Compute $lim_{n to infty}(1+frac{n-1}{n+1})^{frac{n+1}{n-1}}$












2












$begingroup$


Compute $lim_{n to infty}(1+frac{n-1}{n+1})^{frac{n+1}{n-1}}$



I did: $lim_{nto infty}(1+frac{1}{frac{n+1}{n-1}})^{frac{n+1}{n-1}}=e$.



Why is this incorrect?



Thank you for your help.






limits







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$endgroup$








  • 2




    $begingroup$
    Hint: What is the limit of $(n+1)/(n-1)$ ?
    $endgroup$
    – Martin R
    Jan 23 at 12:40






  • 2




    $begingroup$
    Because $(n-1)/(n+1) notto +infty$, and you cannot use that significant well-known limit about $mathrm e$.
    $endgroup$
    – xbh
    Jan 23 at 12:41










  • $begingroup$
    limit of $(n+1)/(n-1)=1$, thank you!
    $endgroup$
    – J.Doe
    Jan 23 at 12:41


















2












$begingroup$


Compute $lim_{n to infty}(1+frac{n-1}{n+1})^{frac{n+1}{n-1}}$



I did: $lim_{nto infty}(1+frac{1}{frac{n+1}{n-1}})^{frac{n+1}{n-1}}=e$.



Why is this incorrect?



Thank you for your help.






limits







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Hint: What is the limit of $(n+1)/(n-1)$ ?
    $endgroup$
    – Martin R
    Jan 23 at 12:40






  • 2




    $begingroup$
    Because $(n-1)/(n+1) notto +infty$, and you cannot use that significant well-known limit about $mathrm e$.
    $endgroup$
    – xbh
    Jan 23 at 12:41










  • $begingroup$
    limit of $(n+1)/(n-1)=1$, thank you!
    $endgroup$
    – J.Doe
    Jan 23 at 12:41
















2












2








2





$begingroup$


Compute $lim_{n to infty}(1+frac{n-1}{n+1})^{frac{n+1}{n-1}}$



I did: $lim_{nto infty}(1+frac{1}{frac{n+1}{n-1}})^{frac{n+1}{n-1}}=e$.



Why is this incorrect?



Thank you for your help.






limits







share|cite|improve this question











$endgroup$




Compute $lim_{n to infty}(1+frac{n-1}{n+1})^{frac{n+1}{n-1}}$



I did: $lim_{nto infty}(1+frac{1}{frac{n+1}{n-1}})^{frac{n+1}{n-1}}=e$.



Why is this incorrect?



Thank you for your help.






limits




limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 23 at 12:52









amWhy

1




1










asked Jan 23 at 12:37









J.DoeJ.Doe

899




899








  • 2




    $begingroup$
    Hint: What is the limit of $(n+1)/(n-1)$ ?
    $endgroup$
    – Martin R
    Jan 23 at 12:40






  • 2




    $begingroup$
    Because $(n-1)/(n+1) notto +infty$, and you cannot use that significant well-known limit about $mathrm e$.
    $endgroup$
    – xbh
    Jan 23 at 12:41










  • $begingroup$
    limit of $(n+1)/(n-1)=1$, thank you!
    $endgroup$
    – J.Doe
    Jan 23 at 12:41
















  • 2




    $begingroup$
    Hint: What is the limit of $(n+1)/(n-1)$ ?
    $endgroup$
    – Martin R
    Jan 23 at 12:40






  • 2




    $begingroup$
    Because $(n-1)/(n+1) notto +infty$, and you cannot use that significant well-known limit about $mathrm e$.
    $endgroup$
    – xbh
    Jan 23 at 12:41










  • $begingroup$
    limit of $(n+1)/(n-1)=1$, thank you!
    $endgroup$
    – J.Doe
    Jan 23 at 12:41










2




2




$begingroup$
Hint: What is the limit of $(n+1)/(n-1)$ ?
$endgroup$
– Martin R
Jan 23 at 12:40




$begingroup$
Hint: What is the limit of $(n+1)/(n-1)$ ?
$endgroup$
– Martin R
Jan 23 at 12:40




2




2




$begingroup$
Because $(n-1)/(n+1) notto +infty$, and you cannot use that significant well-known limit about $mathrm e$.
$endgroup$
– xbh
Jan 23 at 12:41




$begingroup$
Because $(n-1)/(n+1) notto +infty$, and you cannot use that significant well-known limit about $mathrm e$.
$endgroup$
– xbh
Jan 23 at 12:41












$begingroup$
limit of $(n+1)/(n-1)=1$, thank you!
$endgroup$
– J.Doe
Jan 23 at 12:41






$begingroup$
limit of $(n+1)/(n-1)=1$, thank you!
$endgroup$
– J.Doe
Jan 23 at 12:41












2 Answers
2






active

oldest

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6












$begingroup$

I hope you solved the problem. Here is a fact that might help you.



Put, $t=frac{n+1}{n-1}$. Here as $nto infty$, $t=frac{n+1}{n-1}=frac{1+frac{1}{n}}{1-frac{1}{n}}to 1$. Putting all these we get, $$lim_{ntoinfty}(1+frac{n-1}{n+1})^{frac{n+1}{n-1}}= lim_{tto 1}(1+frac{1}{t})^t=2$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Since $,displaystylelim_{ntoinfty}frac{n+1}{n-1} neq infty,$ you can't apply that property.
    But, you can rewrite it as
    $$1+frac{n-1}{n+1} = 2left(1-frac{1}{left(n+1right)}right)$$
    so, the limit is
    $$ lim_{ntoinfty},2^{frac{n+1}{n-1}}cdotleft[left(1+frac{1}{-left(n+1right)}right)^{-left(n+1right)}right]^{frac{-1}{n-1}} = 2cdot e^{lim_{ntoinfty}frac{-1}{n-1}} = 2$$






    share|cite|improve this answer









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      2 Answers
      2






      active

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

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      6












      $begingroup$

      I hope you solved the problem. Here is a fact that might help you.



      Put, $t=frac{n+1}{n-1}$. Here as $nto infty$, $t=frac{n+1}{n-1}=frac{1+frac{1}{n}}{1-frac{1}{n}}to 1$. Putting all these we get, $$lim_{ntoinfty}(1+frac{n-1}{n+1})^{frac{n+1}{n-1}}= lim_{tto 1}(1+frac{1}{t})^t=2$$






      share|cite|improve this answer









      $endgroup$


















        6












        $begingroup$

        I hope you solved the problem. Here is a fact that might help you.



        Put, $t=frac{n+1}{n-1}$. Here as $nto infty$, $t=frac{n+1}{n-1}=frac{1+frac{1}{n}}{1-frac{1}{n}}to 1$. Putting all these we get, $$lim_{ntoinfty}(1+frac{n-1}{n+1})^{frac{n+1}{n-1}}= lim_{tto 1}(1+frac{1}{t})^t=2$$






        share|cite|improve this answer









        $endgroup$
















          6












          6








          6





          $begingroup$

          I hope you solved the problem. Here is a fact that might help you.



          Put, $t=frac{n+1}{n-1}$. Here as $nto infty$, $t=frac{n+1}{n-1}=frac{1+frac{1}{n}}{1-frac{1}{n}}to 1$. Putting all these we get, $$lim_{ntoinfty}(1+frac{n-1}{n+1})^{frac{n+1}{n-1}}= lim_{tto 1}(1+frac{1}{t})^t=2$$






          share|cite|improve this answer









          $endgroup$



          I hope you solved the problem. Here is a fact that might help you.



          Put, $t=frac{n+1}{n-1}$. Here as $nto infty$, $t=frac{n+1}{n-1}=frac{1+frac{1}{n}}{1-frac{1}{n}}to 1$. Putting all these we get, $$lim_{ntoinfty}(1+frac{n-1}{n+1})^{frac{n+1}{n-1}}= lim_{tto 1}(1+frac{1}{t})^t=2$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 23 at 12:50









          Sujit BhattacharyyaSujit Bhattacharyya

          1,372519




          1,372519























              0












              $begingroup$

              Since $,displaystylelim_{ntoinfty}frac{n+1}{n-1} neq infty,$ you can't apply that property.
              But, you can rewrite it as
              $$1+frac{n-1}{n+1} = 2left(1-frac{1}{left(n+1right)}right)$$
              so, the limit is
              $$ lim_{ntoinfty},2^{frac{n+1}{n-1}}cdotleft[left(1+frac{1}{-left(n+1right)}right)^{-left(n+1right)}right]^{frac{-1}{n-1}} = 2cdot e^{lim_{ntoinfty}frac{-1}{n-1}} = 2$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Since $,displaystylelim_{ntoinfty}frac{n+1}{n-1} neq infty,$ you can't apply that property.
                But, you can rewrite it as
                $$1+frac{n-1}{n+1} = 2left(1-frac{1}{left(n+1right)}right)$$
                so, the limit is
                $$ lim_{ntoinfty},2^{frac{n+1}{n-1}}cdotleft[left(1+frac{1}{-left(n+1right)}right)^{-left(n+1right)}right]^{frac{-1}{n-1}} = 2cdot e^{lim_{ntoinfty}frac{-1}{n-1}} = 2$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Since $,displaystylelim_{ntoinfty}frac{n+1}{n-1} neq infty,$ you can't apply that property.
                  But, you can rewrite it as
                  $$1+frac{n-1}{n+1} = 2left(1-frac{1}{left(n+1right)}right)$$
                  so, the limit is
                  $$ lim_{ntoinfty},2^{frac{n+1}{n-1}}cdotleft[left(1+frac{1}{-left(n+1right)}right)^{-left(n+1right)}right]^{frac{-1}{n-1}} = 2cdot e^{lim_{ntoinfty}frac{-1}{n-1}} = 2$$






                  share|cite|improve this answer









                  $endgroup$



                  Since $,displaystylelim_{ntoinfty}frac{n+1}{n-1} neq infty,$ you can't apply that property.
                  But, you can rewrite it as
                  $$1+frac{n-1}{n+1} = 2left(1-frac{1}{left(n+1right)}right)$$
                  so, the limit is
                  $$ lim_{ntoinfty},2^{frac{n+1}{n-1}}cdotleft[left(1+frac{1}{-left(n+1right)}right)^{-left(n+1right)}right]^{frac{-1}{n-1}} = 2cdot e^{lim_{ntoinfty}frac{-1}{n-1}} = 2$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 23 at 13:09









                  CarlIOCarlIO

                  957




                  957






























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