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} }} =$



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



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real-analysis limits radicals limits-without-lhopital







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  • $begingroup$
    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







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  • $begingroup$
    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







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$endgroup$




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



$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






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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






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oldest

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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












    $begingroup$

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






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    $endgroup$













    • $begingroup$
      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}






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    $endgroup$













    • $begingroup$
      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











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    4 Answers
    4






    active

    oldest

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    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














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            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


















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