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Complicated limit calculation



The 2019 Stack Overflow Developer Survey Results Are InProblematic limit calculationLimit calculationDetermine the limit for $ageq-1$Find the limit $L=lim_{nto infty} (-1)^{n}sinleft(pisqrt{n^2+n}right)$Limit involving complicated integralLimit (Calculus)Impossible Limit CalculationHow can we solve this limit: $lim _{xto infty :}left(frac{tanleft(xright)}{x}right)$?Computation of a complicated limitHow to calculate this complicated limit?












1












$begingroup$


Find the limit of $$left[(x^m + 1)^{1/n} - (x^m - 1)^{1/n}right]x^{(mn-m)/n}$$ when $xtoinfty$ and $m,n$ are natural numbers.



Thanks in advance!






calculus limits







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Find the limit of $$left[(x^m + 1)^{1/n} - (x^m - 1)^{1/n}right]x^{(mn-m)/n}$$ when $xtoinfty$ and $m,n$ are natural numbers.



    Thanks in advance!






    calculus limits







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Find the limit of $$left[(x^m + 1)^{1/n} - (x^m - 1)^{1/n}right]x^{(mn-m)/n}$$ when $xtoinfty$ and $m,n$ are natural numbers.



      Thanks in advance!






      calculus limits







      share|cite|improve this question











      $endgroup$




      Find the limit of $$left[(x^m + 1)^{1/n} - (x^m - 1)^{1/n}right]x^{(mn-m)/n}$$ when $xtoinfty$ and $m,n$ are natural numbers.



      Thanks in advance!






      calculus limits




      calculus limits






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 21 '12 at 16:31









      Nameless

      10.5k12255




      10.5k12255










      asked Dec 21 '12 at 16:08









      SmithnsonSmithnson

      255




      255






















          3 Answers
          3






          active

          oldest

          votes


















          3












          $begingroup$

          An even quicker way to the answer comes from observing that



          $$left ( x^m + 1 right )^{frac{1}{n}} - left (x^m - 1 right)^{frac{1}{n}} = x^{frac{m}{n}} left [ left ( 1 + frac{1}{x^m} right )^{frac{1}{n}} - left (1 - frac{1}{x^m} right)^{frac{1}{n}} right ]$$



          $$ approx x^{frac{m}{n}} left [ left ( 1 + frac{1}{n x^m} right ) - left (1 - frac{1}{n x^m} right) right ]$$



          $$ = frac{2 x^{frac{m}{n}}}{n x^m} $$



          The value $2/n$ follows from the factor on the right of the original expression.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Hi. Where did the second line come from?
            $endgroup$
            – Smithnson
            Dec 21 '12 at 16:46






          • 1




            $begingroup$
            Taylor expansion of $(1+epsilon)^{alpha} approx (1+ alpha epsilon)$ for | $epsilon | ll 1$.
            $endgroup$
            – Ron Gordon
            Dec 21 '12 at 16:55





















          1












          $begingroup$

          First, observe that
          $$lim_{xto +infty}left[(x^m + 1)^{1/n} - (x^m - 1)^{1/n}right]x^{(mn-m)/n}
          =lim_{xto +infty}x^{frac mn}left[(1 + frac1{x^m})^{1/n} - (1- frac1{x^m})^{1/n}right]x^{(mn-m)/n}=
          lim_{xto +infty}left[(1+ frac1{x^m})^{1/n} - (1- frac1{x^m})^{1/n}right]x^{m}
          $$
          Now, since
          $$alpha^{1/n}-beta^{1/n}=frac{alpha-beta}{alpha^{{(n-1)}/n}+
          alpha^{{(n-2)}/n}beta+...+alphabeta^{{(n-2)}/n}+beta^{{(n-1)}/n}}$$
          the limit becomes
          $$lim_{xto +infty}left[frac{1+ frac1{x^m}- 1+ frac1{x^m}}{(1+ frac1{x^m})^{{(n-1)}/n}+...+(1+ frac1{x^m})^{{(n-1)}/n}}right]x^m=
          lim_{xto +infty}left[frac{2}{(1+ frac1{x^m})^{{(n-1)}/n}+...+(1+ frac1{x^m})^{{(n-1)}/n}}right]$$
          The denominator tends to $1+1+...+1=n$ therefore
          $$lim_{xto +infty}left[frac2{(1+ frac1{x^m})^{{(n-1)}/n}+...+(1+ frac1{x^m})^{{(n-1)}/n}}right]=frac2n$$






          share|cite|improve this answer











          $endgroup$





















            0












            $begingroup$

            Substitute $y=x^m$. Then after rework the limit is that of



            $$lim_{ytoinfty}y,((1+y^{-1})^{1/n}-(1-y^{-1})^{1/n}).$$



            Then




            • by Taylor, $$lim_{ytoinfty}y,left(1+frac{y^{-1}}n-1+frac{y^{-1}}n+o(y^{-1})right)=frac2n,$$


            • or multiplying by the conjugate multinomial



            $$lim_{ytoinfty}yfrac{1+y^{-1}-1+y^{-1}}{displaystylesum_{k=0}^{n-1}(1+y^{-1})^{k/n}}=frac2n,$$




            • or by L'Hospital with the variable $y^{-1}$,


            $$lim_{ytoinfty}frac{(1+y^{-1})^{1/n}-(1-y^{-1})^{1/n}}{y^{-1}}=lim_{ytoinfty}frac{(1+y^{-1})^{1/n-1}+(1-y^{-1})^{1/n-1}}n=frac2n.$$






            share|cite|improve this answer









            $endgroup$














              Your Answer





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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              3












              $begingroup$

              An even quicker way to the answer comes from observing that



              $$left ( x^m + 1 right )^{frac{1}{n}} - left (x^m - 1 right)^{frac{1}{n}} = x^{frac{m}{n}} left [ left ( 1 + frac{1}{x^m} right )^{frac{1}{n}} - left (1 - frac{1}{x^m} right)^{frac{1}{n}} right ]$$



              $$ approx x^{frac{m}{n}} left [ left ( 1 + frac{1}{n x^m} right ) - left (1 - frac{1}{n x^m} right) right ]$$



              $$ = frac{2 x^{frac{m}{n}}}{n x^m} $$



              The value $2/n$ follows from the factor on the right of the original expression.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Hi. Where did the second line come from?
                $endgroup$
                – Smithnson
                Dec 21 '12 at 16:46






              • 1




                $begingroup$
                Taylor expansion of $(1+epsilon)^{alpha} approx (1+ alpha epsilon)$ for | $epsilon | ll 1$.
                $endgroup$
                – Ron Gordon
                Dec 21 '12 at 16:55


















              3












              $begingroup$

              An even quicker way to the answer comes from observing that



              $$left ( x^m + 1 right )^{frac{1}{n}} - left (x^m - 1 right)^{frac{1}{n}} = x^{frac{m}{n}} left [ left ( 1 + frac{1}{x^m} right )^{frac{1}{n}} - left (1 - frac{1}{x^m} right)^{frac{1}{n}} right ]$$



              $$ approx x^{frac{m}{n}} left [ left ( 1 + frac{1}{n x^m} right ) - left (1 - frac{1}{n x^m} right) right ]$$



              $$ = frac{2 x^{frac{m}{n}}}{n x^m} $$



              The value $2/n$ follows from the factor on the right of the original expression.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Hi. Where did the second line come from?
                $endgroup$
                – Smithnson
                Dec 21 '12 at 16:46






              • 1




                $begingroup$
                Taylor expansion of $(1+epsilon)^{alpha} approx (1+ alpha epsilon)$ for | $epsilon | ll 1$.
                $endgroup$
                – Ron Gordon
                Dec 21 '12 at 16:55
















              3












              3








              3





              $begingroup$

              An even quicker way to the answer comes from observing that



              $$left ( x^m + 1 right )^{frac{1}{n}} - left (x^m - 1 right)^{frac{1}{n}} = x^{frac{m}{n}} left [ left ( 1 + frac{1}{x^m} right )^{frac{1}{n}} - left (1 - frac{1}{x^m} right)^{frac{1}{n}} right ]$$



              $$ approx x^{frac{m}{n}} left [ left ( 1 + frac{1}{n x^m} right ) - left (1 - frac{1}{n x^m} right) right ]$$



              $$ = frac{2 x^{frac{m}{n}}}{n x^m} $$



              The value $2/n$ follows from the factor on the right of the original expression.






              share|cite|improve this answer









              $endgroup$



              An even quicker way to the answer comes from observing that



              $$left ( x^m + 1 right )^{frac{1}{n}} - left (x^m - 1 right)^{frac{1}{n}} = x^{frac{m}{n}} left [ left ( 1 + frac{1}{x^m} right )^{frac{1}{n}} - left (1 - frac{1}{x^m} right)^{frac{1}{n}} right ]$$



              $$ approx x^{frac{m}{n}} left [ left ( 1 + frac{1}{n x^m} right ) - left (1 - frac{1}{n x^m} right) right ]$$



              $$ = frac{2 x^{frac{m}{n}}}{n x^m} $$



              The value $2/n$ follows from the factor on the right of the original expression.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 21 '12 at 16:33









              Ron GordonRon Gordon

              123k14156267




              123k14156267












              • $begingroup$
                Hi. Where did the second line come from?
                $endgroup$
                – Smithnson
                Dec 21 '12 at 16:46






              • 1




                $begingroup$
                Taylor expansion of $(1+epsilon)^{alpha} approx (1+ alpha epsilon)$ for | $epsilon | ll 1$.
                $endgroup$
                – Ron Gordon
                Dec 21 '12 at 16:55




















              • $begingroup$
                Hi. Where did the second line come from?
                $endgroup$
                – Smithnson
                Dec 21 '12 at 16:46






              • 1




                $begingroup$
                Taylor expansion of $(1+epsilon)^{alpha} approx (1+ alpha epsilon)$ for | $epsilon | ll 1$.
                $endgroup$
                – Ron Gordon
                Dec 21 '12 at 16:55


















              $begingroup$
              Hi. Where did the second line come from?
              $endgroup$
              – Smithnson
              Dec 21 '12 at 16:46




              $begingroup$
              Hi. Where did the second line come from?
              $endgroup$
              – Smithnson
              Dec 21 '12 at 16:46




              1




              1




              $begingroup$
              Taylor expansion of $(1+epsilon)^{alpha} approx (1+ alpha epsilon)$ for | $epsilon | ll 1$.
              $endgroup$
              – Ron Gordon
              Dec 21 '12 at 16:55






              $begingroup$
              Taylor expansion of $(1+epsilon)^{alpha} approx (1+ alpha epsilon)$ for | $epsilon | ll 1$.
              $endgroup$
              – Ron Gordon
              Dec 21 '12 at 16:55













              1












              $begingroup$

              First, observe that
              $$lim_{xto +infty}left[(x^m + 1)^{1/n} - (x^m - 1)^{1/n}right]x^{(mn-m)/n}
              =lim_{xto +infty}x^{frac mn}left[(1 + frac1{x^m})^{1/n} - (1- frac1{x^m})^{1/n}right]x^{(mn-m)/n}=
              lim_{xto +infty}left[(1+ frac1{x^m})^{1/n} - (1- frac1{x^m})^{1/n}right]x^{m}
              $$
              Now, since
              $$alpha^{1/n}-beta^{1/n}=frac{alpha-beta}{alpha^{{(n-1)}/n}+
              alpha^{{(n-2)}/n}beta+...+alphabeta^{{(n-2)}/n}+beta^{{(n-1)}/n}}$$
              the limit becomes
              $$lim_{xto +infty}left[frac{1+ frac1{x^m}- 1+ frac1{x^m}}{(1+ frac1{x^m})^{{(n-1)}/n}+...+(1+ frac1{x^m})^{{(n-1)}/n}}right]x^m=
              lim_{xto +infty}left[frac{2}{(1+ frac1{x^m})^{{(n-1)}/n}+...+(1+ frac1{x^m})^{{(n-1)}/n}}right]$$
              The denominator tends to $1+1+...+1=n$ therefore
              $$lim_{xto +infty}left[frac2{(1+ frac1{x^m})^{{(n-1)}/n}+...+(1+ frac1{x^m})^{{(n-1)}/n}}right]=frac2n$$






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                First, observe that
                $$lim_{xto +infty}left[(x^m + 1)^{1/n} - (x^m - 1)^{1/n}right]x^{(mn-m)/n}
                =lim_{xto +infty}x^{frac mn}left[(1 + frac1{x^m})^{1/n} - (1- frac1{x^m})^{1/n}right]x^{(mn-m)/n}=
                lim_{xto +infty}left[(1+ frac1{x^m})^{1/n} - (1- frac1{x^m})^{1/n}right]x^{m}
                $$
                Now, since
                $$alpha^{1/n}-beta^{1/n}=frac{alpha-beta}{alpha^{{(n-1)}/n}+
                alpha^{{(n-2)}/n}beta+...+alphabeta^{{(n-2)}/n}+beta^{{(n-1)}/n}}$$
                the limit becomes
                $$lim_{xto +infty}left[frac{1+ frac1{x^m}- 1+ frac1{x^m}}{(1+ frac1{x^m})^{{(n-1)}/n}+...+(1+ frac1{x^m})^{{(n-1)}/n}}right]x^m=
                lim_{xto +infty}left[frac{2}{(1+ frac1{x^m})^{{(n-1)}/n}+...+(1+ frac1{x^m})^{{(n-1)}/n}}right]$$
                The denominator tends to $1+1+...+1=n$ therefore
                $$lim_{xto +infty}left[frac2{(1+ frac1{x^m})^{{(n-1)}/n}+...+(1+ frac1{x^m})^{{(n-1)}/n}}right]=frac2n$$






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  First, observe that
                  $$lim_{xto +infty}left[(x^m + 1)^{1/n} - (x^m - 1)^{1/n}right]x^{(mn-m)/n}
                  =lim_{xto +infty}x^{frac mn}left[(1 + frac1{x^m})^{1/n} - (1- frac1{x^m})^{1/n}right]x^{(mn-m)/n}=
                  lim_{xto +infty}left[(1+ frac1{x^m})^{1/n} - (1- frac1{x^m})^{1/n}right]x^{m}
                  $$
                  Now, since
                  $$alpha^{1/n}-beta^{1/n}=frac{alpha-beta}{alpha^{{(n-1)}/n}+
                  alpha^{{(n-2)}/n}beta+...+alphabeta^{{(n-2)}/n}+beta^{{(n-1)}/n}}$$
                  the limit becomes
                  $$lim_{xto +infty}left[frac{1+ frac1{x^m}- 1+ frac1{x^m}}{(1+ frac1{x^m})^{{(n-1)}/n}+...+(1+ frac1{x^m})^{{(n-1)}/n}}right]x^m=
                  lim_{xto +infty}left[frac{2}{(1+ frac1{x^m})^{{(n-1)}/n}+...+(1+ frac1{x^m})^{{(n-1)}/n}}right]$$
                  The denominator tends to $1+1+...+1=n$ therefore
                  $$lim_{xto +infty}left[frac2{(1+ frac1{x^m})^{{(n-1)}/n}+...+(1+ frac1{x^m})^{{(n-1)}/n}}right]=frac2n$$






                  share|cite|improve this answer











                  $endgroup$



                  First, observe that
                  $$lim_{xto +infty}left[(x^m + 1)^{1/n} - (x^m - 1)^{1/n}right]x^{(mn-m)/n}
                  =lim_{xto +infty}x^{frac mn}left[(1 + frac1{x^m})^{1/n} - (1- frac1{x^m})^{1/n}right]x^{(mn-m)/n}=
                  lim_{xto +infty}left[(1+ frac1{x^m})^{1/n} - (1- frac1{x^m})^{1/n}right]x^{m}
                  $$
                  Now, since
                  $$alpha^{1/n}-beta^{1/n}=frac{alpha-beta}{alpha^{{(n-1)}/n}+
                  alpha^{{(n-2)}/n}beta+...+alphabeta^{{(n-2)}/n}+beta^{{(n-1)}/n}}$$
                  the limit becomes
                  $$lim_{xto +infty}left[frac{1+ frac1{x^m}- 1+ frac1{x^m}}{(1+ frac1{x^m})^{{(n-1)}/n}+...+(1+ frac1{x^m})^{{(n-1)}/n}}right]x^m=
                  lim_{xto +infty}left[frac{2}{(1+ frac1{x^m})^{{(n-1)}/n}+...+(1+ frac1{x^m})^{{(n-1)}/n}}right]$$
                  The denominator tends to $1+1+...+1=n$ therefore
                  $$lim_{xto +infty}left[frac2{(1+ frac1{x^m})^{{(n-1)}/n}+...+(1+ frac1{x^m})^{{(n-1)}/n}}right]=frac2n$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Mar 21 at 14:15









                  Community♦

                  1




                  1










                  answered Dec 21 '12 at 16:23









                  NamelessNameless

                  10.5k12255




                  10.5k12255























                      0












                      $begingroup$

                      Substitute $y=x^m$. Then after rework the limit is that of



                      $$lim_{ytoinfty}y,((1+y^{-1})^{1/n}-(1-y^{-1})^{1/n}).$$



                      Then




                      • by Taylor, $$lim_{ytoinfty}y,left(1+frac{y^{-1}}n-1+frac{y^{-1}}n+o(y^{-1})right)=frac2n,$$


                      • or multiplying by the conjugate multinomial



                      $$lim_{ytoinfty}yfrac{1+y^{-1}-1+y^{-1}}{displaystylesum_{k=0}^{n-1}(1+y^{-1})^{k/n}}=frac2n,$$




                      • or by L'Hospital with the variable $y^{-1}$,


                      $$lim_{ytoinfty}frac{(1+y^{-1})^{1/n}-(1-y^{-1})^{1/n}}{y^{-1}}=lim_{ytoinfty}frac{(1+y^{-1})^{1/n-1}+(1-y^{-1})^{1/n-1}}n=frac2n.$$






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        Substitute $y=x^m$. Then after rework the limit is that of



                        $$lim_{ytoinfty}y,((1+y^{-1})^{1/n}-(1-y^{-1})^{1/n}).$$



                        Then




                        • by Taylor, $$lim_{ytoinfty}y,left(1+frac{y^{-1}}n-1+frac{y^{-1}}n+o(y^{-1})right)=frac2n,$$


                        • or multiplying by the conjugate multinomial



                        $$lim_{ytoinfty}yfrac{1+y^{-1}-1+y^{-1}}{displaystylesum_{k=0}^{n-1}(1+y^{-1})^{k/n}}=frac2n,$$




                        • or by L'Hospital with the variable $y^{-1}$,


                        $$lim_{ytoinfty}frac{(1+y^{-1})^{1/n}-(1-y^{-1})^{1/n}}{y^{-1}}=lim_{ytoinfty}frac{(1+y^{-1})^{1/n-1}+(1-y^{-1})^{1/n-1}}n=frac2n.$$






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Substitute $y=x^m$. Then after rework the limit is that of



                          $$lim_{ytoinfty}y,((1+y^{-1})^{1/n}-(1-y^{-1})^{1/n}).$$



                          Then




                          • by Taylor, $$lim_{ytoinfty}y,left(1+frac{y^{-1}}n-1+frac{y^{-1}}n+o(y^{-1})right)=frac2n,$$


                          • or multiplying by the conjugate multinomial



                          $$lim_{ytoinfty}yfrac{1+y^{-1}-1+y^{-1}}{displaystylesum_{k=0}^{n-1}(1+y^{-1})^{k/n}}=frac2n,$$




                          • or by L'Hospital with the variable $y^{-1}$,


                          $$lim_{ytoinfty}frac{(1+y^{-1})^{1/n}-(1-y^{-1})^{1/n}}{y^{-1}}=lim_{ytoinfty}frac{(1+y^{-1})^{1/n-1}+(1-y^{-1})^{1/n-1}}n=frac2n.$$






                          share|cite|improve this answer









                          $endgroup$



                          Substitute $y=x^m$. Then after rework the limit is that of



                          $$lim_{ytoinfty}y,((1+y^{-1})^{1/n}-(1-y^{-1})^{1/n}).$$



                          Then




                          • by Taylor, $$lim_{ytoinfty}y,left(1+frac{y^{-1}}n-1+frac{y^{-1}}n+o(y^{-1})right)=frac2n,$$


                          • or multiplying by the conjugate multinomial



                          $$lim_{ytoinfty}yfrac{1+y^{-1}-1+y^{-1}}{displaystylesum_{k=0}^{n-1}(1+y^{-1})^{k/n}}=frac2n,$$




                          • or by L'Hospital with the variable $y^{-1}$,


                          $$lim_{ytoinfty}frac{(1+y^{-1})^{1/n}-(1-y^{-1})^{1/n}}{y^{-1}}=lim_{ytoinfty}frac{(1+y^{-1})^{1/n-1}+(1-y^{-1})^{1/n-1}}n=frac2n.$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Mar 21 at 14:29









                          Yves DaoustYves Daoust

                          133k676231




                          133k676231






























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