Proving a Polynomial Limit where x → 0

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Prove



$$lim limits_xto 0 x^3+x^2+x = 0$$
Note
$|f(x)-L| = |x^3+x^2+x|$



Assume
$ |x-c|<delta implies |x| < delta$



$implies |x^3+x^2+x|<deltacdot|x^2+x+1|$



Assume $|x| < 1 implies -1 < x < 1 implies 0 < x+1 < 2$



And I am not sure where to go from there, since I can't multiply the inequality by $x$ in order to get $x^2$, because $x$ could be negative or positive.




limits epsilon-delta




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    up vote
    4
    down vote

    favorite
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    Prove



    $$lim limits_xto 0 x^3+x^2+x = 0$$
    Note
    $|f(x)-L| = |x^3+x^2+x|$



    Assume
    $ |x-c|<delta implies |x| < delta$



    $implies |x^3+x^2+x|<deltacdot|x^2+x+1|$



    Assume $|x| < 1 implies -1 < x < 1 implies 0 < x+1 < 2$



    And I am not sure where to go from there, since I can't multiply the inequality by $x$ in order to get $x^2$, because $x$ could be negative or positive.




    limits epsilon-delta




    share|cite|improve this question























      up vote
      4
      down vote

      favorite
      1









      up vote
      4
      down vote

      favorite
      1






      1





      Prove



      $$lim limits_xto 0 x^3+x^2+x = 0$$
      Note
      $|f(x)-L| = |x^3+x^2+x|$



      Assume
      $ |x-c|<delta implies |x| < delta$



      $implies |x^3+x^2+x|<deltacdot|x^2+x+1|$



      Assume $|x| < 1 implies -1 < x < 1 implies 0 < x+1 < 2$



      And I am not sure where to go from there, since I can't multiply the inequality by $x$ in order to get $x^2$, because $x$ could be negative or positive.




      limits epsilon-delta




      share|cite|improve this question













      Prove



      $$lim limits_xto 0 x^3+x^2+x = 0$$
      Note
      $|f(x)-L| = |x^3+x^2+x|$



      Assume
      $ |x-c|<delta implies |x| < delta$



      $implies |x^3+x^2+x|<deltacdot|x^2+x+1|$



      Assume $|x| < 1 implies -1 < x < 1 implies 0 < x+1 < 2$



      And I am not sure where to go from there, since I can't multiply the inequality by $x$ in order to get $x^2$, because $x$ could be negative or positive.





      limits epsilon-delta





      share|cite|improve this question












      share|cite|improve this question




      share|cite|improve this question








      edited 2 days ago









      José Carlos Santos

      112k1696172




      112k1696172









      asked 2 days ago









      Vpie649

      915




      915




















          4 Answers
          4






          active

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          up vote
          2
          down vote



          accepted










          You just need a "good enough" estimate of $|x^2+x+1|$ for suitably small $|x|$



          Note that for $|x|lt 1$ you have $|x^2+x+1|le |x^2|+|x|+1lt 3$



          So if $deltalt 1$ you have $|x^3+x^2+x|lt 3delta$ and you can make this as small as you like.






          share|cite|improve this answer




























            up vote
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            Take $varepsilon>0$ and pick $delta=minleft1,fracvarepsilon3right$. Then, $|x|<deltaimplies|x|<1implies|x|^2,|x|^3<|x|$. On the other hand, $$|x|<fracvarepsilon3implies|x^3+x^2+x|leqslant|x|^3+|x|^2+|x|<fracvarepsilon3+fracvarepsilon3+fracvarepsilon3=varepsilon.$$






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              up vote
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              Hint: Use the triangle inequality $|a+b|leq |a|+|b|$. You can conclude $|x^2+x+1|leq |x|^2+|x|+1$ and you don't have to care about the sign of $x$.






              share|cite|improve this answer




























                up vote
                1
                down vote













                As an alternative by squeeze theorem assuming $|x|<1$



                $$0le |x^3+x^2+x|=|x||x^2+x+1|le 3|x| to 0$$






                share|cite|improve this answer





















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                  4 Answers
                  4






                  active

                  oldest

                  votes








                  4 Answers
                  4






                  active

                  oldest

                  votes









                  active

                  oldest

                  votes






                  active

                  oldest

                  votes








                  up vote
                  2
                  down vote



                  accepted










                  You just need a "good enough" estimate of $|x^2+x+1|$ for suitably small $|x|$



                  Note that for $|x|lt 1$ you have $|x^2+x+1|le |x^2|+|x|+1lt 3$



                  So if $deltalt 1$ you have $|x^3+x^2+x|lt 3delta$ and you can make this as small as you like.






                  share|cite|improve this answer

























                    up vote
                    2
                    down vote



                    accepted










                    You just need a "good enough" estimate of $|x^2+x+1|$ for suitably small $|x|$



                    Note that for $|x|lt 1$ you have $|x^2+x+1|le |x^2|+|x|+1lt 3$



                    So if $deltalt 1$ you have $|x^3+x^2+x|lt 3delta$ and you can make this as small as you like.






                    share|cite|improve this answer























                      up vote
                      2
                      down vote



                      accepted







                      up vote
                      2
                      down vote



                      accepted






                      You just need a "good enough" estimate of $|x^2+x+1|$ for suitably small $|x|$



                      Note that for $|x|lt 1$ you have $|x^2+x+1|le |x^2|+|x|+1lt 3$



                      So if $deltalt 1$ you have $|x^3+x^2+x|lt 3delta$ and you can make this as small as you like.






                      share|cite|improve this answer













                      You just need a "good enough" estimate of $|x^2+x+1|$ for suitably small $|x|$



                      Note that for $|x|lt 1$ you have $|x^2+x+1|le |x^2|+|x|+1lt 3$



                      So if $deltalt 1$ you have $|x^3+x^2+x|lt 3delta$ and you can make this as small as you like.







                      share|cite|improve this answer













                      share|cite|improve this answer



                      share|cite|improve this answer











                      answered 2 days ago









                      Mark Bennet

                      76.3k773170




                      76.3k773170




















                          up vote
                          4
                          down vote













                          Take $varepsilon>0$ and pick $delta=minleft1,fracvarepsilon3right$. Then, $|x|<deltaimplies|x|<1implies|x|^2,|x|^3<|x|$. On the other hand, $$|x|<fracvarepsilon3implies|x^3+x^2+x|leqslant|x|^3+|x|^2+|x|<fracvarepsilon3+fracvarepsilon3+fracvarepsilon3=varepsilon.$$






                          share|cite|improve this answer

























                            up vote
                            4
                            down vote













                            Take $varepsilon>0$ and pick $delta=minleft1,fracvarepsilon3right$. Then, $|x|<deltaimplies|x|<1implies|x|^2,|x|^3<|x|$. On the other hand, $$|x|<fracvarepsilon3implies|x^3+x^2+x|leqslant|x|^3+|x|^2+|x|<fracvarepsilon3+fracvarepsilon3+fracvarepsilon3=varepsilon.$$






                            share|cite|improve this answer























                              up vote
                              4
                              down vote










                              up vote
                              4
                              down vote









                              Take $varepsilon>0$ and pick $delta=minleft1,fracvarepsilon3right$. Then, $|x|<deltaimplies|x|<1implies|x|^2,|x|^3<|x|$. On the other hand, $$|x|<fracvarepsilon3implies|x^3+x^2+x|leqslant|x|^3+|x|^2+|x|<fracvarepsilon3+fracvarepsilon3+fracvarepsilon3=varepsilon.$$






                              share|cite|improve this answer













                              Take $varepsilon>0$ and pick $delta=minleft1,fracvarepsilon3right$. Then, $|x|<deltaimplies|x|<1implies|x|^2,|x|^3<|x|$. On the other hand, $$|x|<fracvarepsilon3implies|x^3+x^2+x|leqslant|x|^3+|x|^2+|x|<fracvarepsilon3+fracvarepsilon3+fracvarepsilon3=varepsilon.$$







                              share|cite|improve this answer













                              share|cite|improve this answer



                              share|cite|improve this answer











                              answered 2 days ago









                              José Carlos Santos

                              112k1696172




                              112k1696172




















                                  up vote
                                  2
                                  down vote













                                  Hint: Use the triangle inequality $|a+b|leq |a|+|b|$. You can conclude $|x^2+x+1|leq |x|^2+|x|+1$ and you don't have to care about the sign of $x$.






                                  share|cite|improve this answer

























                                    up vote
                                    2
                                    down vote













                                    Hint: Use the triangle inequality $|a+b|leq |a|+|b|$. You can conclude $|x^2+x+1|leq |x|^2+|x|+1$ and you don't have to care about the sign of $x$.






                                    share|cite|improve this answer























                                      up vote
                                      2
                                      down vote










                                      up vote
                                      2
                                      down vote









                                      Hint: Use the triangle inequality $|a+b|leq |a|+|b|$. You can conclude $|x^2+x+1|leq |x|^2+|x|+1$ and you don't have to care about the sign of $x$.






                                      share|cite|improve this answer













                                      Hint: Use the triangle inequality $|a+b|leq |a|+|b|$. You can conclude $|x^2+x+1|leq |x|^2+|x|+1$ and you don't have to care about the sign of $x$.







                                      share|cite|improve this answer













                                      share|cite|improve this answer



                                      share|cite|improve this answer











                                      answered 2 days ago









                                      Mundron Schmidt

                                      7,1182727




                                      7,1182727




















                                          up vote
                                          1
                                          down vote













                                          As an alternative by squeeze theorem assuming $|x|<1$



                                          $$0le |x^3+x^2+x|=|x||x^2+x+1|le 3|x| to 0$$






                                          share|cite|improve this answer

























                                            up vote
                                            1
                                            down vote













                                            As an alternative by squeeze theorem assuming $|x|<1$



                                            $$0le |x^3+x^2+x|=|x||x^2+x+1|le 3|x| to 0$$






                                            share|cite|improve this answer























                                              up vote
                                              1
                                              down vote










                                              up vote
                                              1
                                              down vote









                                              As an alternative by squeeze theorem assuming $|x|<1$



                                              $$0le |x^3+x^2+x|=|x||x^2+x+1|le 3|x| to 0$$






                                              share|cite|improve this answer













                                              As an alternative by squeeze theorem assuming $|x|<1$



                                              $$0le |x^3+x^2+x|=|x||x^2+x+1|le 3|x| to 0$$







                                              share|cite|improve this answer













                                              share|cite|improve this answer



                                              share|cite|improve this answer











                                              answered 2 days ago









                                              gimusi

                                              63.6k73480




                                              63.6k73480






















                                                   

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