Limit $lim_{ntoinfty} n^2left(sqrt{1+frac{1}{n}}+sqrt{1-frac{1}{n}}-2right)$Calculate: $lim_{n...

What does "Scientists rise up against statistical significance" mean? (Comment in Nature)

US tourist/student visa

What are some good ways to treat frozen vegetables such that they behave like fresh vegetables when stir frying them?

When were female captains banned from Starfleet?

How to preserve electronics (computers, iPads and phones) for hundreds of years

Giving feedback to someone without sounding prejudiced

What is the highest possible scrabble score for placing a single tile

Does the Linux kernel need a file system to run?

Is my low blitz game drawing rate at www.chess.com an indicator that I am weak in chess?

Is there a RAID 0 Equivalent for RAM?

Has any country ever had 2 former presidents in jail simultaneously?

What is the difference between lands and mana?

How to explain what's wrong with this application of the chain rule?

Is there a nicer/politer/more positive alternative for "negates"?

"It doesn't matter" or "it won't matter"?

Why do ¬, ∀ and ∃ have the same precedence?

Delete multiple columns using awk or sed

How much theory knowledge is actually used while playing?

Can I cause damage to electrical appliances by unplugging them when they are turned on?

How do I tell my boss that I'm quitting soon, especially given that a colleague just left this week

How much of a Devil Fruit must be consumed to gain the power?

Why do Radio Buttons not fill the entire outer circle?

How to get directions in deep space?

Why is the Sun approximated as a black body at ~ 5800 K?



Limit $lim_{ntoinfty} n^2left(sqrt{1+frac{1}{n}}+sqrt{1-frac{1}{n}}-2right)$


Calculate: $lim_{n rightarrowinfty}left(frac{3^{-n}sin(3^{(1-n)})}{tan(3^{1-2n})} right)$Calculate:$lim_{x rightarrowinfty}left(frac{(2+x)^{40}(4+x)^{5}}{(2-x)^{45}} right)$Calculate:$lim_{x rightarrow (-1)^{+}}left(frac{sqrt{pi}-sqrt{cos^{-1}x}}{sqrt{x+1}} right)$Show $xsqrt{n} - n lnleft(1+frac{x}{sqrt{n}}right) to frac{x^2}{2}$Evaluating the limit of $lim_{xtoinfty}(sqrt{frac{x^3}{x+2}}-x)$.Finding the limit $lim_{ntoinfty} frac{nleft(sqrt[n]{n}-1right)}{log n}$Find $lim_{xto infty}left(x^2lnleft(cos frac{4}{x}right)right)$Find the limit $lim_{ntoinfty}left(sqrt{n^2+n+1}-leftlfloorsqrt{n^2+n+1}rightrfloorright)$Find $limlimits_{ntoinfty}{nleft(left(1+frac{1}{n}right)^n-eright)}$By using the definition of $e$, prove that $lim_{xtoinfty} left(frac{3+2x}{5+2x}right)^x = frac1e$













1












$begingroup$


Greetings I am trying to solve $$lim_{ntoinfty} n^2left(sqrt{1+frac{1}{n}}+sqrt{1-frac{1}{n}}-2right)$$ Using binomial series is pretty easy: $$lim_{ntoinfty}n^2left(1+frac{1}{2n}-frac{1}{8n^2}+mathcal{O}left(frac{1}{n^3}right)+1-frac{1}{2n}-frac{1}{8n^2}+mathcal{O}left(frac{1}{n^3}right)-2right)=lim_{ntoinfty}n^2left(-frac{1}{8n^2}+mathcal{O}left(frac{1}{n^3}right)-frac{1}{8n^2}+mathcal{O}left(frac{1}{n^3}right)right)=-frac{1}{4}$$ The problem is that I need to solve this using only highschool tools, but I cant seem too take it down. My other try was to use L'Hospital rule but I feel like it just complicate things. Maybe there is even an elegant way, could you give me some help with this?






limits







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    rationalize twice.
    $endgroup$
    – Takahiro Waki
    Jul 8 '18 at 12:55






  • 4




    $begingroup$
    or $=n^2left(sqrt{1+frac{1}{n}}-1-frac{1}{2n}right)+n^2left(sqrt{1-frac{1}{n}}-1+frac{1}{2n}right)=-frac{1}{4}frac{1}{sqrt{1+frac{1}{n}}+1+frac{1}{2n}}-frac{1}{4}frac{1}{sqrt{1-frac{1}{n}}-1+frac{1}{2n}}$ It wouldn't be the first time that a proposed solution for high school looks clever, while the idea is easy to get using further knowledge.
    $endgroup$
    – user566930
    Jul 8 '18 at 13:03








  • 1




    $begingroup$
    @Dahaka LOL. What I was saying. Adding and subtracting $frac{1}{2n}$ came from your Taylor expansion.
    $endgroup$
    – user566930
    Jul 8 '18 at 13:07
















1












$begingroup$


Greetings I am trying to solve $$lim_{ntoinfty} n^2left(sqrt{1+frac{1}{n}}+sqrt{1-frac{1}{n}}-2right)$$ Using binomial series is pretty easy: $$lim_{ntoinfty}n^2left(1+frac{1}{2n}-frac{1}{8n^2}+mathcal{O}left(frac{1}{n^3}right)+1-frac{1}{2n}-frac{1}{8n^2}+mathcal{O}left(frac{1}{n^3}right)-2right)=lim_{ntoinfty}n^2left(-frac{1}{8n^2}+mathcal{O}left(frac{1}{n^3}right)-frac{1}{8n^2}+mathcal{O}left(frac{1}{n^3}right)right)=-frac{1}{4}$$ The problem is that I need to solve this using only highschool tools, but I cant seem too take it down. My other try was to use L'Hospital rule but I feel like it just complicate things. Maybe there is even an elegant way, could you give me some help with this?






limits







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    rationalize twice.
    $endgroup$
    – Takahiro Waki
    Jul 8 '18 at 12:55






  • 4




    $begingroup$
    or $=n^2left(sqrt{1+frac{1}{n}}-1-frac{1}{2n}right)+n^2left(sqrt{1-frac{1}{n}}-1+frac{1}{2n}right)=-frac{1}{4}frac{1}{sqrt{1+frac{1}{n}}+1+frac{1}{2n}}-frac{1}{4}frac{1}{sqrt{1-frac{1}{n}}-1+frac{1}{2n}}$ It wouldn't be the first time that a proposed solution for high school looks clever, while the idea is easy to get using further knowledge.
    $endgroup$
    – user566930
    Jul 8 '18 at 13:03








  • 1




    $begingroup$
    @Dahaka LOL. What I was saying. Adding and subtracting $frac{1}{2n}$ came from your Taylor expansion.
    $endgroup$
    – user566930
    Jul 8 '18 at 13:07














1












1








1





$begingroup$


Greetings I am trying to solve $$lim_{ntoinfty} n^2left(sqrt{1+frac{1}{n}}+sqrt{1-frac{1}{n}}-2right)$$ Using binomial series is pretty easy: $$lim_{ntoinfty}n^2left(1+frac{1}{2n}-frac{1}{8n^2}+mathcal{O}left(frac{1}{n^3}right)+1-frac{1}{2n}-frac{1}{8n^2}+mathcal{O}left(frac{1}{n^3}right)-2right)=lim_{ntoinfty}n^2left(-frac{1}{8n^2}+mathcal{O}left(frac{1}{n^3}right)-frac{1}{8n^2}+mathcal{O}left(frac{1}{n^3}right)right)=-frac{1}{4}$$ The problem is that I need to solve this using only highschool tools, but I cant seem too take it down. My other try was to use L'Hospital rule but I feel like it just complicate things. Maybe there is even an elegant way, could you give me some help with this?






limits







share|cite|improve this question









$endgroup$




Greetings I am trying to solve $$lim_{ntoinfty} n^2left(sqrt{1+frac{1}{n}}+sqrt{1-frac{1}{n}}-2right)$$ Using binomial series is pretty easy: $$lim_{ntoinfty}n^2left(1+frac{1}{2n}-frac{1}{8n^2}+mathcal{O}left(frac{1}{n^3}right)+1-frac{1}{2n}-frac{1}{8n^2}+mathcal{O}left(frac{1}{n^3}right)-2right)=lim_{ntoinfty}n^2left(-frac{1}{8n^2}+mathcal{O}left(frac{1}{n^3}right)-frac{1}{8n^2}+mathcal{O}left(frac{1}{n^3}right)right)=-frac{1}{4}$$ The problem is that I need to solve this using only highschool tools, but I cant seem too take it down. My other try was to use L'Hospital rule but I feel like it just complicate things. Maybe there is even an elegant way, could you give me some help with this?






limits




limits






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jul 8 '18 at 12:48









ZackyZacky

7,89011061




7,89011061








  • 1




    $begingroup$
    rationalize twice.
    $endgroup$
    – Takahiro Waki
    Jul 8 '18 at 12:55






  • 4




    $begingroup$
    or $=n^2left(sqrt{1+frac{1}{n}}-1-frac{1}{2n}right)+n^2left(sqrt{1-frac{1}{n}}-1+frac{1}{2n}right)=-frac{1}{4}frac{1}{sqrt{1+frac{1}{n}}+1+frac{1}{2n}}-frac{1}{4}frac{1}{sqrt{1-frac{1}{n}}-1+frac{1}{2n}}$ It wouldn't be the first time that a proposed solution for high school looks clever, while the idea is easy to get using further knowledge.
    $endgroup$
    – user566930
    Jul 8 '18 at 13:03








  • 1




    $begingroup$
    @Dahaka LOL. What I was saying. Adding and subtracting $frac{1}{2n}$ came from your Taylor expansion.
    $endgroup$
    – user566930
    Jul 8 '18 at 13:07














  • 1




    $begingroup$
    rationalize twice.
    $endgroup$
    – Takahiro Waki
    Jul 8 '18 at 12:55






  • 4




    $begingroup$
    or $=n^2left(sqrt{1+frac{1}{n}}-1-frac{1}{2n}right)+n^2left(sqrt{1-frac{1}{n}}-1+frac{1}{2n}right)=-frac{1}{4}frac{1}{sqrt{1+frac{1}{n}}+1+frac{1}{2n}}-frac{1}{4}frac{1}{sqrt{1-frac{1}{n}}-1+frac{1}{2n}}$ It wouldn't be the first time that a proposed solution for high school looks clever, while the idea is easy to get using further knowledge.
    $endgroup$
    – user566930
    Jul 8 '18 at 13:03








  • 1




    $begingroup$
    @Dahaka LOL. What I was saying. Adding and subtracting $frac{1}{2n}$ came from your Taylor expansion.
    $endgroup$
    – user566930
    Jul 8 '18 at 13:07








1




1




$begingroup$
rationalize twice.
$endgroup$
– Takahiro Waki
Jul 8 '18 at 12:55




$begingroup$
rationalize twice.
$endgroup$
– Takahiro Waki
Jul 8 '18 at 12:55




4




4




$begingroup$
or $=n^2left(sqrt{1+frac{1}{n}}-1-frac{1}{2n}right)+n^2left(sqrt{1-frac{1}{n}}-1+frac{1}{2n}right)=-frac{1}{4}frac{1}{sqrt{1+frac{1}{n}}+1+frac{1}{2n}}-frac{1}{4}frac{1}{sqrt{1-frac{1}{n}}-1+frac{1}{2n}}$ It wouldn't be the first time that a proposed solution for high school looks clever, while the idea is easy to get using further knowledge.
$endgroup$
– user566930
Jul 8 '18 at 13:03






$begingroup$
or $=n^2left(sqrt{1+frac{1}{n}}-1-frac{1}{2n}right)+n^2left(sqrt{1-frac{1}{n}}-1+frac{1}{2n}right)=-frac{1}{4}frac{1}{sqrt{1+frac{1}{n}}+1+frac{1}{2n}}-frac{1}{4}frac{1}{sqrt{1-frac{1}{n}}-1+frac{1}{2n}}$ It wouldn't be the first time that a proposed solution for high school looks clever, while the idea is easy to get using further knowledge.
$endgroup$
– user566930
Jul 8 '18 at 13:03






1




1




$begingroup$
@Dahaka LOL. What I was saying. Adding and subtracting $frac{1}{2n}$ came from your Taylor expansion.
$endgroup$
– user566930
Jul 8 '18 at 13:07




$begingroup$
@Dahaka LOL. What I was saying. Adding and subtracting $frac{1}{2n}$ came from your Taylor expansion.
$endgroup$
– user566930
Jul 8 '18 at 13:07










3 Answers
3






active

oldest

votes


















3












$begingroup$

Hint: multiplying numerator and denominator by $sqrt{1+1/n}+sqrt{1-1/n}+2$ we get
$$2 n^2 frac{sqrt{1-1/n^2}-1}{sqrt{1+1/n}+sqrt{1-1/n}+2}$$ and then do the same with $$sqrt{1-1/n^2}+1$$
you will get



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






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    The most of this Problems can be solved by that way
    $endgroup$
    – Dr. Sonnhard Graubner
    Jul 8 '18 at 13:09





















2












$begingroup$

Making $delta = frac 1n$ You can arrange it as



$$
lim_{deltato 0}left(frac{frac{sqrt{1+delta}-1}{delta}-frac{sqrt{1-delta}-1}{delta}}{delta}right) = left(frac{d^2}{dx^2}sqrt xright)_{x=1} = -frac 14
$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Here's a slightly different approach using multiplication by conjugates.



    For notational simplicity, let $u=1/n$, so that $uto0$ as $ntoinfty$. Then



    $$n^2left(sqrt{1+{1over n}}+sqrt{1-{1over n}}-2right)={sqrt{1+u}-1over u^2}+{sqrt{1-u}-1over u^2}\
    ={1over u(sqrt{1+u}+1)}-{1over u(sqrt{1-u}+1)}\={sqrt{1-u}-sqrt{1+u}over u(sqrt{1+u}+1)(sqrt{1-u}+1)}\={-2over(sqrt{1+u}+1)(sqrt{1-u}+1)(sqrt{1-u}+sqrt{1+u})}to{-2over(1+1)(1+1)(1+1)}=-{1over4}$$






    share|cite|improve this answer









    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2844601%2flimit-lim-n-to-infty-n2-left-sqrt1-frac1n-sqrt1-frac1n-2-ri%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      Hint: multiplying numerator and denominator by $sqrt{1+1/n}+sqrt{1-1/n}+2$ we get
      $$2 n^2 frac{sqrt{1-1/n^2}-1}{sqrt{1+1/n}+sqrt{1-1/n}+2}$$ and then do the same with $$sqrt{1-1/n^2}+1$$
      you will get



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






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        The most of this Problems can be solved by that way
        $endgroup$
        – Dr. Sonnhard Graubner
        Jul 8 '18 at 13:09


















      3












      $begingroup$

      Hint: multiplying numerator and denominator by $sqrt{1+1/n}+sqrt{1-1/n}+2$ we get
      $$2 n^2 frac{sqrt{1-1/n^2}-1}{sqrt{1+1/n}+sqrt{1-1/n}+2}$$ and then do the same with $$sqrt{1-1/n^2}+1$$
      you will get



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






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        The most of this Problems can be solved by that way
        $endgroup$
        – Dr. Sonnhard Graubner
        Jul 8 '18 at 13:09
















      3












      3








      3





      $begingroup$

      Hint: multiplying numerator and denominator by $sqrt{1+1/n}+sqrt{1-1/n}+2$ we get
      $$2 n^2 frac{sqrt{1-1/n^2}-1}{sqrt{1+1/n}+sqrt{1-1/n}+2}$$ and then do the same with $$sqrt{1-1/n^2}+1$$
      you will get



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






      share|cite|improve this answer











      $endgroup$



      Hint: multiplying numerator and denominator by $sqrt{1+1/n}+sqrt{1-1/n}+2$ we get
      $$2 n^2 frac{sqrt{1-1/n^2}-1}{sqrt{1+1/n}+sqrt{1-1/n}+2}$$ and then do the same with $$sqrt{1-1/n^2}+1$$
      you will get



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







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Mar 13 at 5:25









      Rócherz

      2,9863821




      2,9863821










      answered Jul 8 '18 at 13:03









      Dr. Sonnhard GraubnerDr. Sonnhard Graubner

      77.9k42866




      77.9k42866








      • 1




        $begingroup$
        The most of this Problems can be solved by that way
        $endgroup$
        – Dr. Sonnhard Graubner
        Jul 8 '18 at 13:09
















      • 1




        $begingroup$
        The most of this Problems can be solved by that way
        $endgroup$
        – Dr. Sonnhard Graubner
        Jul 8 '18 at 13:09










      1




      1




      $begingroup$
      The most of this Problems can be solved by that way
      $endgroup$
      – Dr. Sonnhard Graubner
      Jul 8 '18 at 13:09






      $begingroup$
      The most of this Problems can be solved by that way
      $endgroup$
      – Dr. Sonnhard Graubner
      Jul 8 '18 at 13:09













      2












      $begingroup$

      Making $delta = frac 1n$ You can arrange it as



      $$
      lim_{deltato 0}left(frac{frac{sqrt{1+delta}-1}{delta}-frac{sqrt{1-delta}-1}{delta}}{delta}right) = left(frac{d^2}{dx^2}sqrt xright)_{x=1} = -frac 14
      $$






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Making $delta = frac 1n$ You can arrange it as



        $$
        lim_{deltato 0}left(frac{frac{sqrt{1+delta}-1}{delta}-frac{sqrt{1-delta}-1}{delta}}{delta}right) = left(frac{d^2}{dx^2}sqrt xright)_{x=1} = -frac 14
        $$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Making $delta = frac 1n$ You can arrange it as



          $$
          lim_{deltato 0}left(frac{frac{sqrt{1+delta}-1}{delta}-frac{sqrt{1-delta}-1}{delta}}{delta}right) = left(frac{d^2}{dx^2}sqrt xright)_{x=1} = -frac 14
          $$






          share|cite|improve this answer









          $endgroup$



          Making $delta = frac 1n$ You can arrange it as



          $$
          lim_{deltato 0}left(frac{frac{sqrt{1+delta}-1}{delta}-frac{sqrt{1-delta}-1}{delta}}{delta}right) = left(frac{d^2}{dx^2}sqrt xright)_{x=1} = -frac 14
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jul 8 '18 at 13:16









          CesareoCesareo

          9,4563517




          9,4563517























              0












              $begingroup$

              Here's a slightly different approach using multiplication by conjugates.



              For notational simplicity, let $u=1/n$, so that $uto0$ as $ntoinfty$. Then



              $$n^2left(sqrt{1+{1over n}}+sqrt{1-{1over n}}-2right)={sqrt{1+u}-1over u^2}+{sqrt{1-u}-1over u^2}\
              ={1over u(sqrt{1+u}+1)}-{1over u(sqrt{1-u}+1)}\={sqrt{1-u}-sqrt{1+u}over u(sqrt{1+u}+1)(sqrt{1-u}+1)}\={-2over(sqrt{1+u}+1)(sqrt{1-u}+1)(sqrt{1-u}+sqrt{1+u})}to{-2over(1+1)(1+1)(1+1)}=-{1over4}$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Here's a slightly different approach using multiplication by conjugates.



                For notational simplicity, let $u=1/n$, so that $uto0$ as $ntoinfty$. Then



                $$n^2left(sqrt{1+{1over n}}+sqrt{1-{1over n}}-2right)={sqrt{1+u}-1over u^2}+{sqrt{1-u}-1over u^2}\
                ={1over u(sqrt{1+u}+1)}-{1over u(sqrt{1-u}+1)}\={sqrt{1-u}-sqrt{1+u}over u(sqrt{1+u}+1)(sqrt{1-u}+1)}\={-2over(sqrt{1+u}+1)(sqrt{1-u}+1)(sqrt{1-u}+sqrt{1+u})}to{-2over(1+1)(1+1)(1+1)}=-{1over4}$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Here's a slightly different approach using multiplication by conjugates.



                  For notational simplicity, let $u=1/n$, so that $uto0$ as $ntoinfty$. Then



                  $$n^2left(sqrt{1+{1over n}}+sqrt{1-{1over n}}-2right)={sqrt{1+u}-1over u^2}+{sqrt{1-u}-1over u^2}\
                  ={1over u(sqrt{1+u}+1)}-{1over u(sqrt{1-u}+1)}\={sqrt{1-u}-sqrt{1+u}over u(sqrt{1+u}+1)(sqrt{1-u}+1)}\={-2over(sqrt{1+u}+1)(sqrt{1-u}+1)(sqrt{1-u}+sqrt{1+u})}to{-2over(1+1)(1+1)(1+1)}=-{1over4}$$






                  share|cite|improve this answer









                  $endgroup$



                  Here's a slightly different approach using multiplication by conjugates.



                  For notational simplicity, let $u=1/n$, so that $uto0$ as $ntoinfty$. Then



                  $$n^2left(sqrt{1+{1over n}}+sqrt{1-{1over n}}-2right)={sqrt{1+u}-1over u^2}+{sqrt{1-u}-1over u^2}\
                  ={1over u(sqrt{1+u}+1)}-{1over u(sqrt{1-u}+1)}\={sqrt{1-u}-sqrt{1+u}over u(sqrt{1+u}+1)(sqrt{1-u}+1)}\={-2over(sqrt{1+u}+1)(sqrt{1-u}+1)(sqrt{1-u}+sqrt{1+u})}to{-2over(1+1)(1+1)(1+1)}=-{1over4}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jul 8 '18 at 15:47









                  Barry CipraBarry Cipra

                  60.3k654127




                  60.3k654127






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2844601%2flimit-lim-n-to-infty-n2-left-sqrt1-frac1n-sqrt1-frac1n-2-ri%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      六本木駅

                      Image
                      六本木駅 7番出入口（東京ミッドタウン横） （2008年7月） ろっぽんぎ Roppongi （六本木ヒルズ・東京ミッドタウン前*） 東京都港区六本木六丁目1-25（東京メトロ） 北緯35度39分46秒東経139度43分53秒 座標: 北緯35度39分46秒 東経139度43分53秒 東京都港区赤坂九丁目7-39（東京都交通局） 北緯35度39分48秒東経139度43分55.6秒 所属事業者 東京地下鉄（東京メトロ） 東京都交通局（都営地下鉄） 電報略号 ロホ（東京メトロ） 六（東京都交通局、駅名略称） 駅構造 地下駅 ホーム 計4面4線 （日比谷線 2面2線 大江戸線 上下二層2面2線） 乗降人員 -統計年度- （東京メトロ）134,902人/日 （東京都交通局）100,587人/日 -2017年- 開業年月日 1964年（昭和39年）3月25日 乗入路線 2 路線 所属路線 ● 東京メトロ日比谷線 駅番号 ○ H 04 キロ程 16.1km（北千住起点） ◄ H 03 広尾 (1.7km) (1.5km) 神谷町 H 05 ► 所属路線 ● 都営地下鉄大江戸線 駅番号 ○ E 23 キロ程 23.2km（都庁前起点） ◄ E 22 麻布十番 (1.1km) (1.3km) 青山一丁目 E 24 ► 備考 * 『東京ミッドタウン前』表記は大江戸線のみ テンプレートを表示 六本木駅 （ろっぽんぎえき）は、東京都港区にある、東京地下鉄（東京メトロ）・東京都交通局（都営地下鉄）の駅である。所在地は東京メトロが六本木六丁目、東京都交通局が赤坂九丁目となっている。 東京メトロの日比谷線と、都営地下鉄の大江戸線が乗り入れている。駅番号は日比谷線が H 04 、大江戸線が E 23 である。 目次 1 歴史 2 駅構造 2.1 東京メトロ 2.1.1 のりば 2.1.2 改札内設備 2.2 都営地下鉄 2.2.1 のりば 2.2.2 改札内設備 3 利用状況 3.1 年度...
                      Read more

                      Integral that is continuous and looks like it converges to a geometric seriesTesting if a geometric series converges by taking limit to infinitySummation of arithmetic-geometric series of higher orderGeometric series with polynomial exponentHow to Recognize a Geometric SeriesShowing an integral equality with series over the integersDiscontinuity of a series of continuous functionsReasons why a Series ConvergesSum of infinite geometric series with two terms in summationUsing geometric series for computing IntegralsLimit of geometric series sum when $r = 1$

                      Image
                      Why did Rep. Omar conclude her criticism of US troops with the phrase "NotTodaySatan"? Contradiction proof for inequality of P and NP? What is meant by electrons moving to lower potential energy? Negative Resistance Zonal Statistics is returning null values in ArcGIS Desktop? Why do distances seem to matter in the Foundation world? Injection into a proper class and choice without regularity A faster way to compute the largest prime factor "Whatever a Russian does, they end up making the Kalashnikov gun"? Are there any similar proverbs in English? Do I need to watch Ant-Man and the Wasp and Captain Marvel before watching Avengers: Endgame? How bug prioritization works in agile projects vs non agile Is this a typo in Section 1.8.1 Mathematics for Computer Science? Drawing a german abacus as in the books of Adam Ries How to have a sharp product image? Creating a chemical industry from a medieval tech level without petroleum Partitioning va...
                      Read more

                      Joseph Lister

                      Image
                      Joseph Lister Fødd 5. april 1827 Upton House, Newham Død 10. februar 1912 Kent Nasjonalitet Det sameinte kongeriket Storbritannia og Irland Område kirurgi Yrke lege, universitetslærar, kirurg Institusjonar King’s College London University of Edinburgh University of Glasgow Alma mater University College London Medlem Royal Society Kungliga Vetenskapsakademien Det ungarske vitskapsakademiet Det greske filologiske selskap i Konstantinopel grunnlagt 1861 American Academy of Arts and Sciences Kungliga Vetenskaps- och Vitterhetssamhället i Göteborg Deutsche Akademie der Naturforscher Leopoldina Académie nationale de médecine Royal Academy of Medicine of Belgium Joseph Lister (fødd 5. april 1827 i Upton i Essex, død 10. februar 1912 i Walmer) var ein britisk kirurg. Han er kjend for å ha oppdaga at karbolsyre påverkar bakteriane. Ved å rense sår og å leggje på desifiserte banda...
                      Read more