Find the limit of $left(1+ frac{2}{n}right)^{n^{2}} exp(-2n)$ as $n to infty$.












2












$begingroup$



Find the limit of $left(1+ frac{2}{n}right)^{n^{2}} exp(-2n)$ as $n to infty$.




By expansion -
$$limlimits_{n to infty} left[1+(n^{2})(2/n) + (n^{2})(n^{2}-1)/2 dots ]/[1+2n+(2n)^{3}/3! dotsright]$$



I didn't get any result.
By applying limit directly, I'm getting indeterminate form.
How to find this limit?






limits







share|cite|improve this question











$endgroup$

















    2












    $begingroup$



    Find the limit of $left(1+ frac{2}{n}right)^{n^{2}} exp(-2n)$ as $n to infty$.




    By expansion -
    $$limlimits_{n to infty} left[1+(n^{2})(2/n) + (n^{2})(n^{2}-1)/2 dots ]/[1+2n+(2n)^{3}/3! dotsright]$$



    I didn't get any result.
    By applying limit directly, I'm getting indeterminate form.
    How to find this limit?






    limits







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$



      Find the limit of $left(1+ frac{2}{n}right)^{n^{2}} exp(-2n)$ as $n to infty$.




      By expansion -
      $$limlimits_{n to infty} left[1+(n^{2})(2/n) + (n^{2})(n^{2}-1)/2 dots ]/[1+2n+(2n)^{3}/3! dotsright]$$



      I didn't get any result.
      By applying limit directly, I'm getting indeterminate form.
      How to find this limit?






      limits







      share|cite|improve this question











      $endgroup$





      Find the limit of $left(1+ frac{2}{n}right)^{n^{2}} exp(-2n)$ as $n to infty$.




      By expansion -
      $$limlimits_{n to infty} left[1+(n^{2})(2/n) + (n^{2})(n^{2}-1)/2 dots ]/[1+2n+(2n)^{3}/3! dotsright]$$



      I didn't get any result.
      By applying limit directly, I'm getting indeterminate form.
      How to find this limit?






      limits




      limits






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 6 at 10:40









      rtybase

      10.7k21533




      10.7k21533










      asked Jan 6 at 10:05









      MathsaddictMathsaddict

      3299




      3299






















          1 Answer
          1






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          2












          $begingroup$

          Hint. Note that
          $$left(1+ frac{2}{n}right)^{n^{2}} exp(-2n)=expleft(n^2logleft(1+ frac{2}{n}right)-2nright).$$
          Now, by using the expansion $log(1+t)=t-frac{t^2}{2}+o(t^2)$ at $t=0$, we have that
          $$logleft(1+ frac{2}{n}right)=frac{2}{n}-frac{2}{n^2}+o(1/n^2).$$
          Can you take it from here?






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Okay, I got exp(-2). Thanks!
            $endgroup$
            – Mathsaddict
            Jan 6 at 10:19






          • 1




            $begingroup$
            Yes, that's it!
            $endgroup$
            – Robert Z
            Jan 6 at 10:34










          • $begingroup$
            maybe $O(1/n^3)$?
            $endgroup$
            – John Joy
            Jan 6 at 11:08






          • 1




            $begingroup$
            @JohnJoy Yes, here we can replace $o(1/n^2)$ with $O(1/n^3)$, but $o(1/n^2)$ suffices.
            $endgroup$
            – Robert Z
            Jan 6 at 13:31











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          1 Answer
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          1 Answer
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          active

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          active

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          2












          $begingroup$

          Hint. Note that
          $$left(1+ frac{2}{n}right)^{n^{2}} exp(-2n)=expleft(n^2logleft(1+ frac{2}{n}right)-2nright).$$
          Now, by using the expansion $log(1+t)=t-frac{t^2}{2}+o(t^2)$ at $t=0$, we have that
          $$logleft(1+ frac{2}{n}right)=frac{2}{n}-frac{2}{n^2}+o(1/n^2).$$
          Can you take it from here?






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Okay, I got exp(-2). Thanks!
            $endgroup$
            – Mathsaddict
            Jan 6 at 10:19






          • 1




            $begingroup$
            Yes, that's it!
            $endgroup$
            – Robert Z
            Jan 6 at 10:34










          • $begingroup$
            maybe $O(1/n^3)$?
            $endgroup$
            – John Joy
            Jan 6 at 11:08






          • 1




            $begingroup$
            @JohnJoy Yes, here we can replace $o(1/n^2)$ with $O(1/n^3)$, but $o(1/n^2)$ suffices.
            $endgroup$
            – Robert Z
            Jan 6 at 13:31
















          2












          $begingroup$

          Hint. Note that
          $$left(1+ frac{2}{n}right)^{n^{2}} exp(-2n)=expleft(n^2logleft(1+ frac{2}{n}right)-2nright).$$
          Now, by using the expansion $log(1+t)=t-frac{t^2}{2}+o(t^2)$ at $t=0$, we have that
          $$logleft(1+ frac{2}{n}right)=frac{2}{n}-frac{2}{n^2}+o(1/n^2).$$
          Can you take it from here?






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Okay, I got exp(-2). Thanks!
            $endgroup$
            – Mathsaddict
            Jan 6 at 10:19






          • 1




            $begingroup$
            Yes, that's it!
            $endgroup$
            – Robert Z
            Jan 6 at 10:34










          • $begingroup$
            maybe $O(1/n^3)$?
            $endgroup$
            – John Joy
            Jan 6 at 11:08






          • 1




            $begingroup$
            @JohnJoy Yes, here we can replace $o(1/n^2)$ with $O(1/n^3)$, but $o(1/n^2)$ suffices.
            $endgroup$
            – Robert Z
            Jan 6 at 13:31














          2












          2








          2





          $begingroup$

          Hint. Note that
          $$left(1+ frac{2}{n}right)^{n^{2}} exp(-2n)=expleft(n^2logleft(1+ frac{2}{n}right)-2nright).$$
          Now, by using the expansion $log(1+t)=t-frac{t^2}{2}+o(t^2)$ at $t=0$, we have that
          $$logleft(1+ frac{2}{n}right)=frac{2}{n}-frac{2}{n^2}+o(1/n^2).$$
          Can you take it from here?






          share|cite|improve this answer











          $endgroup$



          Hint. Note that
          $$left(1+ frac{2}{n}right)^{n^{2}} exp(-2n)=expleft(n^2logleft(1+ frac{2}{n}right)-2nright).$$
          Now, by using the expansion $log(1+t)=t-frac{t^2}{2}+o(t^2)$ at $t=0$, we have that
          $$logleft(1+ frac{2}{n}right)=frac{2}{n}-frac{2}{n^2}+o(1/n^2).$$
          Can you take it from here?







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 6 at 10:15

























          answered Jan 6 at 10:07









          Robert ZRobert Z

          95.4k1064135




          95.4k1064135








          • 1




            $begingroup$
            Okay, I got exp(-2). Thanks!
            $endgroup$
            – Mathsaddict
            Jan 6 at 10:19






          • 1




            $begingroup$
            Yes, that's it!
            $endgroup$
            – Robert Z
            Jan 6 at 10:34










          • $begingroup$
            maybe $O(1/n^3)$?
            $endgroup$
            – John Joy
            Jan 6 at 11:08






          • 1




            $begingroup$
            @JohnJoy Yes, here we can replace $o(1/n^2)$ with $O(1/n^3)$, but $o(1/n^2)$ suffices.
            $endgroup$
            – Robert Z
            Jan 6 at 13:31














          • 1




            $begingroup$
            Okay, I got exp(-2). Thanks!
            $endgroup$
            – Mathsaddict
            Jan 6 at 10:19






          • 1




            $begingroup$
            Yes, that's it!
            $endgroup$
            – Robert Z
            Jan 6 at 10:34










          • $begingroup$
            maybe $O(1/n^3)$?
            $endgroup$
            – John Joy
            Jan 6 at 11:08






          • 1




            $begingroup$
            @JohnJoy Yes, here we can replace $o(1/n^2)$ with $O(1/n^3)$, but $o(1/n^2)$ suffices.
            $endgroup$
            – Robert Z
            Jan 6 at 13:31








          1




          1




          $begingroup$
          Okay, I got exp(-2). Thanks!
          $endgroup$
          – Mathsaddict
          Jan 6 at 10:19




          $begingroup$
          Okay, I got exp(-2). Thanks!
          $endgroup$
          – Mathsaddict
          Jan 6 at 10:19




          1




          1




          $begingroup$
          Yes, that's it!
          $endgroup$
          – Robert Z
          Jan 6 at 10:34




          $begingroup$
          Yes, that's it!
          $endgroup$
          – Robert Z
          Jan 6 at 10:34












          $begingroup$
          maybe $O(1/n^3)$?
          $endgroup$
          – John Joy
          Jan 6 at 11:08




          $begingroup$
          maybe $O(1/n^3)$?
          $endgroup$
          – John Joy
          Jan 6 at 11:08




          1




          1




          $begingroup$
          @JohnJoy Yes, here we can replace $o(1/n^2)$ with $O(1/n^3)$, but $o(1/n^2)$ suffices.
          $endgroup$
          – Robert Z
          Jan 6 at 13:31




          $begingroup$
          @JohnJoy Yes, here we can replace $o(1/n^2)$ with $O(1/n^3)$, but $o(1/n^2)$ suffices.
          $endgroup$
          – Robert Z
          Jan 6 at 13:31


















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