How to evaluate $lim_{n to infty} frac{(1+sqrt 2)^n+(1-sqrt 2)^n}{(1+sqrt 2)^n-(1-sqrt 2)^n}$?











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Evaluate $$lim_{n to infty} frac{(1+sqrt 2)^n+(1-sqrt 2)^n}{(1+sqrt 2)^n-(1-sqrt 2)^n}.$$




I tried to expand using Newton's Binomial Theorem, but it didn't work.










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  • 8




    Divide throughout by $(1+sqrt{2})^n$ and observe that $left|dfrac{1-sqrt{2}}{1+sqrt{2}}right| < 1$.
    – Muralidharan
    Nov 15 at 16:22










  • @Muralidharan you might post your comment as an answer :)
    – Nosrati
    Nov 15 at 16:27










  • observe that $|1-sqrt{2}|<1$
    – Vasya
    Nov 15 at 16:51















up vote
2
down vote

favorite













Evaluate $$lim_{n to infty} frac{(1+sqrt 2)^n+(1-sqrt 2)^n}{(1+sqrt 2)^n-(1-sqrt 2)^n}.$$




I tried to expand using Newton's Binomial Theorem, but it didn't work.










share|cite|improve this question




















  • 8




    Divide throughout by $(1+sqrt{2})^n$ and observe that $left|dfrac{1-sqrt{2}}{1+sqrt{2}}right| < 1$.
    – Muralidharan
    Nov 15 at 16:22










  • @Muralidharan you might post your comment as an answer :)
    – Nosrati
    Nov 15 at 16:27










  • observe that $|1-sqrt{2}|<1$
    – Vasya
    Nov 15 at 16:51













up vote
2
down vote

favorite









up vote
2
down vote

favorite












Evaluate $$lim_{n to infty} frac{(1+sqrt 2)^n+(1-sqrt 2)^n}{(1+sqrt 2)^n-(1-sqrt 2)^n}.$$




I tried to expand using Newton's Binomial Theorem, but it didn't work.










share|cite|improve this question
















Evaluate $$lim_{n to infty} frac{(1+sqrt 2)^n+(1-sqrt 2)^n}{(1+sqrt 2)^n-(1-sqrt 2)^n}.$$




I tried to expand using Newton's Binomial Theorem, but it didn't work.







calculus sequences-and-series limits






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edited Nov 15 at 16:41









amWhy

191k27223438




191k27223438










asked Nov 15 at 16:17









user69503

566




566








  • 8




    Divide throughout by $(1+sqrt{2})^n$ and observe that $left|dfrac{1-sqrt{2}}{1+sqrt{2}}right| < 1$.
    – Muralidharan
    Nov 15 at 16:22










  • @Muralidharan you might post your comment as an answer :)
    – Nosrati
    Nov 15 at 16:27










  • observe that $|1-sqrt{2}|<1$
    – Vasya
    Nov 15 at 16:51














  • 8




    Divide throughout by $(1+sqrt{2})^n$ and observe that $left|dfrac{1-sqrt{2}}{1+sqrt{2}}right| < 1$.
    – Muralidharan
    Nov 15 at 16:22










  • @Muralidharan you might post your comment as an answer :)
    – Nosrati
    Nov 15 at 16:27










  • observe that $|1-sqrt{2}|<1$
    – Vasya
    Nov 15 at 16:51








8




8




Divide throughout by $(1+sqrt{2})^n$ and observe that $left|dfrac{1-sqrt{2}}{1+sqrt{2}}right| < 1$.
– Muralidharan
Nov 15 at 16:22




Divide throughout by $(1+sqrt{2})^n$ and observe that $left|dfrac{1-sqrt{2}}{1+sqrt{2}}right| < 1$.
– Muralidharan
Nov 15 at 16:22












@Muralidharan you might post your comment as an answer :)
– Nosrati
Nov 15 at 16:27




@Muralidharan you might post your comment as an answer :)
– Nosrati
Nov 15 at 16:27












observe that $|1-sqrt{2}|<1$
– Vasya
Nov 15 at 16:51




observe that $|1-sqrt{2}|<1$
– Vasya
Nov 15 at 16:51










3 Answers
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$lim_{n to infty} frac{(1+sqrt 2)^n+(1-sqrt 2)^n}{(1+sqrt 2)^n-(1-sqrt 2)^n}$
$= lim_{nto infty }frac{1+a^n}{1-a^n} $ such that
$a=frac{(1-2^{frac{1}{2}})}{(1+2^{frac{1}{2}})} $ , $|a|<1$. So
lim is equal to $frac{1-0}{1+0}=1$






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  • 3




    Dadrahm.Maybe you could add |a| <1.
    – Peter Szilas
    Nov 15 at 16:49










  • Thank you . . .
    – Dadrahm
    Nov 15 at 17:43










  • Dahdram.A pleasure +.
    – Peter Szilas
    Nov 15 at 17:55


















up vote
1
down vote













Roughly, $|1-sqrt 2| lt 1$, so a high power of it will go to $0$. $1+sqrt 2 gt 1$, so a high power of it will be large and positive. We can ignore the two small terms and be left with the fixed ratio $1$. Depending on what theorems you have proved about limits that may be enough.






share|cite|improve this answer




























    up vote
    1
    down vote













    We have that




    • $|1-sqrt 2|<1 implies (1-sqrt 2)^n to 0$


    therefore



    $$frac{(1+sqrt 2)^n+(1-sqrt 2)^n}{(1+sqrt 2)^n-(1-sqrt 2)^n}sim frac{(1+sqrt 2)^n}{(1+sqrt 2)^n}=1$$



    or more rigoursly



    $$frac{(1+sqrt 2)^n+(1-sqrt 2)^n}{(1+sqrt 2)^n-(1-sqrt 2)^n}= frac{(1+sqrt 2)^n}{(1+sqrt 2)^n}frac{1+frac{(1-sqrt 2)^n}{(1+sqrt 2)^n}}{1-frac{(1-sqrt 2)^n}{(1+sqrt 2)^n}}to frac{1+0}{1-0}$$






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      3 Answers
      3






      active

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      3 Answers
      3






      active

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      active

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      up vote
      5
      down vote













      $lim_{n to infty} frac{(1+sqrt 2)^n+(1-sqrt 2)^n}{(1+sqrt 2)^n-(1-sqrt 2)^n}$
      $= lim_{nto infty }frac{1+a^n}{1-a^n} $ such that
      $a=frac{(1-2^{frac{1}{2}})}{(1+2^{frac{1}{2}})} $ , $|a|<1$. So
      lim is equal to $frac{1-0}{1+0}=1$






      share|cite|improve this answer



















      • 3




        Dadrahm.Maybe you could add |a| <1.
        – Peter Szilas
        Nov 15 at 16:49










      • Thank you . . .
        – Dadrahm
        Nov 15 at 17:43










      • Dahdram.A pleasure +.
        – Peter Szilas
        Nov 15 at 17:55















      up vote
      5
      down vote













      $lim_{n to infty} frac{(1+sqrt 2)^n+(1-sqrt 2)^n}{(1+sqrt 2)^n-(1-sqrt 2)^n}$
      $= lim_{nto infty }frac{1+a^n}{1-a^n} $ such that
      $a=frac{(1-2^{frac{1}{2}})}{(1+2^{frac{1}{2}})} $ , $|a|<1$. So
      lim is equal to $frac{1-0}{1+0}=1$






      share|cite|improve this answer



















      • 3




        Dadrahm.Maybe you could add |a| <1.
        – Peter Szilas
        Nov 15 at 16:49










      • Thank you . . .
        – Dadrahm
        Nov 15 at 17:43










      • Dahdram.A pleasure +.
        – Peter Szilas
        Nov 15 at 17:55













      up vote
      5
      down vote










      up vote
      5
      down vote









      $lim_{n to infty} frac{(1+sqrt 2)^n+(1-sqrt 2)^n}{(1+sqrt 2)^n-(1-sqrt 2)^n}$
      $= lim_{nto infty }frac{1+a^n}{1-a^n} $ such that
      $a=frac{(1-2^{frac{1}{2}})}{(1+2^{frac{1}{2}})} $ , $|a|<1$. So
      lim is equal to $frac{1-0}{1+0}=1$






      share|cite|improve this answer














      $lim_{n to infty} frac{(1+sqrt 2)^n+(1-sqrt 2)^n}{(1+sqrt 2)^n-(1-sqrt 2)^n}$
      $= lim_{nto infty }frac{1+a^n}{1-a^n} $ such that
      $a=frac{(1-2^{frac{1}{2}})}{(1+2^{frac{1}{2}})} $ , $|a|<1$. So
      lim is equal to $frac{1-0}{1+0}=1$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Nov 15 at 17:42

























      answered Nov 15 at 16:42









      Dadrahm

      3448




      3448








      • 3




        Dadrahm.Maybe you could add |a| <1.
        – Peter Szilas
        Nov 15 at 16:49










      • Thank you . . .
        – Dadrahm
        Nov 15 at 17:43










      • Dahdram.A pleasure +.
        – Peter Szilas
        Nov 15 at 17:55














      • 3




        Dadrahm.Maybe you could add |a| <1.
        – Peter Szilas
        Nov 15 at 16:49










      • Thank you . . .
        – Dadrahm
        Nov 15 at 17:43










      • Dahdram.A pleasure +.
        – Peter Szilas
        Nov 15 at 17:55








      3




      3




      Dadrahm.Maybe you could add |a| <1.
      – Peter Szilas
      Nov 15 at 16:49




      Dadrahm.Maybe you could add |a| <1.
      – Peter Szilas
      Nov 15 at 16:49












      Thank you . . .
      – Dadrahm
      Nov 15 at 17:43




      Thank you . . .
      – Dadrahm
      Nov 15 at 17:43












      Dahdram.A pleasure +.
      – Peter Szilas
      Nov 15 at 17:55




      Dahdram.A pleasure +.
      – Peter Szilas
      Nov 15 at 17:55










      up vote
      1
      down vote













      Roughly, $|1-sqrt 2| lt 1$, so a high power of it will go to $0$. $1+sqrt 2 gt 1$, so a high power of it will be large and positive. We can ignore the two small terms and be left with the fixed ratio $1$. Depending on what theorems you have proved about limits that may be enough.






      share|cite|improve this answer

























        up vote
        1
        down vote













        Roughly, $|1-sqrt 2| lt 1$, so a high power of it will go to $0$. $1+sqrt 2 gt 1$, so a high power of it will be large and positive. We can ignore the two small terms and be left with the fixed ratio $1$. Depending on what theorems you have proved about limits that may be enough.






        share|cite|improve this answer























          up vote
          1
          down vote










          up vote
          1
          down vote









          Roughly, $|1-sqrt 2| lt 1$, so a high power of it will go to $0$. $1+sqrt 2 gt 1$, so a high power of it will be large and positive. We can ignore the two small terms and be left with the fixed ratio $1$. Depending on what theorems you have proved about limits that may be enough.






          share|cite|improve this answer












          Roughly, $|1-sqrt 2| lt 1$, so a high power of it will go to $0$. $1+sqrt 2 gt 1$, so a high power of it will be large and positive. We can ignore the two small terms and be left with the fixed ratio $1$. Depending on what theorems you have proved about limits that may be enough.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 15 at 16:54









          Ross Millikan

          287k23195364




          287k23195364






















              up vote
              1
              down vote













              We have that




              • $|1-sqrt 2|<1 implies (1-sqrt 2)^n to 0$


              therefore



              $$frac{(1+sqrt 2)^n+(1-sqrt 2)^n}{(1+sqrt 2)^n-(1-sqrt 2)^n}sim frac{(1+sqrt 2)^n}{(1+sqrt 2)^n}=1$$



              or more rigoursly



              $$frac{(1+sqrt 2)^n+(1-sqrt 2)^n}{(1+sqrt 2)^n-(1-sqrt 2)^n}= frac{(1+sqrt 2)^n}{(1+sqrt 2)^n}frac{1+frac{(1-sqrt 2)^n}{(1+sqrt 2)^n}}{1-frac{(1-sqrt 2)^n}{(1+sqrt 2)^n}}to frac{1+0}{1-0}$$






              share|cite|improve this answer

























                up vote
                1
                down vote













                We have that




                • $|1-sqrt 2|<1 implies (1-sqrt 2)^n to 0$


                therefore



                $$frac{(1+sqrt 2)^n+(1-sqrt 2)^n}{(1+sqrt 2)^n-(1-sqrt 2)^n}sim frac{(1+sqrt 2)^n}{(1+sqrt 2)^n}=1$$



                or more rigoursly



                $$frac{(1+sqrt 2)^n+(1-sqrt 2)^n}{(1+sqrt 2)^n-(1-sqrt 2)^n}= frac{(1+sqrt 2)^n}{(1+sqrt 2)^n}frac{1+frac{(1-sqrt 2)^n}{(1+sqrt 2)^n}}{1-frac{(1-sqrt 2)^n}{(1+sqrt 2)^n}}to frac{1+0}{1-0}$$






                share|cite|improve this answer























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  We have that




                  • $|1-sqrt 2|<1 implies (1-sqrt 2)^n to 0$


                  therefore



                  $$frac{(1+sqrt 2)^n+(1-sqrt 2)^n}{(1+sqrt 2)^n-(1-sqrt 2)^n}sim frac{(1+sqrt 2)^n}{(1+sqrt 2)^n}=1$$



                  or more rigoursly



                  $$frac{(1+sqrt 2)^n+(1-sqrt 2)^n}{(1+sqrt 2)^n-(1-sqrt 2)^n}= frac{(1+sqrt 2)^n}{(1+sqrt 2)^n}frac{1+frac{(1-sqrt 2)^n}{(1+sqrt 2)^n}}{1-frac{(1-sqrt 2)^n}{(1+sqrt 2)^n}}to frac{1+0}{1-0}$$






                  share|cite|improve this answer












                  We have that




                  • $|1-sqrt 2|<1 implies (1-sqrt 2)^n to 0$


                  therefore



                  $$frac{(1+sqrt 2)^n+(1-sqrt 2)^n}{(1+sqrt 2)^n-(1-sqrt 2)^n}sim frac{(1+sqrt 2)^n}{(1+sqrt 2)^n}=1$$



                  or more rigoursly



                  $$frac{(1+sqrt 2)^n+(1-sqrt 2)^n}{(1+sqrt 2)^n-(1-sqrt 2)^n}= frac{(1+sqrt 2)^n}{(1+sqrt 2)^n}frac{1+frac{(1-sqrt 2)^n}{(1+sqrt 2)^n}}{1-frac{(1-sqrt 2)^n}{(1+sqrt 2)^n}}to frac{1+0}{1-0}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 15 at 17:17









                  gimusi

                  87.7k74393




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