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












2














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



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
















2














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



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














2












2








2







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



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



$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 at 13:43









Martin Sleziak

44.6k7115270




44.6k7115270










asked Dec 8 at 9:43









Angella

111




111












  • See math.stackexchange.com/questions/3025375/…
    – lab bhattacharjee
    Dec 8 at 9:47


















  • See math.stackexchange.com/questions/3025375/…
    – lab bhattacharjee
    Dec 8 at 9:47
















See math.stackexchange.com/questions/3025375/…
– lab bhattacharjee
Dec 8 at 9:47




See math.stackexchange.com/questions/3025375/…
– lab bhattacharjee
Dec 8 at 9:47










4 Answers
4






active

oldest

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6














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






share|cite|improve this answer





























    3














    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























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



















    2














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






    share|cite|improve this answer























    • Are you sure, Yiorgos?
      – Michael Rozenberg
      Dec 8 at 10:01










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










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



















    1














    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





















    • Could you precise? How is $x=cos t$ with $tto 0$ if $xto infty$?
      – user376343
      Dec 8 at 10:59










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










    • When $t to 0$ then $cos t to 1$ therefore I find you substitution mysterious...
      – user376343
      Dec 8 at 11:17










    • Of course it is :)
      – Nosrati
      Dec 8 at 11:20










    • @user376343, a letter makes a difference here: $cos t$ and $cot t$.
      – farruhota
      Dec 8 at 11:35











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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6














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






    share|cite|improve this answer


























      6














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






      share|cite|improve this answer
























        6












        6








        6






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






        share|cite|improve this answer












        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 at 9:44









        Michael Rozenberg

        95.4k1588184




        95.4k1588184























            3














            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























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
















            3














            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























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














            3












            3








            3






            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














            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 at 11:07

























            answered Dec 8 at 9:45









            gimusi

            1




            1












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


















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
















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




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











            2














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






            share|cite|improve this answer























            • Are you sure, Yiorgos?
              – Michael Rozenberg
              Dec 8 at 10:01










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










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
















            2














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






            share|cite|improve this answer























            • Are you sure, Yiorgos?
              – Michael Rozenberg
              Dec 8 at 10:01










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










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














            2












            2








            2






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






            share|cite|improve this answer














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







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 8 at 10:07

























            answered Dec 8 at 10:00









            Yiorgos S. Smyrlis

            62.4k1383162




            62.4k1383162












            • Are you sure, Yiorgos?
              – Michael Rozenberg
              Dec 8 at 10:01










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










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


















            • Are you sure, Yiorgos?
              – Michael Rozenberg
              Dec 8 at 10:01










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










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
















            Are you sure, Yiorgos?
            – Michael Rozenberg
            Dec 8 at 10:01




            Are you sure, Yiorgos?
            – Michael Rozenberg
            Dec 8 at 10:01












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




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












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




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











            1














            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





















            • Could you precise? How is $x=cos t$ with $tto 0$ if $xto infty$?
              – user376343
              Dec 8 at 10:59










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










            • When $t to 0$ then $cos t to 1$ therefore I find you substitution mysterious...
              – user376343
              Dec 8 at 11:17










            • Of course it is :)
              – Nosrati
              Dec 8 at 11:20










            • @user376343, a letter makes a difference here: $cos t$ and $cot t$.
              – farruhota
              Dec 8 at 11:35
















            1














            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





















            • Could you precise? How is $x=cos t$ with $tto 0$ if $xto infty$?
              – user376343
              Dec 8 at 10:59










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










            • When $t to 0$ then $cos t to 1$ therefore I find you substitution mysterious...
              – user376343
              Dec 8 at 11:17










            • Of course it is :)
              – Nosrati
              Dec 8 at 11:20










            • @user376343, a letter makes a difference here: $cos t$ and $cot t$.
              – farruhota
              Dec 8 at 11:35














            1












            1








            1






            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












            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 at 10:20









            Nosrati

            26.4k62353




            26.4k62353












            • Could you precise? How is $x=cos t$ with $tto 0$ if $xto infty$?
              – user376343
              Dec 8 at 10:59










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










            • When $t to 0$ then $cos t to 1$ therefore I find you substitution mysterious...
              – user376343
              Dec 8 at 11:17










            • Of course it is :)
              – Nosrati
              Dec 8 at 11:20










            • @user376343, a letter makes a difference here: $cos t$ and $cot t$.
              – farruhota
              Dec 8 at 11:35


















            • Could you precise? How is $x=cos t$ with $tto 0$ if $xto infty$?
              – user376343
              Dec 8 at 10:59










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










            • When $t to 0$ then $cos t to 1$ therefore I find you substitution mysterious...
              – user376343
              Dec 8 at 11:17










            • Of course it is :)
              – Nosrati
              Dec 8 at 11:20










            • @user376343, a letter makes a difference here: $cos t$ and $cot t$.
              – farruhota
              Dec 8 at 11:35
















            Could you precise? How is $x=cos t$ with $tto 0$ if $xto infty$?
            – user376343
            Dec 8 at 10:59




            Could you precise? How is $x=cos t$ with $tto 0$ if $xto infty$?
            – user376343
            Dec 8 at 10:59












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




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












            When $t to 0$ then $cos t to 1$ therefore I find you substitution mysterious...
            – user376343
            Dec 8 at 11:17




            When $t to 0$ then $cos t to 1$ therefore I find you substitution mysterious...
            – user376343
            Dec 8 at 11:17












            Of course it is :)
            – Nosrati
            Dec 8 at 11:20




            Of course it is :)
            – Nosrati
            Dec 8 at 11:20












            @user376343, a letter makes a difference here: $cos t$ and $cot t$.
            – farruhota
            Dec 8 at 11:35




            @user376343, a letter makes a difference here: $cos t$ and $cot t$.
            – farruhota
            Dec 8 at 11:35


















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