Find limit $limlimits _{xrightarrow infty }left(sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} -sqrt{2x^{4}}right)$












2












$begingroup$


$displaystyle limlimits _{xrightarrow infty }left(sqrt{x^{4} +x^{2}sqrt{x^{4} +1}}displaystyle -sqrt{2x^{4}}right)$



$displaystyle limlimits _{xrightarrow infty }left(sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} -sqrt{2x^{4}}right)$$displaystyle =displaystyle limlimits _{xrightarrow infty }dfrac{left(sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} -sqrt{2x^{4}}right)left(sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4} }right)}{sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4} }} =$



$displaystyle =limlimits _{xrightarrow infty }dfrac{x^{2}sqrt{x^{4} +1} -x^{4}}{sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4} }}$



What is the next step should be? Please help!






real-analysis limits radicals limits-without-lhopital







share|cite|improve this question











$endgroup$












  • $begingroup$
    See math.stackexchange.com/questions/3025375/…
    $endgroup$
    – lab bhattacharjee
    Dec 8 '18 at 9:47
















2












$begingroup$


$displaystyle limlimits _{xrightarrow infty }left(sqrt{x^{4} +x^{2}sqrt{x^{4} +1}}displaystyle -sqrt{2x^{4}}right)$



$displaystyle limlimits _{xrightarrow infty }left(sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} -sqrt{2x^{4}}right)$$displaystyle =displaystyle limlimits _{xrightarrow infty }dfrac{left(sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} -sqrt{2x^{4}}right)left(sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4} }right)}{sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4} }} =$



$displaystyle =limlimits _{xrightarrow infty }dfrac{x^{2}sqrt{x^{4} +1} -x^{4}}{sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4} }}$



What is the next step should be? Please help!






real-analysis limits radicals limits-without-lhopital







share|cite|improve this question











$endgroup$












  • $begingroup$
    See math.stackexchange.com/questions/3025375/…
    $endgroup$
    – lab bhattacharjee
    Dec 8 '18 at 9:47














2












2








2





$begingroup$


$displaystyle limlimits _{xrightarrow infty }left(sqrt{x^{4} +x^{2}sqrt{x^{4} +1}}displaystyle -sqrt{2x^{4}}right)$



$displaystyle limlimits _{xrightarrow infty }left(sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} -sqrt{2x^{4}}right)$$displaystyle =displaystyle limlimits _{xrightarrow infty }dfrac{left(sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} -sqrt{2x^{4}}right)left(sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4} }right)}{sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4} }} =$



$displaystyle =limlimits _{xrightarrow infty }dfrac{x^{2}sqrt{x^{4} +1} -x^{4}}{sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4} }}$



What is the next step should be? Please help!






real-analysis limits radicals limits-without-lhopital







share|cite|improve this question











$endgroup$




$displaystyle limlimits _{xrightarrow infty }left(sqrt{x^{4} +x^{2}sqrt{x^{4} +1}}displaystyle -sqrt{2x^{4}}right)$



$displaystyle limlimits _{xrightarrow infty }left(sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} -sqrt{2x^{4}}right)$$displaystyle =displaystyle limlimits _{xrightarrow infty }dfrac{left(sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} -sqrt{2x^{4}}right)left(sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4} }right)}{sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4} }} =$



$displaystyle =limlimits _{xrightarrow infty }dfrac{x^{2}sqrt{x^{4} +1} -x^{4}}{sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4} }}$



What is the next step should be? Please help!






real-analysis limits radicals limits-without-lhopital




real-analysis limits radicals limits-without-lhopital






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 13:43









Martin Sleziak

44.8k10118272




44.8k10118272










asked Dec 8 '18 at 9:43









Angella Angella

111




111












  • $begingroup$
    See math.stackexchange.com/questions/3025375/…
    $endgroup$
    – lab bhattacharjee
    Dec 8 '18 at 9:47


















  • $begingroup$
    See math.stackexchange.com/questions/3025375/…
    $endgroup$
    – lab bhattacharjee
    Dec 8 '18 at 9:47
















$begingroup$
See math.stackexchange.com/questions/3025375/…
$endgroup$
– lab bhattacharjee
Dec 8 '18 at 9:47




$begingroup$
See math.stackexchange.com/questions/3025375/…
$endgroup$
– lab bhattacharjee
Dec 8 '18 at 9:47










4 Answers
4






active

oldest

votes


















6












$begingroup$

Now, use that
$$sqrt{x^4+1}-x^2=frac{1}{sqrt{x^4+1}+x^2}.$$






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    Use the trick twice for the numerator to obtain



    $$dfrac{x^{2}sqrt{x^{4} +1} -x^{4}}{sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4} }}=dfrac{x^{2}sqrt{x^{4} +1} -x^{4}}{sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4} }}cdot dfrac{x^{2}sqrt{x^{4} +1} +x^{4}}{x^{2}sqrt{x^{4} +1} +x^{4}}=$$



    $$=dfrac{x^4}{left(sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4} }right)left(x^{2}sqrt{x^{4} +1} +x^{4}right)}simfrac{1}{4sqrt 2 x^2}to 0$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      @user376343 Opsss yes of course I fix the typo! Thanks
      $endgroup$
      – gimusi
      Dec 8 '18 at 11:07



















    2












    $begingroup$

    $$
    sqrt{x^{4} +x^{2}sqrt{x^{4} +1}}-sqrt{2x^4}=
    dfrac{x^{2}sqrt{x^{4} +1} -x^{4}}{sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4}}}=
    dfrac{x^{2}sqrt{x^{4} +1} -x^{4}}{sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4} }}cdotfrac{x^{2}sqrt{x^{4} +1} +x^{4}}{x^{2}sqrt{x^{4} +1} +x^{4}}\=frac{x^4}{big(sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4}}big)big(x^{2}sqrt{x^{4} +1}+x^4big)}\=frac{1}{big(sqrt{1 +sqrt{1+x^{-4}}} +sqrt{2}big)big(sqrt{x^{4}+1}+x^2big)},to,0
    $$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Are you sure, Yiorgos?
      $endgroup$
      – Michael Rozenberg
      Dec 8 '18 at 10:01










    • $begingroup$
      You have divided twice by $x^4$.
      $endgroup$
      – gimusi
      Dec 8 '18 at 10:05










    • $begingroup$
      @gimusi You are right! - Corrected it!
      $endgroup$
      – Yiorgos S. Smyrlis
      Dec 8 '18 at 10:07



















    1












    $begingroup$

    Another approach is using substitution $x=cot t$:
    begin{align}
    lim_{xtoinfty}left(sqrt{x^{4} +x^{2}sqrt{x^{4} +1}}-sqrt{2x^{4}}right)
    &=lim_{tto0}dfrac{sqrt{cos^2t+cos t}-sqrt{2cos^2t}}{sin t} \
    &=lim_{tto0}dfrac{cos t(1-cos t)}{sin t(sqrt{cos^2t+cos t}+sqrt{2cos^2t})} \
    &=lim_{tto0}dfrac{cos t}{sqrt{cos^2t+cos t}+sqrt{2cos^2t}}dfrac{2sin^2frac{t}{2}}{frac{t^2}{2}}dfrac{t}{sin t}dfrac{t}{2} \
    &=color{blue}{0}
    end{align}






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Could you precise? How is $x=cos t$ with $tto 0$ if $xto infty$?
      $endgroup$
      – user376343
      Dec 8 '18 at 10:59










    • $begingroup$
      @user376343 It's impossible when $xtoinfty$ then $tto0$ with $x=cos t$.
      $endgroup$
      – Nosrati
      Dec 8 '18 at 11:03










    • $begingroup$
      When $t to 0$ then $cos t to 1$ therefore I find you substitution mysterious...
      $endgroup$
      – user376343
      Dec 8 '18 at 11:17










    • $begingroup$
      Of course it is :)
      $endgroup$
      – Nosrati
      Dec 8 '18 at 11:20










    • $begingroup$
      @user376343, a letter makes a difference here: $cos t$ and $cot t$.
      $endgroup$
      – farruhota
      Dec 8 '18 at 11:35











    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%2f3030896%2ffind-limit-lim-limits-x-rightarrow-infty-left-sqrtx4-x2-sqrtx%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6












    $begingroup$

    Now, use that
    $$sqrt{x^4+1}-x^2=frac{1}{sqrt{x^4+1}+x^2}.$$






    share|cite|improve this answer









    $endgroup$


















      6












      $begingroup$

      Now, use that
      $$sqrt{x^4+1}-x^2=frac{1}{sqrt{x^4+1}+x^2}.$$






      share|cite|improve this answer









      $endgroup$
















        6












        6








        6





        $begingroup$

        Now, use that
        $$sqrt{x^4+1}-x^2=frac{1}{sqrt{x^4+1}+x^2}.$$






        share|cite|improve this answer









        $endgroup$



        Now, use that
        $$sqrt{x^4+1}-x^2=frac{1}{sqrt{x^4+1}+x^2}.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 8 '18 at 9:44









        Michael RozenbergMichael Rozenberg

        103k1891195




        103k1891195























            3












            $begingroup$

            Use the trick twice for the numerator to obtain



            $$dfrac{x^{2}sqrt{x^{4} +1} -x^{4}}{sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4} }}=dfrac{x^{2}sqrt{x^{4} +1} -x^{4}}{sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4} }}cdot dfrac{x^{2}sqrt{x^{4} +1} +x^{4}}{x^{2}sqrt{x^{4} +1} +x^{4}}=$$



            $$=dfrac{x^4}{left(sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4} }right)left(x^{2}sqrt{x^{4} +1} +x^{4}right)}simfrac{1}{4sqrt 2 x^2}to 0$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              @user376343 Opsss yes of course I fix the typo! Thanks
              $endgroup$
              – gimusi
              Dec 8 '18 at 11:07
















            3












            $begingroup$

            Use the trick twice for the numerator to obtain



            $$dfrac{x^{2}sqrt{x^{4} +1} -x^{4}}{sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4} }}=dfrac{x^{2}sqrt{x^{4} +1} -x^{4}}{sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4} }}cdot dfrac{x^{2}sqrt{x^{4} +1} +x^{4}}{x^{2}sqrt{x^{4} +1} +x^{4}}=$$



            $$=dfrac{x^4}{left(sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4} }right)left(x^{2}sqrt{x^{4} +1} +x^{4}right)}simfrac{1}{4sqrt 2 x^2}to 0$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              @user376343 Opsss yes of course I fix the typo! Thanks
              $endgroup$
              – gimusi
              Dec 8 '18 at 11:07














            3












            3








            3





            $begingroup$

            Use the trick twice for the numerator to obtain



            $$dfrac{x^{2}sqrt{x^{4} +1} -x^{4}}{sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4} }}=dfrac{x^{2}sqrt{x^{4} +1} -x^{4}}{sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4} }}cdot dfrac{x^{2}sqrt{x^{4} +1} +x^{4}}{x^{2}sqrt{x^{4} +1} +x^{4}}=$$



            $$=dfrac{x^4}{left(sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4} }right)left(x^{2}sqrt{x^{4} +1} +x^{4}right)}simfrac{1}{4sqrt 2 x^2}to 0$$






            share|cite|improve this answer











            $endgroup$



            Use the trick twice for the numerator to obtain



            $$dfrac{x^{2}sqrt{x^{4} +1} -x^{4}}{sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4} }}=dfrac{x^{2}sqrt{x^{4} +1} -x^{4}}{sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4} }}cdot dfrac{x^{2}sqrt{x^{4} +1} +x^{4}}{x^{2}sqrt{x^{4} +1} +x^{4}}=$$



            $$=dfrac{x^4}{left(sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4} }right)left(x^{2}sqrt{x^{4} +1} +x^{4}right)}simfrac{1}{4sqrt 2 x^2}to 0$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 8 '18 at 11:07

























            answered Dec 8 '18 at 9:45









            gimusigimusi

            92.8k84494




            92.8k84494












            • $begingroup$
              @user376343 Opsss yes of course I fix the typo! Thanks
              $endgroup$
              – gimusi
              Dec 8 '18 at 11:07


















            • $begingroup$
              @user376343 Opsss yes of course I fix the typo! Thanks
              $endgroup$
              – gimusi
              Dec 8 '18 at 11:07
















            $begingroup$
            @user376343 Opsss yes of course I fix the typo! Thanks
            $endgroup$
            – gimusi
            Dec 8 '18 at 11:07




            $begingroup$
            @user376343 Opsss yes of course I fix the typo! Thanks
            $endgroup$
            – gimusi
            Dec 8 '18 at 11:07











            2












            $begingroup$

            $$
            sqrt{x^{4} +x^{2}sqrt{x^{4} +1}}-sqrt{2x^4}=
            dfrac{x^{2}sqrt{x^{4} +1} -x^{4}}{sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4}}}=
            dfrac{x^{2}sqrt{x^{4} +1} -x^{4}}{sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4} }}cdotfrac{x^{2}sqrt{x^{4} +1} +x^{4}}{x^{2}sqrt{x^{4} +1} +x^{4}}\=frac{x^4}{big(sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4}}big)big(x^{2}sqrt{x^{4} +1}+x^4big)}\=frac{1}{big(sqrt{1 +sqrt{1+x^{-4}}} +sqrt{2}big)big(sqrt{x^{4}+1}+x^2big)},to,0
            $$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Are you sure, Yiorgos?
              $endgroup$
              – Michael Rozenberg
              Dec 8 '18 at 10:01










            • $begingroup$
              You have divided twice by $x^4$.
              $endgroup$
              – gimusi
              Dec 8 '18 at 10:05










            • $begingroup$
              @gimusi You are right! - Corrected it!
              $endgroup$
              – Yiorgos S. Smyrlis
              Dec 8 '18 at 10:07
















            2












            $begingroup$

            $$
            sqrt{x^{4} +x^{2}sqrt{x^{4} +1}}-sqrt{2x^4}=
            dfrac{x^{2}sqrt{x^{4} +1} -x^{4}}{sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4}}}=
            dfrac{x^{2}sqrt{x^{4} +1} -x^{4}}{sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4} }}cdotfrac{x^{2}sqrt{x^{4} +1} +x^{4}}{x^{2}sqrt{x^{4} +1} +x^{4}}\=frac{x^4}{big(sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4}}big)big(x^{2}sqrt{x^{4} +1}+x^4big)}\=frac{1}{big(sqrt{1 +sqrt{1+x^{-4}}} +sqrt{2}big)big(sqrt{x^{4}+1}+x^2big)},to,0
            $$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Are you sure, Yiorgos?
              $endgroup$
              – Michael Rozenberg
              Dec 8 '18 at 10:01










            • $begingroup$
              You have divided twice by $x^4$.
              $endgroup$
              – gimusi
              Dec 8 '18 at 10:05










            • $begingroup$
              @gimusi You are right! - Corrected it!
              $endgroup$
              – Yiorgos S. Smyrlis
              Dec 8 '18 at 10:07














            2












            2








            2





            $begingroup$

            $$
            sqrt{x^{4} +x^{2}sqrt{x^{4} +1}}-sqrt{2x^4}=
            dfrac{x^{2}sqrt{x^{4} +1} -x^{4}}{sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4}}}=
            dfrac{x^{2}sqrt{x^{4} +1} -x^{4}}{sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4} }}cdotfrac{x^{2}sqrt{x^{4} +1} +x^{4}}{x^{2}sqrt{x^{4} +1} +x^{4}}\=frac{x^4}{big(sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4}}big)big(x^{2}sqrt{x^{4} +1}+x^4big)}\=frac{1}{big(sqrt{1 +sqrt{1+x^{-4}}} +sqrt{2}big)big(sqrt{x^{4}+1}+x^2big)},to,0
            $$






            share|cite|improve this answer











            $endgroup$



            $$
            sqrt{x^{4} +x^{2}sqrt{x^{4} +1}}-sqrt{2x^4}=
            dfrac{x^{2}sqrt{x^{4} +1} -x^{4}}{sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4}}}=
            dfrac{x^{2}sqrt{x^{4} +1} -x^{4}}{sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4} }}cdotfrac{x^{2}sqrt{x^{4} +1} +x^{4}}{x^{2}sqrt{x^{4} +1} +x^{4}}\=frac{x^4}{big(sqrt{x^{4} +x^{2}sqrt{x^{4} +1}} +sqrt{2x^{4}}big)big(x^{2}sqrt{x^{4} +1}+x^4big)}\=frac{1}{big(sqrt{1 +sqrt{1+x^{-4}}} +sqrt{2}big)big(sqrt{x^{4}+1}+x^2big)},to,0
            $$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 8 '18 at 10:07

























            answered Dec 8 '18 at 10:00









            Yiorgos S. SmyrlisYiorgos S. Smyrlis

            63.3k1385163




            63.3k1385163












            • $begingroup$
              Are you sure, Yiorgos?
              $endgroup$
              – Michael Rozenberg
              Dec 8 '18 at 10:01










            • $begingroup$
              You have divided twice by $x^4$.
              $endgroup$
              – gimusi
              Dec 8 '18 at 10:05










            • $begingroup$
              @gimusi You are right! - Corrected it!
              $endgroup$
              – Yiorgos S. Smyrlis
              Dec 8 '18 at 10:07


















            • $begingroup$
              Are you sure, Yiorgos?
              $endgroup$
              – Michael Rozenberg
              Dec 8 '18 at 10:01










            • $begingroup$
              You have divided twice by $x^4$.
              $endgroup$
              – gimusi
              Dec 8 '18 at 10:05










            • $begingroup$
              @gimusi You are right! - Corrected it!
              $endgroup$
              – Yiorgos S. Smyrlis
              Dec 8 '18 at 10:07
















            $begingroup$
            Are you sure, Yiorgos?
            $endgroup$
            – Michael Rozenberg
            Dec 8 '18 at 10:01




            $begingroup$
            Are you sure, Yiorgos?
            $endgroup$
            – Michael Rozenberg
            Dec 8 '18 at 10:01












            $begingroup$
            You have divided twice by $x^4$.
            $endgroup$
            – gimusi
            Dec 8 '18 at 10:05




            $begingroup$
            You have divided twice by $x^4$.
            $endgroup$
            – gimusi
            Dec 8 '18 at 10:05












            $begingroup$
            @gimusi You are right! - Corrected it!
            $endgroup$
            – Yiorgos S. Smyrlis
            Dec 8 '18 at 10:07




            $begingroup$
            @gimusi You are right! - Corrected it!
            $endgroup$
            – Yiorgos S. Smyrlis
            Dec 8 '18 at 10:07











            1












            $begingroup$

            Another approach is using substitution $x=cot t$:
            begin{align}
            lim_{xtoinfty}left(sqrt{x^{4} +x^{2}sqrt{x^{4} +1}}-sqrt{2x^{4}}right)
            &=lim_{tto0}dfrac{sqrt{cos^2t+cos t}-sqrt{2cos^2t}}{sin t} \
            &=lim_{tto0}dfrac{cos t(1-cos t)}{sin t(sqrt{cos^2t+cos t}+sqrt{2cos^2t})} \
            &=lim_{tto0}dfrac{cos t}{sqrt{cos^2t+cos t}+sqrt{2cos^2t}}dfrac{2sin^2frac{t}{2}}{frac{t^2}{2}}dfrac{t}{sin t}dfrac{t}{2} \
            &=color{blue}{0}
            end{align}






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Could you precise? How is $x=cos t$ with $tto 0$ if $xto infty$?
              $endgroup$
              – user376343
              Dec 8 '18 at 10:59










            • $begingroup$
              @user376343 It's impossible when $xtoinfty$ then $tto0$ with $x=cos t$.
              $endgroup$
              – Nosrati
              Dec 8 '18 at 11:03










            • $begingroup$
              When $t to 0$ then $cos t to 1$ therefore I find you substitution mysterious...
              $endgroup$
              – user376343
              Dec 8 '18 at 11:17










            • $begingroup$
              Of course it is :)
              $endgroup$
              – Nosrati
              Dec 8 '18 at 11:20










            • $begingroup$
              @user376343, a letter makes a difference here: $cos t$ and $cot t$.
              $endgroup$
              – farruhota
              Dec 8 '18 at 11:35
















            1












            $begingroup$

            Another approach is using substitution $x=cot t$:
            begin{align}
            lim_{xtoinfty}left(sqrt{x^{4} +x^{2}sqrt{x^{4} +1}}-sqrt{2x^{4}}right)
            &=lim_{tto0}dfrac{sqrt{cos^2t+cos t}-sqrt{2cos^2t}}{sin t} \
            &=lim_{tto0}dfrac{cos t(1-cos t)}{sin t(sqrt{cos^2t+cos t}+sqrt{2cos^2t})} \
            &=lim_{tto0}dfrac{cos t}{sqrt{cos^2t+cos t}+sqrt{2cos^2t}}dfrac{2sin^2frac{t}{2}}{frac{t^2}{2}}dfrac{t}{sin t}dfrac{t}{2} \
            &=color{blue}{0}
            end{align}






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Could you precise? How is $x=cos t$ with $tto 0$ if $xto infty$?
              $endgroup$
              – user376343
              Dec 8 '18 at 10:59










            • $begingroup$
              @user376343 It's impossible when $xtoinfty$ then $tto0$ with $x=cos t$.
              $endgroup$
              – Nosrati
              Dec 8 '18 at 11:03










            • $begingroup$
              When $t to 0$ then $cos t to 1$ therefore I find you substitution mysterious...
              $endgroup$
              – user376343
              Dec 8 '18 at 11:17










            • $begingroup$
              Of course it is :)
              $endgroup$
              – Nosrati
              Dec 8 '18 at 11:20










            • $begingroup$
              @user376343, a letter makes a difference here: $cos t$ and $cot t$.
              $endgroup$
              – farruhota
              Dec 8 '18 at 11:35














            1












            1








            1





            $begingroup$

            Another approach is using substitution $x=cot t$:
            begin{align}
            lim_{xtoinfty}left(sqrt{x^{4} +x^{2}sqrt{x^{4} +1}}-sqrt{2x^{4}}right)
            &=lim_{tto0}dfrac{sqrt{cos^2t+cos t}-sqrt{2cos^2t}}{sin t} \
            &=lim_{tto0}dfrac{cos t(1-cos t)}{sin t(sqrt{cos^2t+cos t}+sqrt{2cos^2t})} \
            &=lim_{tto0}dfrac{cos t}{sqrt{cos^2t+cos t}+sqrt{2cos^2t}}dfrac{2sin^2frac{t}{2}}{frac{t^2}{2}}dfrac{t}{sin t}dfrac{t}{2} \
            &=color{blue}{0}
            end{align}






            share|cite|improve this answer









            $endgroup$



            Another approach is using substitution $x=cot t$:
            begin{align}
            lim_{xtoinfty}left(sqrt{x^{4} +x^{2}sqrt{x^{4} +1}}-sqrt{2x^{4}}right)
            &=lim_{tto0}dfrac{sqrt{cos^2t+cos t}-sqrt{2cos^2t}}{sin t} \
            &=lim_{tto0}dfrac{cos t(1-cos t)}{sin t(sqrt{cos^2t+cos t}+sqrt{2cos^2t})} \
            &=lim_{tto0}dfrac{cos t}{sqrt{cos^2t+cos t}+sqrt{2cos^2t}}dfrac{2sin^2frac{t}{2}}{frac{t^2}{2}}dfrac{t}{sin t}dfrac{t}{2} \
            &=color{blue}{0}
            end{align}







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 8 '18 at 10:20









            NosratiNosrati

            26.5k62354




            26.5k62354












            • $begingroup$
              Could you precise? How is $x=cos t$ with $tto 0$ if $xto infty$?
              $endgroup$
              – user376343
              Dec 8 '18 at 10:59










            • $begingroup$
              @user376343 It's impossible when $xtoinfty$ then $tto0$ with $x=cos t$.
              $endgroup$
              – Nosrati
              Dec 8 '18 at 11:03










            • $begingroup$
              When $t to 0$ then $cos t to 1$ therefore I find you substitution mysterious...
              $endgroup$
              – user376343
              Dec 8 '18 at 11:17










            • $begingroup$
              Of course it is :)
              $endgroup$
              – Nosrati
              Dec 8 '18 at 11:20










            • $begingroup$
              @user376343, a letter makes a difference here: $cos t$ and $cot t$.
              $endgroup$
              – farruhota
              Dec 8 '18 at 11:35


















            • $begingroup$
              Could you precise? How is $x=cos t$ with $tto 0$ if $xto infty$?
              $endgroup$
              – user376343
              Dec 8 '18 at 10:59










            • $begingroup$
              @user376343 It's impossible when $xtoinfty$ then $tto0$ with $x=cos t$.
              $endgroup$
              – Nosrati
              Dec 8 '18 at 11:03










            • $begingroup$
              When $t to 0$ then $cos t to 1$ therefore I find you substitution mysterious...
              $endgroup$
              – user376343
              Dec 8 '18 at 11:17










            • $begingroup$
              Of course it is :)
              $endgroup$
              – Nosrati
              Dec 8 '18 at 11:20










            • $begingroup$
              @user376343, a letter makes a difference here: $cos t$ and $cot t$.
              $endgroup$
              – farruhota
              Dec 8 '18 at 11:35
















            $begingroup$
            Could you precise? How is $x=cos t$ with $tto 0$ if $xto infty$?
            $endgroup$
            – user376343
            Dec 8 '18 at 10:59




            $begingroup$
            Could you precise? How is $x=cos t$ with $tto 0$ if $xto infty$?
            $endgroup$
            – user376343
            Dec 8 '18 at 10:59












            $begingroup$
            @user376343 It's impossible when $xtoinfty$ then $tto0$ with $x=cos t$.
            $endgroup$
            – Nosrati
            Dec 8 '18 at 11:03




            $begingroup$
            @user376343 It's impossible when $xtoinfty$ then $tto0$ with $x=cos t$.
            $endgroup$
            – Nosrati
            Dec 8 '18 at 11:03












            $begingroup$
            When $t to 0$ then $cos t to 1$ therefore I find you substitution mysterious...
            $endgroup$
            – user376343
            Dec 8 '18 at 11:17




            $begingroup$
            When $t to 0$ then $cos t to 1$ therefore I find you substitution mysterious...
            $endgroup$
            – user376343
            Dec 8 '18 at 11:17












            $begingroup$
            Of course it is :)
            $endgroup$
            – Nosrati
            Dec 8 '18 at 11:20




            $begingroup$
            Of course it is :)
            $endgroup$
            – Nosrati
            Dec 8 '18 at 11:20












            $begingroup$
            @user376343, a letter makes a difference here: $cos t$ and $cot t$.
            $endgroup$
            – farruhota
            Dec 8 '18 at 11:35




            $begingroup$
            @user376343, a letter makes a difference here: $cos t$ and $cot t$.
            $endgroup$
            – farruhota
            Dec 8 '18 at 11:35


















            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%2f3030896%2ffind-limit-lim-limits-x-rightarrow-infty-left-sqrtx4-x2-sqrtx%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

            Plaza Victoria

            Image
            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
            Read more

            In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

            Image
            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
            Read more

            How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

            Image
            1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
            Read more