Finding $limlimits_{n→∞}n^3(sqrt{n^2+sqrt{n^4+1}}-nsqrt2)$












3












$begingroup$


What is$$lim_{n→∞}n^3(sqrt{n^2+sqrt{n^4+1}}-nsqrt2)?$$So it is$$lim_{n→∞}frac{n^3(sqrt{n^2+sqrt{n^4+1}})^2-(nsqrt{2})^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}=lim_{n→∞}frac{n^3(n^2+sqrt{n^4+1}-2n^2)}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}.$$
I do not know what to do next, because my resuts is $∞$ but the answer from book is $dfrac{1}{4sqrt{2}}$.






calculus limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    try putting n=1/x.then apply binomial expansions.
    $endgroup$
    – maveric
    Dec 4 '18 at 10:07












  • $begingroup$
    @maveric That's of course an effective method but we don't really need that on this case.
    $endgroup$
    – gimusi
    Dec 4 '18 at 10:08










  • $begingroup$
    How can be your result $infty$ from here $$lim_{n→∞}frac{n^3(n^2+sqrt{n^4+1}-2n^2)}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}$$ Note that at the numerator we have an indeterminate form $infty-infty$ therefore you can't conclude form here.
    $endgroup$
    – gimusi
    Dec 4 '18 at 10:14












  • $begingroup$
    @gimusi. Using the binomial expansion is a good way if you want to know how is approached the limit (this is just building the Taylor expansion). I hope and wish that we could have a discussion about this kind of problems one of these days. Cheers.
    $endgroup$
    – Claude Leibovici
    Dec 4 '18 at 10:37












  • $begingroup$
    @ClaudeLeibovici Yes of course, I also like Taylor's very much but in that case I think that it suffices operate by agebraic methods also I think thet the OP is looking for such kind of solution. Anyway the suggestion is good as an alternative approach! Cheers
    $endgroup$
    – gimusi
    Dec 4 '18 at 10:39
















3












$begingroup$


What is$$lim_{n→∞}n^3(sqrt{n^2+sqrt{n^4+1}}-nsqrt2)?$$So it is$$lim_{n→∞}frac{n^3(sqrt{n^2+sqrt{n^4+1}})^2-(nsqrt{2})^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}=lim_{n→∞}frac{n^3(n^2+sqrt{n^4+1}-2n^2)}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}.$$
I do not know what to do next, because my resuts is $∞$ but the answer from book is $dfrac{1}{4sqrt{2}}$.






calculus limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    try putting n=1/x.then apply binomial expansions.
    $endgroup$
    – maveric
    Dec 4 '18 at 10:07












  • $begingroup$
    @maveric That's of course an effective method but we don't really need that on this case.
    $endgroup$
    – gimusi
    Dec 4 '18 at 10:08










  • $begingroup$
    How can be your result $infty$ from here $$lim_{n→∞}frac{n^3(n^2+sqrt{n^4+1}-2n^2)}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}$$ Note that at the numerator we have an indeterminate form $infty-infty$ therefore you can't conclude form here.
    $endgroup$
    – gimusi
    Dec 4 '18 at 10:14












  • $begingroup$
    @gimusi. Using the binomial expansion is a good way if you want to know how is approached the limit (this is just building the Taylor expansion). I hope and wish that we could have a discussion about this kind of problems one of these days. Cheers.
    $endgroup$
    – Claude Leibovici
    Dec 4 '18 at 10:37












  • $begingroup$
    @ClaudeLeibovici Yes of course, I also like Taylor's very much but in that case I think that it suffices operate by agebraic methods also I think thet the OP is looking for such kind of solution. Anyway the suggestion is good as an alternative approach! Cheers
    $endgroup$
    – gimusi
    Dec 4 '18 at 10:39














3












3








3





$begingroup$


What is$$lim_{n→∞}n^3(sqrt{n^2+sqrt{n^4+1}}-nsqrt2)?$$So it is$$lim_{n→∞}frac{n^3(sqrt{n^2+sqrt{n^4+1}})^2-(nsqrt{2})^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}=lim_{n→∞}frac{n^3(n^2+sqrt{n^4+1}-2n^2)}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}.$$
I do not know what to do next, because my resuts is $∞$ but the answer from book is $dfrac{1}{4sqrt{2}}$.






calculus limits







share|cite|improve this question











$endgroup$




What is$$lim_{n→∞}n^3(sqrt{n^2+sqrt{n^4+1}}-nsqrt2)?$$So it is$$lim_{n→∞}frac{n^3(sqrt{n^2+sqrt{n^4+1}})^2-(nsqrt{2})^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}=lim_{n→∞}frac{n^3(n^2+sqrt{n^4+1}-2n^2)}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}.$$
I do not know what to do next, because my resuts is $∞$ but the answer from book is $dfrac{1}{4sqrt{2}}$.






calculus limits




calculus limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 10:18









Jyrki Lahtonen

109k13170376




109k13170376










asked Dec 4 '18 at 9:59









mona1lisamona1lisa

337




337












  • $begingroup$
    try putting n=1/x.then apply binomial expansions.
    $endgroup$
    – maveric
    Dec 4 '18 at 10:07












  • $begingroup$
    @maveric That's of course an effective method but we don't really need that on this case.
    $endgroup$
    – gimusi
    Dec 4 '18 at 10:08










  • $begingroup$
    How can be your result $infty$ from here $$lim_{n→∞}frac{n^3(n^2+sqrt{n^4+1}-2n^2)}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}$$ Note that at the numerator we have an indeterminate form $infty-infty$ therefore you can't conclude form here.
    $endgroup$
    – gimusi
    Dec 4 '18 at 10:14












  • $begingroup$
    @gimusi. Using the binomial expansion is a good way if you want to know how is approached the limit (this is just building the Taylor expansion). I hope and wish that we could have a discussion about this kind of problems one of these days. Cheers.
    $endgroup$
    – Claude Leibovici
    Dec 4 '18 at 10:37












  • $begingroup$
    @ClaudeLeibovici Yes of course, I also like Taylor's very much but in that case I think that it suffices operate by agebraic methods also I think thet the OP is looking for such kind of solution. Anyway the suggestion is good as an alternative approach! Cheers
    $endgroup$
    – gimusi
    Dec 4 '18 at 10:39


















  • $begingroup$
    try putting n=1/x.then apply binomial expansions.
    $endgroup$
    – maveric
    Dec 4 '18 at 10:07












  • $begingroup$
    @maveric That's of course an effective method but we don't really need that on this case.
    $endgroup$
    – gimusi
    Dec 4 '18 at 10:08










  • $begingroup$
    How can be your result $infty$ from here $$lim_{n→∞}frac{n^3(n^2+sqrt{n^4+1}-2n^2)}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}$$ Note that at the numerator we have an indeterminate form $infty-infty$ therefore you can't conclude form here.
    $endgroup$
    – gimusi
    Dec 4 '18 at 10:14












  • $begingroup$
    @gimusi. Using the binomial expansion is a good way if you want to know how is approached the limit (this is just building the Taylor expansion). I hope and wish that we could have a discussion about this kind of problems one of these days. Cheers.
    $endgroup$
    – Claude Leibovici
    Dec 4 '18 at 10:37












  • $begingroup$
    @ClaudeLeibovici Yes of course, I also like Taylor's very much but in that case I think that it suffices operate by agebraic methods also I think thet the OP is looking for such kind of solution. Anyway the suggestion is good as an alternative approach! Cheers
    $endgroup$
    – gimusi
    Dec 4 '18 at 10:39
















$begingroup$
try putting n=1/x.then apply binomial expansions.
$endgroup$
– maveric
Dec 4 '18 at 10:07






$begingroup$
try putting n=1/x.then apply binomial expansions.
$endgroup$
– maveric
Dec 4 '18 at 10:07














$begingroup$
@maveric That's of course an effective method but we don't really need that on this case.
$endgroup$
– gimusi
Dec 4 '18 at 10:08




$begingroup$
@maveric That's of course an effective method but we don't really need that on this case.
$endgroup$
– gimusi
Dec 4 '18 at 10:08












$begingroup$
How can be your result $infty$ from here $$lim_{n→∞}frac{n^3(n^2+sqrt{n^4+1}-2n^2)}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}$$ Note that at the numerator we have an indeterminate form $infty-infty$ therefore you can't conclude form here.
$endgroup$
– gimusi
Dec 4 '18 at 10:14






$begingroup$
How can be your result $infty$ from here $$lim_{n→∞}frac{n^3(n^2+sqrt{n^4+1}-2n^2)}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}$$ Note that at the numerator we have an indeterminate form $infty-infty$ therefore you can't conclude form here.
$endgroup$
– gimusi
Dec 4 '18 at 10:14














$begingroup$
@gimusi. Using the binomial expansion is a good way if you want to know how is approached the limit (this is just building the Taylor expansion). I hope and wish that we could have a discussion about this kind of problems one of these days. Cheers.
$endgroup$
– Claude Leibovici
Dec 4 '18 at 10:37






$begingroup$
@gimusi. Using the binomial expansion is a good way if you want to know how is approached the limit (this is just building the Taylor expansion). I hope and wish that we could have a discussion about this kind of problems one of these days. Cheers.
$endgroup$
– Claude Leibovici
Dec 4 '18 at 10:37














$begingroup$
@ClaudeLeibovici Yes of course, I also like Taylor's very much but in that case I think that it suffices operate by agebraic methods also I think thet the OP is looking for such kind of solution. Anyway the suggestion is good as an alternative approach! Cheers
$endgroup$
– gimusi
Dec 4 '18 at 10:39




$begingroup$
@ClaudeLeibovici Yes of course, I also like Taylor's very much but in that case I think that it suffices operate by agebraic methods also I think thet the OP is looking for such kind of solution. Anyway the suggestion is good as an alternative approach! Cheers
$endgroup$
– gimusi
Dec 4 '18 at 10:39










5 Answers
5






active

oldest

votes


















4












$begingroup$

HINT



You only need to apply the trick twice, indeed we have that



$$sqrt{n^2+sqrt{n^4+1}}-nsqrt{2}=(sqrt{n^2+sqrt{n^4+1}}-nsqrt{2})cdotfrac{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}=$$$$=frac{n^2+sqrt{n^4+1}-2n^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}$$



and



$$frac{sqrt{n^4+1}-n^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}=frac{sqrt{n^4+1}-n^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}cdot frac{sqrt{n^4+1}+n^2}{sqrt{n^4+1}+n^2}=$$$$=frac{1}{(sqrt{n^2+sqrt{n^4+1}}+nsqrt{2})(sqrt{n^4+1}+n^2)}$$



Can you conclude form here?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yes, thanks i understand !
    $endgroup$
    – mona1lisa
    Dec 4 '18 at 19:13










  • $begingroup$
    That’s nice! Well done, you are welcome! Bye
    $endgroup$
    – gimusi
    Dec 4 '18 at 19:17



















2












$begingroup$

Let $1/n=h$



$$lim_{hto0^+}dfrac{sqrt{1+sqrt{1+h^4}}-sqrt2}{h^4}$$



$$=lim_{hto0^+}dfrac{1+sqrt{1+h^4}-2}{h^4}cdotlim_{hto0^+}dfrac1{sqrt{1+sqrt{1+h^4}}+sqrt2}$$



$$=lim_{hto0^+}dfrac{1+h^4-1}{h^4}cdotlim_{hto0^+}dfrac1{sqrt{1+h^4}+1}cdotlim_{hto0^+}dfrac1{sqrt{1+sqrt{1+h^4}}+sqrt2}$$



$$=dfrac1{(sqrt1+1)(sqrt{1+sqrt1}+sqrt2)}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Nice way to simplify the manipulation. (+1)
    $endgroup$
    – gimusi
    Dec 4 '18 at 10:12










  • $begingroup$
    Sorry, but the final result is wrong
    $endgroup$
    – egreg
    Dec 4 '18 at 10:28










  • $begingroup$
    @egreg, Thanks for your feedback. Rectfied.
    $endgroup$
    – lab bhattacharjee
    Dec 4 '18 at 10:33










  • $begingroup$
    Related : math.stackexchange.com/questions/524288/…
    $endgroup$
    – lab bhattacharjee
    Dec 7 '18 at 5:10



















1












$begingroup$

The expedite way:



$$sqrt{1+sqrt{1+n^{-4}}}=sqrt{1+1+dfrac12n^{-4}+o(n^{-4})}=sqrt2sqrt{1+dfrac14n^{-4}+o(n^{-4})}=sqrt2left(1+dfrac18n^{-4}+o(n^{-4})right)$$



and the limit is



$$frac{sqrt2}8.$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    First replace $1/x=h$ to find



    $$L=lim_{n→∞}n^3(sqrt{n^2+sqrt{n^4+1}}-nsqrt2)=lim_{hto0^+}dfrac{sqrt{1+sqrt{1+h^4}}-sqrt2}{h^4}$$



    Let $sqrt{1+sqrt{1+h^4}}=y+sqrt2implies1+sqrt{1+h^4}=(sqrt2+y)^2=2+2sqrt2y+y^2$



    $$1+h^4=(1+y(2sqrt2+y))^2=1+2y(2sqrt2+y)+y^2(2sqrt2+y)^2$$



    $$L=lim_{yto0}dfrac{1+2y(2sqrt2+y)+y^2(2sqrt2+y)^2-1}y=?$$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Formally substitute $n=1/t$; if the function you get has a limit for $tto0^+$, then it is the same as the limit you are looking for. So consider
      $$
      lim_{tto0^+}frac{1}{t^3}left(
      sqrt{frac{1}{t^2}+sqrt{frac{1}{t^4}+1}}-frac{sqrt{2}}{t}
      right)=
      lim_{tto0^+}frac{sqrt{1+sqrt{1+t^4}}-sqrt{2}}{t^4}
      $$
      Now the dependency is only on $t^4$, so the limit is the same as
      $$
      lim_{uto0^+}frac{sqrt{1+sqrt{1+u}}-sqrt{2}}{u}
      $$
      which is the derivative at $0$ of $f(u)=sqrt{1+sqrt{1+u}}$.




      Since $$f'(u)=frac{1}{2sqrt{1+sqrt{1+u}}}frac{1}{2sqrt{1+u}}$$ we have $$f'(0)=frac{1}{4sqrt{2}}$$







      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        As always I don't agree that you give a full answer. In my opinion it is not useful for the asker nor for the site. I don't downvote because it is of course a correct and good answer. But virtually I give you a "-1".
        $endgroup$
        – gimusi
        Dec 4 '18 at 10:31













      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%2f3025375%2ffinding-lim-limits-n%25e2%2586%2592%25e2%2588%259en3-sqrtn2-sqrtn41-n-sqrt2%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      5 Answers
      5






      active

      oldest

      votes








      5 Answers
      5






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      HINT



      You only need to apply the trick twice, indeed we have that



      $$sqrt{n^2+sqrt{n^4+1}}-nsqrt{2}=(sqrt{n^2+sqrt{n^4+1}}-nsqrt{2})cdotfrac{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}=$$$$=frac{n^2+sqrt{n^4+1}-2n^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}$$



      and



      $$frac{sqrt{n^4+1}-n^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}=frac{sqrt{n^4+1}-n^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}cdot frac{sqrt{n^4+1}+n^2}{sqrt{n^4+1}+n^2}=$$$$=frac{1}{(sqrt{n^2+sqrt{n^4+1}}+nsqrt{2})(sqrt{n^4+1}+n^2)}$$



      Can you conclude form here?






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Yes, thanks i understand !
        $endgroup$
        – mona1lisa
        Dec 4 '18 at 19:13










      • $begingroup$
        That’s nice! Well done, you are welcome! Bye
        $endgroup$
        – gimusi
        Dec 4 '18 at 19:17
















      4












      $begingroup$

      HINT



      You only need to apply the trick twice, indeed we have that



      $$sqrt{n^2+sqrt{n^4+1}}-nsqrt{2}=(sqrt{n^2+sqrt{n^4+1}}-nsqrt{2})cdotfrac{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}=$$$$=frac{n^2+sqrt{n^4+1}-2n^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}$$



      and



      $$frac{sqrt{n^4+1}-n^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}=frac{sqrt{n^4+1}-n^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}cdot frac{sqrt{n^4+1}+n^2}{sqrt{n^4+1}+n^2}=$$$$=frac{1}{(sqrt{n^2+sqrt{n^4+1}}+nsqrt{2})(sqrt{n^4+1}+n^2)}$$



      Can you conclude form here?






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Yes, thanks i understand !
        $endgroup$
        – mona1lisa
        Dec 4 '18 at 19:13










      • $begingroup$
        That’s nice! Well done, you are welcome! Bye
        $endgroup$
        – gimusi
        Dec 4 '18 at 19:17














      4












      4








      4





      $begingroup$

      HINT



      You only need to apply the trick twice, indeed we have that



      $$sqrt{n^2+sqrt{n^4+1}}-nsqrt{2}=(sqrt{n^2+sqrt{n^4+1}}-nsqrt{2})cdotfrac{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}=$$$$=frac{n^2+sqrt{n^4+1}-2n^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}$$



      and



      $$frac{sqrt{n^4+1}-n^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}=frac{sqrt{n^4+1}-n^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}cdot frac{sqrt{n^4+1}+n^2}{sqrt{n^4+1}+n^2}=$$$$=frac{1}{(sqrt{n^2+sqrt{n^4+1}}+nsqrt{2})(sqrt{n^4+1}+n^2)}$$



      Can you conclude form here?






      share|cite|improve this answer











      $endgroup$



      HINT



      You only need to apply the trick twice, indeed we have that



      $$sqrt{n^2+sqrt{n^4+1}}-nsqrt{2}=(sqrt{n^2+sqrt{n^4+1}}-nsqrt{2})cdotfrac{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}=$$$$=frac{n^2+sqrt{n^4+1}-2n^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}$$



      and



      $$frac{sqrt{n^4+1}-n^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}=frac{sqrt{n^4+1}-n^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}cdot frac{sqrt{n^4+1}+n^2}{sqrt{n^4+1}+n^2}=$$$$=frac{1}{(sqrt{n^2+sqrt{n^4+1}}+nsqrt{2})(sqrt{n^4+1}+n^2)}$$



      Can you conclude form here?







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 4 '18 at 10:10

























      answered Dec 4 '18 at 10:05









      gimusigimusi

      92.9k84494




      92.9k84494












      • $begingroup$
        Yes, thanks i understand !
        $endgroup$
        – mona1lisa
        Dec 4 '18 at 19:13










      • $begingroup$
        That’s nice! Well done, you are welcome! Bye
        $endgroup$
        – gimusi
        Dec 4 '18 at 19:17


















      • $begingroup$
        Yes, thanks i understand !
        $endgroup$
        – mona1lisa
        Dec 4 '18 at 19:13










      • $begingroup$
        That’s nice! Well done, you are welcome! Bye
        $endgroup$
        – gimusi
        Dec 4 '18 at 19:17
















      $begingroup$
      Yes, thanks i understand !
      $endgroup$
      – mona1lisa
      Dec 4 '18 at 19:13




      $begingroup$
      Yes, thanks i understand !
      $endgroup$
      – mona1lisa
      Dec 4 '18 at 19:13












      $begingroup$
      That’s nice! Well done, you are welcome! Bye
      $endgroup$
      – gimusi
      Dec 4 '18 at 19:17




      $begingroup$
      That’s nice! Well done, you are welcome! Bye
      $endgroup$
      – gimusi
      Dec 4 '18 at 19:17











      2












      $begingroup$

      Let $1/n=h$



      $$lim_{hto0^+}dfrac{sqrt{1+sqrt{1+h^4}}-sqrt2}{h^4}$$



      $$=lim_{hto0^+}dfrac{1+sqrt{1+h^4}-2}{h^4}cdotlim_{hto0^+}dfrac1{sqrt{1+sqrt{1+h^4}}+sqrt2}$$



      $$=lim_{hto0^+}dfrac{1+h^4-1}{h^4}cdotlim_{hto0^+}dfrac1{sqrt{1+h^4}+1}cdotlim_{hto0^+}dfrac1{sqrt{1+sqrt{1+h^4}}+sqrt2}$$



      $$=dfrac1{(sqrt1+1)(sqrt{1+sqrt1}+sqrt2)}$$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Nice way to simplify the manipulation. (+1)
        $endgroup$
        – gimusi
        Dec 4 '18 at 10:12










      • $begingroup$
        Sorry, but the final result is wrong
        $endgroup$
        – egreg
        Dec 4 '18 at 10:28










      • $begingroup$
        @egreg, Thanks for your feedback. Rectfied.
        $endgroup$
        – lab bhattacharjee
        Dec 4 '18 at 10:33










      • $begingroup$
        Related : math.stackexchange.com/questions/524288/…
        $endgroup$
        – lab bhattacharjee
        Dec 7 '18 at 5:10
















      2












      $begingroup$

      Let $1/n=h$



      $$lim_{hto0^+}dfrac{sqrt{1+sqrt{1+h^4}}-sqrt2}{h^4}$$



      $$=lim_{hto0^+}dfrac{1+sqrt{1+h^4}-2}{h^4}cdotlim_{hto0^+}dfrac1{sqrt{1+sqrt{1+h^4}}+sqrt2}$$



      $$=lim_{hto0^+}dfrac{1+h^4-1}{h^4}cdotlim_{hto0^+}dfrac1{sqrt{1+h^4}+1}cdotlim_{hto0^+}dfrac1{sqrt{1+sqrt{1+h^4}}+sqrt2}$$



      $$=dfrac1{(sqrt1+1)(sqrt{1+sqrt1}+sqrt2)}$$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Nice way to simplify the manipulation. (+1)
        $endgroup$
        – gimusi
        Dec 4 '18 at 10:12










      • $begingroup$
        Sorry, but the final result is wrong
        $endgroup$
        – egreg
        Dec 4 '18 at 10:28










      • $begingroup$
        @egreg, Thanks for your feedback. Rectfied.
        $endgroup$
        – lab bhattacharjee
        Dec 4 '18 at 10:33










      • $begingroup$
        Related : math.stackexchange.com/questions/524288/…
        $endgroup$
        – lab bhattacharjee
        Dec 7 '18 at 5:10














      2












      2








      2





      $begingroup$

      Let $1/n=h$



      $$lim_{hto0^+}dfrac{sqrt{1+sqrt{1+h^4}}-sqrt2}{h^4}$$



      $$=lim_{hto0^+}dfrac{1+sqrt{1+h^4}-2}{h^4}cdotlim_{hto0^+}dfrac1{sqrt{1+sqrt{1+h^4}}+sqrt2}$$



      $$=lim_{hto0^+}dfrac{1+h^4-1}{h^4}cdotlim_{hto0^+}dfrac1{sqrt{1+h^4}+1}cdotlim_{hto0^+}dfrac1{sqrt{1+sqrt{1+h^4}}+sqrt2}$$



      $$=dfrac1{(sqrt1+1)(sqrt{1+sqrt1}+sqrt2)}$$






      share|cite|improve this answer











      $endgroup$



      Let $1/n=h$



      $$lim_{hto0^+}dfrac{sqrt{1+sqrt{1+h^4}}-sqrt2}{h^4}$$



      $$=lim_{hto0^+}dfrac{1+sqrt{1+h^4}-2}{h^4}cdotlim_{hto0^+}dfrac1{sqrt{1+sqrt{1+h^4}}+sqrt2}$$



      $$=lim_{hto0^+}dfrac{1+h^4-1}{h^4}cdotlim_{hto0^+}dfrac1{sqrt{1+h^4}+1}cdotlim_{hto0^+}dfrac1{sqrt{1+sqrt{1+h^4}}+sqrt2}$$



      $$=dfrac1{(sqrt1+1)(sqrt{1+sqrt1}+sqrt2)}$$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 4 '18 at 10:33

























      answered Dec 4 '18 at 10:08









      lab bhattacharjeelab bhattacharjee

      226k15157275




      226k15157275












      • $begingroup$
        Nice way to simplify the manipulation. (+1)
        $endgroup$
        – gimusi
        Dec 4 '18 at 10:12










      • $begingroup$
        Sorry, but the final result is wrong
        $endgroup$
        – egreg
        Dec 4 '18 at 10:28










      • $begingroup$
        @egreg, Thanks for your feedback. Rectfied.
        $endgroup$
        – lab bhattacharjee
        Dec 4 '18 at 10:33










      • $begingroup$
        Related : math.stackexchange.com/questions/524288/…
        $endgroup$
        – lab bhattacharjee
        Dec 7 '18 at 5:10


















      • $begingroup$
        Nice way to simplify the manipulation. (+1)
        $endgroup$
        – gimusi
        Dec 4 '18 at 10:12










      • $begingroup$
        Sorry, but the final result is wrong
        $endgroup$
        – egreg
        Dec 4 '18 at 10:28










      • $begingroup$
        @egreg, Thanks for your feedback. Rectfied.
        $endgroup$
        – lab bhattacharjee
        Dec 4 '18 at 10:33










      • $begingroup$
        Related : math.stackexchange.com/questions/524288/…
        $endgroup$
        – lab bhattacharjee
        Dec 7 '18 at 5:10
















      $begingroup$
      Nice way to simplify the manipulation. (+1)
      $endgroup$
      – gimusi
      Dec 4 '18 at 10:12




      $begingroup$
      Nice way to simplify the manipulation. (+1)
      $endgroup$
      – gimusi
      Dec 4 '18 at 10:12












      $begingroup$
      Sorry, but the final result is wrong
      $endgroup$
      – egreg
      Dec 4 '18 at 10:28




      $begingroup$
      Sorry, but the final result is wrong
      $endgroup$
      – egreg
      Dec 4 '18 at 10:28












      $begingroup$
      @egreg, Thanks for your feedback. Rectfied.
      $endgroup$
      – lab bhattacharjee
      Dec 4 '18 at 10:33




      $begingroup$
      @egreg, Thanks for your feedback. Rectfied.
      $endgroup$
      – lab bhattacharjee
      Dec 4 '18 at 10:33












      $begingroup$
      Related : math.stackexchange.com/questions/524288/…
      $endgroup$
      – lab bhattacharjee
      Dec 7 '18 at 5:10




      $begingroup$
      Related : math.stackexchange.com/questions/524288/…
      $endgroup$
      – lab bhattacharjee
      Dec 7 '18 at 5:10











      1












      $begingroup$

      The expedite way:



      $$sqrt{1+sqrt{1+n^{-4}}}=sqrt{1+1+dfrac12n^{-4}+o(n^{-4})}=sqrt2sqrt{1+dfrac14n^{-4}+o(n^{-4})}=sqrt2left(1+dfrac18n^{-4}+o(n^{-4})right)$$



      and the limit is



      $$frac{sqrt2}8.$$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        The expedite way:



        $$sqrt{1+sqrt{1+n^{-4}}}=sqrt{1+1+dfrac12n^{-4}+o(n^{-4})}=sqrt2sqrt{1+dfrac14n^{-4}+o(n^{-4})}=sqrt2left(1+dfrac18n^{-4}+o(n^{-4})right)$$



        and the limit is



        $$frac{sqrt2}8.$$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          The expedite way:



          $$sqrt{1+sqrt{1+n^{-4}}}=sqrt{1+1+dfrac12n^{-4}+o(n^{-4})}=sqrt2sqrt{1+dfrac14n^{-4}+o(n^{-4})}=sqrt2left(1+dfrac18n^{-4}+o(n^{-4})right)$$



          and the limit is



          $$frac{sqrt2}8.$$






          share|cite|improve this answer









          $endgroup$



          The expedite way:



          $$sqrt{1+sqrt{1+n^{-4}}}=sqrt{1+1+dfrac12n^{-4}+o(n^{-4})}=sqrt2sqrt{1+dfrac14n^{-4}+o(n^{-4})}=sqrt2left(1+dfrac18n^{-4}+o(n^{-4})right)$$



          and the limit is



          $$frac{sqrt2}8.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 4 '18 at 10:31









          Yves DaoustYves Daoust

          129k675227




          129k675227























              0












              $begingroup$

              First replace $1/x=h$ to find



              $$L=lim_{n→∞}n^3(sqrt{n^2+sqrt{n^4+1}}-nsqrt2)=lim_{hto0^+}dfrac{sqrt{1+sqrt{1+h^4}}-sqrt2}{h^4}$$



              Let $sqrt{1+sqrt{1+h^4}}=y+sqrt2implies1+sqrt{1+h^4}=(sqrt2+y)^2=2+2sqrt2y+y^2$



              $$1+h^4=(1+y(2sqrt2+y))^2=1+2y(2sqrt2+y)+y^2(2sqrt2+y)^2$$



              $$L=lim_{yto0}dfrac{1+2y(2sqrt2+y)+y^2(2sqrt2+y)^2-1}y=?$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                First replace $1/x=h$ to find



                $$L=lim_{n→∞}n^3(sqrt{n^2+sqrt{n^4+1}}-nsqrt2)=lim_{hto0^+}dfrac{sqrt{1+sqrt{1+h^4}}-sqrt2}{h^4}$$



                Let $sqrt{1+sqrt{1+h^4}}=y+sqrt2implies1+sqrt{1+h^4}=(sqrt2+y)^2=2+2sqrt2y+y^2$



                $$1+h^4=(1+y(2sqrt2+y))^2=1+2y(2sqrt2+y)+y^2(2sqrt2+y)^2$$



                $$L=lim_{yto0}dfrac{1+2y(2sqrt2+y)+y^2(2sqrt2+y)^2-1}y=?$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  First replace $1/x=h$ to find



                  $$L=lim_{n→∞}n^3(sqrt{n^2+sqrt{n^4+1}}-nsqrt2)=lim_{hto0^+}dfrac{sqrt{1+sqrt{1+h^4}}-sqrt2}{h^4}$$



                  Let $sqrt{1+sqrt{1+h^4}}=y+sqrt2implies1+sqrt{1+h^4}=(sqrt2+y)^2=2+2sqrt2y+y^2$



                  $$1+h^4=(1+y(2sqrt2+y))^2=1+2y(2sqrt2+y)+y^2(2sqrt2+y)^2$$



                  $$L=lim_{yto0}dfrac{1+2y(2sqrt2+y)+y^2(2sqrt2+y)^2-1}y=?$$






                  share|cite|improve this answer









                  $endgroup$



                  First replace $1/x=h$ to find



                  $$L=lim_{n→∞}n^3(sqrt{n^2+sqrt{n^4+1}}-nsqrt2)=lim_{hto0^+}dfrac{sqrt{1+sqrt{1+h^4}}-sqrt2}{h^4}$$



                  Let $sqrt{1+sqrt{1+h^4}}=y+sqrt2implies1+sqrt{1+h^4}=(sqrt2+y)^2=2+2sqrt2y+y^2$



                  $$1+h^4=(1+y(2sqrt2+y))^2=1+2y(2sqrt2+y)+y^2(2sqrt2+y)^2$$



                  $$L=lim_{yto0}dfrac{1+2y(2sqrt2+y)+y^2(2sqrt2+y)^2-1}y=?$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 4 '18 at 10:16









                  lab bhattacharjeelab bhattacharjee

                  226k15157275




                  226k15157275























                      0












                      $begingroup$

                      Formally substitute $n=1/t$; if the function you get has a limit for $tto0^+$, then it is the same as the limit you are looking for. So consider
                      $$
                      lim_{tto0^+}frac{1}{t^3}left(
                      sqrt{frac{1}{t^2}+sqrt{frac{1}{t^4}+1}}-frac{sqrt{2}}{t}
                      right)=
                      lim_{tto0^+}frac{sqrt{1+sqrt{1+t^4}}-sqrt{2}}{t^4}
                      $$
                      Now the dependency is only on $t^4$, so the limit is the same as
                      $$
                      lim_{uto0^+}frac{sqrt{1+sqrt{1+u}}-sqrt{2}}{u}
                      $$
                      which is the derivative at $0$ of $f(u)=sqrt{1+sqrt{1+u}}$.




                      Since $$f'(u)=frac{1}{2sqrt{1+sqrt{1+u}}}frac{1}{2sqrt{1+u}}$$ we have $$f'(0)=frac{1}{4sqrt{2}}$$







                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        As always I don't agree that you give a full answer. In my opinion it is not useful for the asker nor for the site. I don't downvote because it is of course a correct and good answer. But virtually I give you a "-1".
                        $endgroup$
                        – gimusi
                        Dec 4 '18 at 10:31


















                      0












                      $begingroup$

                      Formally substitute $n=1/t$; if the function you get has a limit for $tto0^+$, then it is the same as the limit you are looking for. So consider
                      $$
                      lim_{tto0^+}frac{1}{t^3}left(
                      sqrt{frac{1}{t^2}+sqrt{frac{1}{t^4}+1}}-frac{sqrt{2}}{t}
                      right)=
                      lim_{tto0^+}frac{sqrt{1+sqrt{1+t^4}}-sqrt{2}}{t^4}
                      $$
                      Now the dependency is only on $t^4$, so the limit is the same as
                      $$
                      lim_{uto0^+}frac{sqrt{1+sqrt{1+u}}-sqrt{2}}{u}
                      $$
                      which is the derivative at $0$ of $f(u)=sqrt{1+sqrt{1+u}}$.




                      Since $$f'(u)=frac{1}{2sqrt{1+sqrt{1+u}}}frac{1}{2sqrt{1+u}}$$ we have $$f'(0)=frac{1}{4sqrt{2}}$$







                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        As always I don't agree that you give a full answer. In my opinion it is not useful for the asker nor for the site. I don't downvote because it is of course a correct and good answer. But virtually I give you a "-1".
                        $endgroup$
                        – gimusi
                        Dec 4 '18 at 10:31
















                      0












                      0








                      0





                      $begingroup$

                      Formally substitute $n=1/t$; if the function you get has a limit for $tto0^+$, then it is the same as the limit you are looking for. So consider
                      $$
                      lim_{tto0^+}frac{1}{t^3}left(
                      sqrt{frac{1}{t^2}+sqrt{frac{1}{t^4}+1}}-frac{sqrt{2}}{t}
                      right)=
                      lim_{tto0^+}frac{sqrt{1+sqrt{1+t^4}}-sqrt{2}}{t^4}
                      $$
                      Now the dependency is only on $t^4$, so the limit is the same as
                      $$
                      lim_{uto0^+}frac{sqrt{1+sqrt{1+u}}-sqrt{2}}{u}
                      $$
                      which is the derivative at $0$ of $f(u)=sqrt{1+sqrt{1+u}}$.




                      Since $$f'(u)=frac{1}{2sqrt{1+sqrt{1+u}}}frac{1}{2sqrt{1+u}}$$ we have $$f'(0)=frac{1}{4sqrt{2}}$$







                      share|cite|improve this answer











                      $endgroup$



                      Formally substitute $n=1/t$; if the function you get has a limit for $tto0^+$, then it is the same as the limit you are looking for. So consider
                      $$
                      lim_{tto0^+}frac{1}{t^3}left(
                      sqrt{frac{1}{t^2}+sqrt{frac{1}{t^4}+1}}-frac{sqrt{2}}{t}
                      right)=
                      lim_{tto0^+}frac{sqrt{1+sqrt{1+t^4}}-sqrt{2}}{t^4}
                      $$
                      Now the dependency is only on $t^4$, so the limit is the same as
                      $$
                      lim_{uto0^+}frac{sqrt{1+sqrt{1+u}}-sqrt{2}}{u}
                      $$
                      which is the derivative at $0$ of $f(u)=sqrt{1+sqrt{1+u}}$.




                      Since $$f'(u)=frac{1}{2sqrt{1+sqrt{1+u}}}frac{1}{2sqrt{1+u}}$$ we have $$f'(0)=frac{1}{4sqrt{2}}$$








                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Dec 4 '18 at 12:36

























                      answered Dec 4 '18 at 10:28









                      egregegreg

                      183k1486204




                      183k1486204












                      • $begingroup$
                        As always I don't agree that you give a full answer. In my opinion it is not useful for the asker nor for the site. I don't downvote because it is of course a correct and good answer. But virtually I give you a "-1".
                        $endgroup$
                        – gimusi
                        Dec 4 '18 at 10:31




















                      • $begingroup$
                        As always I don't agree that you give a full answer. In my opinion it is not useful for the asker nor for the site. I don't downvote because it is of course a correct and good answer. But virtually I give you a "-1".
                        $endgroup$
                        – gimusi
                        Dec 4 '18 at 10:31


















                      $begingroup$
                      As always I don't agree that you give a full answer. In my opinion it is not useful for the asker nor for the site. I don't downvote because it is of course a correct and good answer. But virtually I give you a "-1".
                      $endgroup$
                      – gimusi
                      Dec 4 '18 at 10:31






                      $begingroup$
                      As always I don't agree that you give a full answer. In my opinion it is not useful for the asker nor for the site. I don't downvote because it is of course a correct and good answer. But virtually I give you a "-1".
                      $endgroup$
                      – gimusi
                      Dec 4 '18 at 10:31




















                      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%2f3025375%2ffinding-lim-limits-n%25e2%2586%2592%25e2%2588%259en3-sqrtn2-sqrtn41-n-sqrt2%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

                      mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924

                      Image
                      0 I just uploaded a new plugin to my wordpress site and I got this warning: mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924 Usually if I have a problem, I can find answers on google but this one seems pretty particular and I'm not finding any info on it except for this forum. I'm not terribly techie but if someone can tell me how to read this and how to change it I can follow directions pretty well. Thanks All! PS If I don't have the proper tags here could someone add the appropriate tags? php mysql wordpress share | improve this question asked Nov 16 '18 at 20:41 ...
                      Read more

                      How to change which sound is reproduced for terminal bell?

                      Image
                      3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
                      Read more

                      Can I use Tabulator js library in my java Spring + Thymeleaf project?

                      Image
                      0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
                      Read more