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












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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







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  • $begingroup$
    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






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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
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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?






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  • $begingroup$
    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



















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Let $1/n=h$



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



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



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



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






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  • $begingroup$
    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



$$frac{sqrt2}8.$$






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    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













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      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




















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