        ¯              n    −                    y        ¯              n        |    =      |                      x        ¯              n    −                    x        ¯              n    +      1          2      n            |    =      1          2      n        &amp;lt;      1    n                      y        ¯              n    =                    x        ¯              n    −      1          2      n            |                      x        ¯              n    −                    y        ¯              n        |    =      1          2      n        &amp;lt;      1    n        choose                             ϵ          ¯                    =      1            |    f  (            x      ¯        )  −  f  (            y      ¯        )      |    ≥            ϵ      ¯        =  1      |                      x        ¯              n    3    −  (                    x        ¯              n    3    −            3                                    x            ¯                          n        2                    2      n        +            3                                    x            ¯                          n                    4              n        2              −      1          8              n        3              )      |    =      |              3                                    x            ¯                          n        2                    2      n        −            3                                    x            ¯                          n                    4              n        2              +      1          8              n        3                  |        choose                                         x            ¯                          n            =      2              n        2              =      |              3      (      2              n        2                    )        2                    2      n        −            3      (      2              n        2            )              4              n        2              +      1          8              n        3                  |    =      |    6      n    3    −      3    2    +      1          8              n        3                  |    ≥  1      thus           f      (      x      )      =              x        3            ,      (      x      ∈              R            )         is not uniformly continuous, as required

Philip Deleon · Accepted Answer

This is a standard proof.Let us fix  ϵ  =  1Given any   δ  &amp;gt;  0, we choose a real number   x  &amp;gt;  0 such that                    3        δ                  x          2                    2        &amp;gt;  1Obviously,      |    (    x    +                  δ        2              )    −    x    |    =            δ      2        &amp;lt;  δHowever, we have      |                  (        x        +                              δ            2                          )            3        −          x      3        |    =      |                            3          δ                      x            2                          2              +                            3                      δ            2                    x                          2          2                      +                            δ          3                          2          3                      |    ≥                    3        δ                  x          2                    2        &amp;gt;  1Thus, we have shown that  f  (  x  )  =      x    3  is not uniformly continuous on       R  .

I have been assigned this question but the solution given uses a different method so i don't know if

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