$lim limits_{n to infty} left(1-frac1{n^2}right)^{n}$ equals












2














The limit $lim limits_{n to infty} bigg(1-dfrac{1}{n^2}bigg)^{n}$ equals?



$(A) e^{-1}$



$(B) e^{-frac{1}{2}}$



$(C) e^{-2}$



$(D) 1$



Using the expansion $(1+x)^{n}=1+nx+ dfrac{n(n-1)x^{2}}{2!} ...$

I am having large degree of $n$ in denominator therefor i am left with $1$ after applying the limit. Did i do it right ? If i did it right is there any alternative method to solve this problem like a quick version?






limits exponential-function







share|cite|improve this question





























    2














    The limit $lim limits_{n to infty} bigg(1-dfrac{1}{n^2}bigg)^{n}$ equals?



    $(A) e^{-1}$



    $(B) e^{-frac{1}{2}}$



    $(C) e^{-2}$



    $(D) 1$



    Using the expansion $(1+x)^{n}=1+nx+ dfrac{n(n-1)x^{2}}{2!} ...$

    I am having large degree of $n$ in denominator therefor i am left with $1$ after applying the limit. Did i do it right ? If i did it right is there any alternative method to solve this problem like a quick version?






    limits exponential-function







    share|cite|improve this question



























      2












      2








      2


      1





      The limit $lim limits_{n to infty} bigg(1-dfrac{1}{n^2}bigg)^{n}$ equals?



      $(A) e^{-1}$



      $(B) e^{-frac{1}{2}}$



      $(C) e^{-2}$



      $(D) 1$



      Using the expansion $(1+x)^{n}=1+nx+ dfrac{n(n-1)x^{2}}{2!} ...$

      I am having large degree of $n$ in denominator therefor i am left with $1$ after applying the limit. Did i do it right ? If i did it right is there any alternative method to solve this problem like a quick version?






      limits exponential-function







      share|cite|improve this question















      The limit $lim limits_{n to infty} bigg(1-dfrac{1}{n^2}bigg)^{n}$ equals?



      $(A) e^{-1}$



      $(B) e^{-frac{1}{2}}$



      $(C) e^{-2}$



      $(D) 1$



      Using the expansion $(1+x)^{n}=1+nx+ dfrac{n(n-1)x^{2}}{2!} ...$

      I am having large degree of $n$ in denominator therefor i am left with $1$ after applying the limit. Did i do it right ? If i did it right is there any alternative method to solve this problem like a quick version?






      limits exponential-function




      limits exponential-function






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 9 at 11:03









      Martin Sleziak

      44.7k7115270




      44.7k7115270










      asked Feb 5 at 11:01









      Daman deep

      643218




      643218






















          4 Answers
          4






          active

          oldest

          votes


















          7














          Yes, $lim_{ntoinfty}left(1-frac{1}{n^2}right)^n=lim_{ntoinfty}left(1-frac{1}{n}right)^nleft(1+frac{1}{n}right)^n=lim_{ntoinfty}left(left(1+frac{1}{-n}right)^{-n}right)^{-1}lim_{ntoinfty}left(1+frac{1}{n}right)^n=e^{-1}cdot e=1$






          share|cite|improve this answer





















          • It looks so easy now. How did miss that anyways thanks:)
            – Daman deep
            Feb 5 at 11:12






          • 1




            @Damn1o1 I don't know if you were supposed to solve the problem that way, but at least it serves as a good way to check your result that you may have obtained differently.
            – user491874
            Feb 5 at 11:16










          • No it was best I am building up speed for MCQ exam i shouldn't miss these types of catch. I guess they come with experience.
            – Daman deep
            Feb 5 at 11:18



















          2














          $1-frac{1}{n^2} to 1$ and $n to infty$, so:
          $$lim_{n to infty}left(1-frac{1}{n^2}right)^n=expleft(lim_{n to infty}left(1-frac{1}{n^2}-1right)nright)=expleft(lim_{n to infty}left(-frac{1}{n^2}right)nright)=expleft(lim_{n to infty}-frac{1}{n}right)=exp(-0)=exp(0)=1$$






          share|cite|improve this answer





























            0














            Hint:



            $$lim_{ntoinfty}left(1-dfrac1{n^2}right)^n=left(lim_{ntoinfty}left(1+dfrac1{-n^2}right)^{-n^2}right)^{lim_{ntoinfty}-1/n}$$



            Set $x=1$ in About $lim left(1+frac {x}{n}right)^n$






            share|cite|improve this answer























            • How did you do that with power i don't understand.That makes it $n^3$
              – Daman deep
              Feb 5 at 11:06












            • @Damn1o1, Sorry for the typo
              – lab bhattacharjee
              Feb 5 at 11:07












            • Ohh inside part is $e$ and $e^{-frac{1}{n}}$ makes it 1 yes thank you.
              – Daman deep
              Feb 5 at 11:09





















            0














            By Bernoulli: $ left(1-dfrac1{n^2}right)^n ge 1-frac{1}{n}$, hence



            $$1-frac{1}{n} le left(1-dfrac1{n^2}right)^n le 1.$$



            Conclusion ?






            share|cite|improve this answer





















              Your Answer





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              4 Answers
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              4 Answers
              4






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              active

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              active

              oldest

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              7














              Yes, $lim_{ntoinfty}left(1-frac{1}{n^2}right)^n=lim_{ntoinfty}left(1-frac{1}{n}right)^nleft(1+frac{1}{n}right)^n=lim_{ntoinfty}left(left(1+frac{1}{-n}right)^{-n}right)^{-1}lim_{ntoinfty}left(1+frac{1}{n}right)^n=e^{-1}cdot e=1$






              share|cite|improve this answer





















              • It looks so easy now. How did miss that anyways thanks:)
                – Daman deep
                Feb 5 at 11:12






              • 1




                @Damn1o1 I don't know if you were supposed to solve the problem that way, but at least it serves as a good way to check your result that you may have obtained differently.
                – user491874
                Feb 5 at 11:16










              • No it was best I am building up speed for MCQ exam i shouldn't miss these types of catch. I guess they come with experience.
                – Daman deep
                Feb 5 at 11:18
















              7














              Yes, $lim_{ntoinfty}left(1-frac{1}{n^2}right)^n=lim_{ntoinfty}left(1-frac{1}{n}right)^nleft(1+frac{1}{n}right)^n=lim_{ntoinfty}left(left(1+frac{1}{-n}right)^{-n}right)^{-1}lim_{ntoinfty}left(1+frac{1}{n}right)^n=e^{-1}cdot e=1$






              share|cite|improve this answer





















              • It looks so easy now. How did miss that anyways thanks:)
                – Daman deep
                Feb 5 at 11:12






              • 1




                @Damn1o1 I don't know if you were supposed to solve the problem that way, but at least it serves as a good way to check your result that you may have obtained differently.
                – user491874
                Feb 5 at 11:16










              • No it was best I am building up speed for MCQ exam i shouldn't miss these types of catch. I guess they come with experience.
                – Daman deep
                Feb 5 at 11:18














              7












              7








              7






              Yes, $lim_{ntoinfty}left(1-frac{1}{n^2}right)^n=lim_{ntoinfty}left(1-frac{1}{n}right)^nleft(1+frac{1}{n}right)^n=lim_{ntoinfty}left(left(1+frac{1}{-n}right)^{-n}right)^{-1}lim_{ntoinfty}left(1+frac{1}{n}right)^n=e^{-1}cdot e=1$






              share|cite|improve this answer












              Yes, $lim_{ntoinfty}left(1-frac{1}{n^2}right)^n=lim_{ntoinfty}left(1-frac{1}{n}right)^nleft(1+frac{1}{n}right)^n=lim_{ntoinfty}left(left(1+frac{1}{-n}right)^{-n}right)^{-1}lim_{ntoinfty}left(1+frac{1}{n}right)^n=e^{-1}cdot e=1$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Feb 5 at 11:05







              user491874



















              • It looks so easy now. How did miss that anyways thanks:)
                – Daman deep
                Feb 5 at 11:12






              • 1




                @Damn1o1 I don't know if you were supposed to solve the problem that way, but at least it serves as a good way to check your result that you may have obtained differently.
                – user491874
                Feb 5 at 11:16










              • No it was best I am building up speed for MCQ exam i shouldn't miss these types of catch. I guess they come with experience.
                – Daman deep
                Feb 5 at 11:18


















              • It looks so easy now. How did miss that anyways thanks:)
                – Daman deep
                Feb 5 at 11:12






              • 1




                @Damn1o1 I don't know if you were supposed to solve the problem that way, but at least it serves as a good way to check your result that you may have obtained differently.
                – user491874
                Feb 5 at 11:16










              • No it was best I am building up speed for MCQ exam i shouldn't miss these types of catch. I guess they come with experience.
                – Daman deep
                Feb 5 at 11:18
















              It looks so easy now. How did miss that anyways thanks:)
              – Daman deep
              Feb 5 at 11:12




              It looks so easy now. How did miss that anyways thanks:)
              – Daman deep
              Feb 5 at 11:12




              1




              1




              @Damn1o1 I don't know if you were supposed to solve the problem that way, but at least it serves as a good way to check your result that you may have obtained differently.
              – user491874
              Feb 5 at 11:16




              @Damn1o1 I don't know if you were supposed to solve the problem that way, but at least it serves as a good way to check your result that you may have obtained differently.
              – user491874
              Feb 5 at 11:16












              No it was best I am building up speed for MCQ exam i shouldn't miss these types of catch. I guess they come with experience.
              – Daman deep
              Feb 5 at 11:18




              No it was best I am building up speed for MCQ exam i shouldn't miss these types of catch. I guess they come with experience.
              – Daman deep
              Feb 5 at 11:18











              2














              $1-frac{1}{n^2} to 1$ and $n to infty$, so:
              $$lim_{n to infty}left(1-frac{1}{n^2}right)^n=expleft(lim_{n to infty}left(1-frac{1}{n^2}-1right)nright)=expleft(lim_{n to infty}left(-frac{1}{n^2}right)nright)=expleft(lim_{n to infty}-frac{1}{n}right)=exp(-0)=exp(0)=1$$






              share|cite|improve this answer


























                2














                $1-frac{1}{n^2} to 1$ and $n to infty$, so:
                $$lim_{n to infty}left(1-frac{1}{n^2}right)^n=expleft(lim_{n to infty}left(1-frac{1}{n^2}-1right)nright)=expleft(lim_{n to infty}left(-frac{1}{n^2}right)nright)=expleft(lim_{n to infty}-frac{1}{n}right)=exp(-0)=exp(0)=1$$






                share|cite|improve this answer
























                  2












                  2








                  2






                  $1-frac{1}{n^2} to 1$ and $n to infty$, so:
                  $$lim_{n to infty}left(1-frac{1}{n^2}right)^n=expleft(lim_{n to infty}left(1-frac{1}{n^2}-1right)nright)=expleft(lim_{n to infty}left(-frac{1}{n^2}right)nright)=expleft(lim_{n to infty}-frac{1}{n}right)=exp(-0)=exp(0)=1$$






                  share|cite|improve this answer












                  $1-frac{1}{n^2} to 1$ and $n to infty$, so:
                  $$lim_{n to infty}left(1-frac{1}{n^2}right)^n=expleft(lim_{n to infty}left(1-frac{1}{n^2}-1right)nright)=expleft(lim_{n to infty}left(-frac{1}{n^2}right)nright)=expleft(lim_{n to infty}-frac{1}{n}right)=exp(-0)=exp(0)=1$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Feb 5 at 11:20









                  Botond

                  5,5232732




                  5,5232732























                      0














                      Hint:



                      $$lim_{ntoinfty}left(1-dfrac1{n^2}right)^n=left(lim_{ntoinfty}left(1+dfrac1{-n^2}right)^{-n^2}right)^{lim_{ntoinfty}-1/n}$$



                      Set $x=1$ in About $lim left(1+frac {x}{n}right)^n$






                      share|cite|improve this answer























                      • How did you do that with power i don't understand.That makes it $n^3$
                        – Daman deep
                        Feb 5 at 11:06












                      • @Damn1o1, Sorry for the typo
                        – lab bhattacharjee
                        Feb 5 at 11:07












                      • Ohh inside part is $e$ and $e^{-frac{1}{n}}$ makes it 1 yes thank you.
                        – Daman deep
                        Feb 5 at 11:09


















                      0














                      Hint:



                      $$lim_{ntoinfty}left(1-dfrac1{n^2}right)^n=left(lim_{ntoinfty}left(1+dfrac1{-n^2}right)^{-n^2}right)^{lim_{ntoinfty}-1/n}$$



                      Set $x=1$ in About $lim left(1+frac {x}{n}right)^n$






                      share|cite|improve this answer























                      • How did you do that with power i don't understand.That makes it $n^3$
                        – Daman deep
                        Feb 5 at 11:06












                      • @Damn1o1, Sorry for the typo
                        – lab bhattacharjee
                        Feb 5 at 11:07












                      • Ohh inside part is $e$ and $e^{-frac{1}{n}}$ makes it 1 yes thank you.
                        – Daman deep
                        Feb 5 at 11:09
















                      0












                      0








                      0






                      Hint:



                      $$lim_{ntoinfty}left(1-dfrac1{n^2}right)^n=left(lim_{ntoinfty}left(1+dfrac1{-n^2}right)^{-n^2}right)^{lim_{ntoinfty}-1/n}$$



                      Set $x=1$ in About $lim left(1+frac {x}{n}right)^n$






                      share|cite|improve this answer














                      Hint:



                      $$lim_{ntoinfty}left(1-dfrac1{n^2}right)^n=left(lim_{ntoinfty}left(1+dfrac1{-n^2}right)^{-n^2}right)^{lim_{ntoinfty}-1/n}$$



                      Set $x=1$ in About $lim left(1+frac {x}{n}right)^n$







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Feb 5 at 11:07

























                      answered Feb 5 at 11:03









                      lab bhattacharjee

                      223k15156274




                      223k15156274












                      • How did you do that with power i don't understand.That makes it $n^3$
                        – Daman deep
                        Feb 5 at 11:06












                      • @Damn1o1, Sorry for the typo
                        – lab bhattacharjee
                        Feb 5 at 11:07












                      • Ohh inside part is $e$ and $e^{-frac{1}{n}}$ makes it 1 yes thank you.
                        – Daman deep
                        Feb 5 at 11:09




















                      • How did you do that with power i don't understand.That makes it $n^3$
                        – Daman deep
                        Feb 5 at 11:06












                      • @Damn1o1, Sorry for the typo
                        – lab bhattacharjee
                        Feb 5 at 11:07












                      • Ohh inside part is $e$ and $e^{-frac{1}{n}}$ makes it 1 yes thank you.
                        – Daman deep
                        Feb 5 at 11:09


















                      How did you do that with power i don't understand.That makes it $n^3$
                      – Daman deep
                      Feb 5 at 11:06






                      How did you do that with power i don't understand.That makes it $n^3$
                      – Daman deep
                      Feb 5 at 11:06














                      @Damn1o1, Sorry for the typo
                      – lab bhattacharjee
                      Feb 5 at 11:07






                      @Damn1o1, Sorry for the typo
                      – lab bhattacharjee
                      Feb 5 at 11:07














                      Ohh inside part is $e$ and $e^{-frac{1}{n}}$ makes it 1 yes thank you.
                      – Daman deep
                      Feb 5 at 11:09






                      Ohh inside part is $e$ and $e^{-frac{1}{n}}$ makes it 1 yes thank you.
                      – Daman deep
                      Feb 5 at 11:09













                      0














                      By Bernoulli: $ left(1-dfrac1{n^2}right)^n ge 1-frac{1}{n}$, hence



                      $$1-frac{1}{n} le left(1-dfrac1{n^2}right)^n le 1.$$



                      Conclusion ?






                      share|cite|improve this answer


























                        0














                        By Bernoulli: $ left(1-dfrac1{n^2}right)^n ge 1-frac{1}{n}$, hence



                        $$1-frac{1}{n} le left(1-dfrac1{n^2}right)^n le 1.$$



                        Conclusion ?






                        share|cite|improve this answer
























                          0












                          0








                          0






                          By Bernoulli: $ left(1-dfrac1{n^2}right)^n ge 1-frac{1}{n}$, hence



                          $$1-frac{1}{n} le left(1-dfrac1{n^2}right)^n le 1.$$



                          Conclusion ?






                          share|cite|improve this answer












                          By Bernoulli: $ left(1-dfrac1{n^2}right)^n ge 1-frac{1}{n}$, hence



                          $$1-frac{1}{n} le left(1-dfrac1{n^2}right)^n le 1.$$



                          Conclusion ?







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Feb 5 at 12:04









                          Fred

                          44.2k1845




                          44.2k1845






























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