Prove limit of cubic function by definition of limit












0












$begingroup$


I'm a little bit confused by showing the following limit directly from the definition:




$$lim_{xto 4} x^3 = 64$$




$|x^3-64|=|x-4| |x^2+4x+16|$



Since I know that $|x-4|< sigma $, I'm trying to do something like



$|x-4+x^2+3x+20|<sigma +x^2+3x+20$



Than I multiply two inequalities:



$|x-4||x^2+4x+16|<sigma^2 +x^2sigma+3xsigma+20sigma$



Here's the point of my confusion. Since we know that x < 4 will always holds, can we multiply right side of inequality to get the following?



$|x-4||x^2+4x+16|<sigma +16sigma+12sigma+20sigma = 49sigma$



So then we can take $sigma = min{1/2, sigma/49}$



Thank you in advance!






real-analysis limits analysis







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    0












    $begingroup$


    I'm a little bit confused by showing the following limit directly from the definition:




    $$lim_{xto 4} x^3 = 64$$




    $|x^3-64|=|x-4| |x^2+4x+16|$



    Since I know that $|x-4|< sigma $, I'm trying to do something like



    $|x-4+x^2+3x+20|<sigma +x^2+3x+20$



    Than I multiply two inequalities:



    $|x-4||x^2+4x+16|<sigma^2 +x^2sigma+3xsigma+20sigma$



    Here's the point of my confusion. Since we know that x < 4 will always holds, can we multiply right side of inequality to get the following?



    $|x-4||x^2+4x+16|<sigma +16sigma+12sigma+20sigma = 49sigma$



    So then we can take $sigma = min{1/2, sigma/49}$



    Thank you in advance!






    real-analysis limits analysis







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I'm a little bit confused by showing the following limit directly from the definition:




      $$lim_{xto 4} x^3 = 64$$




      $|x^3-64|=|x-4| |x^2+4x+16|$



      Since I know that $|x-4|< sigma $, I'm trying to do something like



      $|x-4+x^2+3x+20|<sigma +x^2+3x+20$



      Than I multiply two inequalities:



      $|x-4||x^2+4x+16|<sigma^2 +x^2sigma+3xsigma+20sigma$



      Here's the point of my confusion. Since we know that x < 4 will always holds, can we multiply right side of inequality to get the following?



      $|x-4||x^2+4x+16|<sigma +16sigma+12sigma+20sigma = 49sigma$



      So then we can take $sigma = min{1/2, sigma/49}$



      Thank you in advance!






      real-analysis limits analysis







      share|cite|improve this question









      $endgroup$




      I'm a little bit confused by showing the following limit directly from the definition:




      $$lim_{xto 4} x^3 = 64$$




      $|x^3-64|=|x-4| |x^2+4x+16|$



      Since I know that $|x-4|< sigma $, I'm trying to do something like



      $|x-4+x^2+3x+20|<sigma +x^2+3x+20$



      Than I multiply two inequalities:



      $|x-4||x^2+4x+16|<sigma^2 +x^2sigma+3xsigma+20sigma$



      Here's the point of my confusion. Since we know that x < 4 will always holds, can we multiply right side of inequality to get the following?



      $|x-4||x^2+4x+16|<sigma +16sigma+12sigma+20sigma = 49sigma$



      So then we can take $sigma = min{1/2, sigma/49}$



      Thank you in advance!






      real-analysis limits analysis




      real-analysis limits analysis






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 16 '18 at 20:56









      Sergey MalinovSergey Malinov

      174




      174






















          2 Answers
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          $begingroup$

          Almost: We know that $4-sigma<x<4+sigma$. So we have that
          $$|x-4||x^2+4x+16|<sigma |x^2+4x+16|.$$
          Since $x^2+4x+16$ has all positive coefficients, $x^2+4x+16$ is increasing when $x>0$. Since $x<4+sigma$ (and $x>4-sigma>0$, assume $sigma < 4$) we have
          $$sigma|x^2+4x+16|<sigma((4+sigma)^2+4(4+sigma)+16)$$
          and thus
          $$|x-4|<sigmaRightarrow |x^3-64|<sigma((4+sigma)^2+4(4+sigma)+16).$$
          Now you can reverse engineer this to make $sigma((4+sigma)^2+4(4+sigma)+16)<epsilon$.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            In a limit, only the behavior of the function in the neighborhood of the point is important. You can't require that $x < 4$ (that would be taking a side limit!), but you can fix $r$ and work as if the whole dominion was $(4 - r, 4 + r)$, since only the behavior very close to 4 is important (much closer than $r$).



            In your case, one could argue something in the lines of



            $ |x - 4||x^2 + 4x + 16| < |x - 4||25 + 20 + 16| = 61|x - 4| < 61delta$



            for every $x in (4 - delta, 4 + delta)$, with $delta$ small enough (in fact, $delta < 1$, such that $x < 5$).






            share|cite|improve this answer









            $endgroup$













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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              5












              $begingroup$

              Almost: We know that $4-sigma<x<4+sigma$. So we have that
              $$|x-4||x^2+4x+16|<sigma |x^2+4x+16|.$$
              Since $x^2+4x+16$ has all positive coefficients, $x^2+4x+16$ is increasing when $x>0$. Since $x<4+sigma$ (and $x>4-sigma>0$, assume $sigma < 4$) we have
              $$sigma|x^2+4x+16|<sigma((4+sigma)^2+4(4+sigma)+16)$$
              and thus
              $$|x-4|<sigmaRightarrow |x^3-64|<sigma((4+sigma)^2+4(4+sigma)+16).$$
              Now you can reverse engineer this to make $sigma((4+sigma)^2+4(4+sigma)+16)<epsilon$.






              share|cite|improve this answer









              $endgroup$


















                5












                $begingroup$

                Almost: We know that $4-sigma<x<4+sigma$. So we have that
                $$|x-4||x^2+4x+16|<sigma |x^2+4x+16|.$$
                Since $x^2+4x+16$ has all positive coefficients, $x^2+4x+16$ is increasing when $x>0$. Since $x<4+sigma$ (and $x>4-sigma>0$, assume $sigma < 4$) we have
                $$sigma|x^2+4x+16|<sigma((4+sigma)^2+4(4+sigma)+16)$$
                and thus
                $$|x-4|<sigmaRightarrow |x^3-64|<sigma((4+sigma)^2+4(4+sigma)+16).$$
                Now you can reverse engineer this to make $sigma((4+sigma)^2+4(4+sigma)+16)<epsilon$.






                share|cite|improve this answer









                $endgroup$
















                  5












                  5








                  5





                  $begingroup$

                  Almost: We know that $4-sigma<x<4+sigma$. So we have that
                  $$|x-4||x^2+4x+16|<sigma |x^2+4x+16|.$$
                  Since $x^2+4x+16$ has all positive coefficients, $x^2+4x+16$ is increasing when $x>0$. Since $x<4+sigma$ (and $x>4-sigma>0$, assume $sigma < 4$) we have
                  $$sigma|x^2+4x+16|<sigma((4+sigma)^2+4(4+sigma)+16)$$
                  and thus
                  $$|x-4|<sigmaRightarrow |x^3-64|<sigma((4+sigma)^2+4(4+sigma)+16).$$
                  Now you can reverse engineer this to make $sigma((4+sigma)^2+4(4+sigma)+16)<epsilon$.






                  share|cite|improve this answer









                  $endgroup$



                  Almost: We know that $4-sigma<x<4+sigma$. So we have that
                  $$|x-4||x^2+4x+16|<sigma |x^2+4x+16|.$$
                  Since $x^2+4x+16$ has all positive coefficients, $x^2+4x+16$ is increasing when $x>0$. Since $x<4+sigma$ (and $x>4-sigma>0$, assume $sigma < 4$) we have
                  $$sigma|x^2+4x+16|<sigma((4+sigma)^2+4(4+sigma)+16)$$
                  and thus
                  $$|x-4|<sigmaRightarrow |x^3-64|<sigma((4+sigma)^2+4(4+sigma)+16).$$
                  Now you can reverse engineer this to make $sigma((4+sigma)^2+4(4+sigma)+16)<epsilon$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 16 '18 at 21:04









                  user627023user627023

                  512




                  512























                      2












                      $begingroup$

                      In a limit, only the behavior of the function in the neighborhood of the point is important. You can't require that $x < 4$ (that would be taking a side limit!), but you can fix $r$ and work as if the whole dominion was $(4 - r, 4 + r)$, since only the behavior very close to 4 is important (much closer than $r$).



                      In your case, one could argue something in the lines of



                      $ |x - 4||x^2 + 4x + 16| < |x - 4||25 + 20 + 16| = 61|x - 4| < 61delta$



                      for every $x in (4 - delta, 4 + delta)$, with $delta$ small enough (in fact, $delta < 1$, such that $x < 5$).






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        In a limit, only the behavior of the function in the neighborhood of the point is important. You can't require that $x < 4$ (that would be taking a side limit!), but you can fix $r$ and work as if the whole dominion was $(4 - r, 4 + r)$, since only the behavior very close to 4 is important (much closer than $r$).



                        In your case, one could argue something in the lines of



                        $ |x - 4||x^2 + 4x + 16| < |x - 4||25 + 20 + 16| = 61|x - 4| < 61delta$



                        for every $x in (4 - delta, 4 + delta)$, with $delta$ small enough (in fact, $delta < 1$, such that $x < 5$).






                        share|cite|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          In a limit, only the behavior of the function in the neighborhood of the point is important. You can't require that $x < 4$ (that would be taking a side limit!), but you can fix $r$ and work as if the whole dominion was $(4 - r, 4 + r)$, since only the behavior very close to 4 is important (much closer than $r$).



                          In your case, one could argue something in the lines of



                          $ |x - 4||x^2 + 4x + 16| < |x - 4||25 + 20 + 16| = 61|x - 4| < 61delta$



                          for every $x in (4 - delta, 4 + delta)$, with $delta$ small enough (in fact, $delta < 1$, such that $x < 5$).






                          share|cite|improve this answer









                          $endgroup$



                          In a limit, only the behavior of the function in the neighborhood of the point is important. You can't require that $x < 4$ (that would be taking a side limit!), but you can fix $r$ and work as if the whole dominion was $(4 - r, 4 + r)$, since only the behavior very close to 4 is important (much closer than $r$).



                          In your case, one could argue something in the lines of



                          $ |x - 4||x^2 + 4x + 16| < |x - 4||25 + 20 + 16| = 61|x - 4| < 61delta$



                          for every $x in (4 - delta, 4 + delta)$, with $delta$ small enough (in fact, $delta < 1$, such that $x < 5$).







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 16 '18 at 21:14









                          M. SantosM. Santos

                          7615




                          7615






























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