How to compute $lim limits_{xto infty} ((2x^3-2x^2+x)e^{1/x}-sqrt{x^6+3})$ through Taylor series












1












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I am trying to compute the below limit through Taylor series:
$lim limits_{xto infty} ((2x^3-2x^2+x)e^{1/x}-sqrt{x^6+3})$



What I have already tried is first of all change the variable x to
$x=1/t$ and the limit to t limits to 0, so I am able to use Maclaurin series.
After that I change $e$ to exponent polynomial up to t=6 + $O(X^6)$
however, I don`t know what can I do with square root.






limits limits-without-lhopital







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  • $begingroup$
    Use $sqrt{3+t}=sqrt 3{sqrt{1+t/3}}$.
    $endgroup$
    – Yves Daoust
    Dec 28 '18 at 19:17
















1












$begingroup$


I am trying to compute the below limit through Taylor series:
$lim limits_{xto infty} ((2x^3-2x^2+x)e^{1/x}-sqrt{x^6+3})$



What I have already tried is first of all change the variable x to
$x=1/t$ and the limit to t limits to 0, so I am able to use Maclaurin series.
After that I change $e$ to exponent polynomial up to t=6 + $O(X^6)$
however, I don`t know what can I do with square root.






limits limits-without-lhopital







share|cite|improve this question









$endgroup$












  • $begingroup$
    Use $sqrt{3+t}=sqrt 3{sqrt{1+t/3}}$.
    $endgroup$
    – Yves Daoust
    Dec 28 '18 at 19:17














1












1








1





$begingroup$


I am trying to compute the below limit through Taylor series:
$lim limits_{xto infty} ((2x^3-2x^2+x)e^{1/x}-sqrt{x^6+3})$



What I have already tried is first of all change the variable x to
$x=1/t$ and the limit to t limits to 0, so I am able to use Maclaurin series.
After that I change $e$ to exponent polynomial up to t=6 + $O(X^6)$
however, I don`t know what can I do with square root.






limits limits-without-lhopital







share|cite|improve this question









$endgroup$




I am trying to compute the below limit through Taylor series:
$lim limits_{xto infty} ((2x^3-2x^2+x)e^{1/x}-sqrt{x^6+3})$



What I have already tried is first of all change the variable x to
$x=1/t$ and the limit to t limits to 0, so I am able to use Maclaurin series.
After that I change $e$ to exponent polynomial up to t=6 + $O(X^6)$
however, I don`t know what can I do with square root.






limits limits-without-lhopital




limits limits-without-lhopital






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share|cite|improve this question











share|cite|improve this question




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asked Dec 28 '18 at 19:14









John DJohn D

366




366












  • $begingroup$
    Use $sqrt{3+t}=sqrt 3{sqrt{1+t/3}}$.
    $endgroup$
    – Yves Daoust
    Dec 28 '18 at 19:17


















  • $begingroup$
    Use $sqrt{3+t}=sqrt 3{sqrt{1+t/3}}$.
    $endgroup$
    – Yves Daoust
    Dec 28 '18 at 19:17
















$begingroup$
Use $sqrt{3+t}=sqrt 3{sqrt{1+t/3}}$.
$endgroup$
– Yves Daoust
Dec 28 '18 at 19:17




$begingroup$
Use $sqrt{3+t}=sqrt 3{sqrt{1+t/3}}$.
$endgroup$
– Yves Daoust
Dec 28 '18 at 19:17










4 Answers
4






active

oldest

votes


















4












$begingroup$

HINT:



$$sqrt{x^6+3}=frac1{t^3}left(1+3t^6right)^{1/2}=frac1{t^3}+frac{frac12cdot3t^6}{t^3}-frac{frac1{2cdot4}cdot(3t^6)^2}{t^3}+frac{frac{1cdot3}{2cdot4cdot6}cdot(3t^6)^3}{t^3}-cdots\(2x^3-2x^2+x)e^{1/x}=left(frac2{t^3}-frac2{t^2}+frac1tright)left(1+t+frac{t^2}{2cdot1}+frac{t^3}{3cdot2cdot1}+cdotsright)$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Could you please explain what did you do in the first row? I`m not sure I understand your development after the first step in the first row. What series is it?
    $endgroup$
    – John D
    Dec 28 '18 at 20:00






  • 1




    $begingroup$
    @JohnD $$sqrt{x^6+3}=sqrt{frac1{t^6}+3}=sqrt{frac1{t^6}(1+3t^6)}=frac1{t^3}(1+3t^6)^{1/2} \ (1+3t^6)^{1/2}=1+frac12(3t^6)-frac1{2cdot4}(3t^6)^2+cdots$$
    $endgroup$
    – TheSimpliFire
    Dec 28 '18 at 20:04












  • $begingroup$
    Sorry, But I have some a confusion. Did you develop this series with Taylor? what is your point which function is derivative? if we derivative once $f(t)=(1+3t^6)^1/2$ we got $1/2*(1+3t^6)^-1/2 * 18t^5$ So it does not work out according to what development you've done. you don`t need to write again the develop, just if you can explain in words what did you do and what series it it? This would be greatly appreciated!
    $endgroup$
    – John D
    Dec 28 '18 at 20:43










  • $begingroup$
    Please see here and here. The binomial series with $alpha=1/2$ is a particular application of Taylor/Maclaurin series.
    $endgroup$
    – TheSimpliFire
    Dec 28 '18 at 20:58





















1












$begingroup$

I think that you don't have to go that far. You know that (for large values of $x$):
$$
e^{1/x}=sum_{k=0}^{infty}{frac{1}{k!x^k}}geq1
$$
so that:
$$
(2x^3-2x^2+x)e^{1/x}-sqrt{x^6+3} geq (2x^3-2x^2+x)-sqrt{x^6+3}geqfrac{x^3}{2}
$$
Thus, the limit goes to $infty$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The OP specifically wants to find the limit using Taylor series.
    $endgroup$
    – TheSimpliFire
    Dec 28 '18 at 19:20



















0












$begingroup$

hint



The square root becomes



$$frac{sqrt{1+3t^6}}{|t^3|}=$$



$$frac{1}{|t^3|}Bigl(1+frac{3t^6}{2}-frac 98t^{12}+t^{12}epsilon(t)Bigr)$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    $y = frac 1x$



    then we have



    $lim_limits{yto 0^+} frac {(2 -2y+ y^2)e^y - sqrt {1+3y^6}}{y^3}$



    Now if you want to do a Taylor expansion...



    $lim_limits{yto 0^+} frac {(2 -2y+ y^2)(1+y+frac 12 y^2+cdots) - (1+frac 32 y^6 - cdots )}{y^3}$



    $lim_limits{yto 0^+} frac {1 + O(y)}{y^3} = infty$






    share|cite|improve this answer











    $endgroup$













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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      HINT:



      $$sqrt{x^6+3}=frac1{t^3}left(1+3t^6right)^{1/2}=frac1{t^3}+frac{frac12cdot3t^6}{t^3}-frac{frac1{2cdot4}cdot(3t^6)^2}{t^3}+frac{frac{1cdot3}{2cdot4cdot6}cdot(3t^6)^3}{t^3}-cdots\(2x^3-2x^2+x)e^{1/x}=left(frac2{t^3}-frac2{t^2}+frac1tright)left(1+t+frac{t^2}{2cdot1}+frac{t^3}{3cdot2cdot1}+cdotsright)$$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Could you please explain what did you do in the first row? I`m not sure I understand your development after the first step in the first row. What series is it?
        $endgroup$
        – John D
        Dec 28 '18 at 20:00






      • 1




        $begingroup$
        @JohnD $$sqrt{x^6+3}=sqrt{frac1{t^6}+3}=sqrt{frac1{t^6}(1+3t^6)}=frac1{t^3}(1+3t^6)^{1/2} \ (1+3t^6)^{1/2}=1+frac12(3t^6)-frac1{2cdot4}(3t^6)^2+cdots$$
        $endgroup$
        – TheSimpliFire
        Dec 28 '18 at 20:04












      • $begingroup$
        Sorry, But I have some a confusion. Did you develop this series with Taylor? what is your point which function is derivative? if we derivative once $f(t)=(1+3t^6)^1/2$ we got $1/2*(1+3t^6)^-1/2 * 18t^5$ So it does not work out according to what development you've done. you don`t need to write again the develop, just if you can explain in words what did you do and what series it it? This would be greatly appreciated!
        $endgroup$
        – John D
        Dec 28 '18 at 20:43










      • $begingroup$
        Please see here and here. The binomial series with $alpha=1/2$ is a particular application of Taylor/Maclaurin series.
        $endgroup$
        – TheSimpliFire
        Dec 28 '18 at 20:58


















      4












      $begingroup$

      HINT:



      $$sqrt{x^6+3}=frac1{t^3}left(1+3t^6right)^{1/2}=frac1{t^3}+frac{frac12cdot3t^6}{t^3}-frac{frac1{2cdot4}cdot(3t^6)^2}{t^3}+frac{frac{1cdot3}{2cdot4cdot6}cdot(3t^6)^3}{t^3}-cdots\(2x^3-2x^2+x)e^{1/x}=left(frac2{t^3}-frac2{t^2}+frac1tright)left(1+t+frac{t^2}{2cdot1}+frac{t^3}{3cdot2cdot1}+cdotsright)$$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Could you please explain what did you do in the first row? I`m not sure I understand your development after the first step in the first row. What series is it?
        $endgroup$
        – John D
        Dec 28 '18 at 20:00






      • 1




        $begingroup$
        @JohnD $$sqrt{x^6+3}=sqrt{frac1{t^6}+3}=sqrt{frac1{t^6}(1+3t^6)}=frac1{t^3}(1+3t^6)^{1/2} \ (1+3t^6)^{1/2}=1+frac12(3t^6)-frac1{2cdot4}(3t^6)^2+cdots$$
        $endgroup$
        – TheSimpliFire
        Dec 28 '18 at 20:04












      • $begingroup$
        Sorry, But I have some a confusion. Did you develop this series with Taylor? what is your point which function is derivative? if we derivative once $f(t)=(1+3t^6)^1/2$ we got $1/2*(1+3t^6)^-1/2 * 18t^5$ So it does not work out according to what development you've done. you don`t need to write again the develop, just if you can explain in words what did you do and what series it it? This would be greatly appreciated!
        $endgroup$
        – John D
        Dec 28 '18 at 20:43










      • $begingroup$
        Please see here and here. The binomial series with $alpha=1/2$ is a particular application of Taylor/Maclaurin series.
        $endgroup$
        – TheSimpliFire
        Dec 28 '18 at 20:58
















      4












      4








      4





      $begingroup$

      HINT:



      $$sqrt{x^6+3}=frac1{t^3}left(1+3t^6right)^{1/2}=frac1{t^3}+frac{frac12cdot3t^6}{t^3}-frac{frac1{2cdot4}cdot(3t^6)^2}{t^3}+frac{frac{1cdot3}{2cdot4cdot6}cdot(3t^6)^3}{t^3}-cdots\(2x^3-2x^2+x)e^{1/x}=left(frac2{t^3}-frac2{t^2}+frac1tright)left(1+t+frac{t^2}{2cdot1}+frac{t^3}{3cdot2cdot1}+cdotsright)$$






      share|cite|improve this answer











      $endgroup$



      HINT:



      $$sqrt{x^6+3}=frac1{t^3}left(1+3t^6right)^{1/2}=frac1{t^3}+frac{frac12cdot3t^6}{t^3}-frac{frac1{2cdot4}cdot(3t^6)^2}{t^3}+frac{frac{1cdot3}{2cdot4cdot6}cdot(3t^6)^3}{t^3}-cdots\(2x^3-2x^2+x)e^{1/x}=left(frac2{t^3}-frac2{t^2}+frac1tright)left(1+t+frac{t^2}{2cdot1}+frac{t^3}{3cdot2cdot1}+cdotsright)$$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 28 '18 at 20:05

























      answered Dec 28 '18 at 19:20









      TheSimpliFireTheSimpliFire

      12.5k62460




      12.5k62460












      • $begingroup$
        Could you please explain what did you do in the first row? I`m not sure I understand your development after the first step in the first row. What series is it?
        $endgroup$
        – John D
        Dec 28 '18 at 20:00






      • 1




        $begingroup$
        @JohnD $$sqrt{x^6+3}=sqrt{frac1{t^6}+3}=sqrt{frac1{t^6}(1+3t^6)}=frac1{t^3}(1+3t^6)^{1/2} \ (1+3t^6)^{1/2}=1+frac12(3t^6)-frac1{2cdot4}(3t^6)^2+cdots$$
        $endgroup$
        – TheSimpliFire
        Dec 28 '18 at 20:04












      • $begingroup$
        Sorry, But I have some a confusion. Did you develop this series with Taylor? what is your point which function is derivative? if we derivative once $f(t)=(1+3t^6)^1/2$ we got $1/2*(1+3t^6)^-1/2 * 18t^5$ So it does not work out according to what development you've done. you don`t need to write again the develop, just if you can explain in words what did you do and what series it it? This would be greatly appreciated!
        $endgroup$
        – John D
        Dec 28 '18 at 20:43










      • $begingroup$
        Please see here and here. The binomial series with $alpha=1/2$ is a particular application of Taylor/Maclaurin series.
        $endgroup$
        – TheSimpliFire
        Dec 28 '18 at 20:58




















      • $begingroup$
        Could you please explain what did you do in the first row? I`m not sure I understand your development after the first step in the first row. What series is it?
        $endgroup$
        – John D
        Dec 28 '18 at 20:00






      • 1




        $begingroup$
        @JohnD $$sqrt{x^6+3}=sqrt{frac1{t^6}+3}=sqrt{frac1{t^6}(1+3t^6)}=frac1{t^3}(1+3t^6)^{1/2} \ (1+3t^6)^{1/2}=1+frac12(3t^6)-frac1{2cdot4}(3t^6)^2+cdots$$
        $endgroup$
        – TheSimpliFire
        Dec 28 '18 at 20:04












      • $begingroup$
        Sorry, But I have some a confusion. Did you develop this series with Taylor? what is your point which function is derivative? if we derivative once $f(t)=(1+3t^6)^1/2$ we got $1/2*(1+3t^6)^-1/2 * 18t^5$ So it does not work out according to what development you've done. you don`t need to write again the develop, just if you can explain in words what did you do and what series it it? This would be greatly appreciated!
        $endgroup$
        – John D
        Dec 28 '18 at 20:43










      • $begingroup$
        Please see here and here. The binomial series with $alpha=1/2$ is a particular application of Taylor/Maclaurin series.
        $endgroup$
        – TheSimpliFire
        Dec 28 '18 at 20:58


















      $begingroup$
      Could you please explain what did you do in the first row? I`m not sure I understand your development after the first step in the first row. What series is it?
      $endgroup$
      – John D
      Dec 28 '18 at 20:00




      $begingroup$
      Could you please explain what did you do in the first row? I`m not sure I understand your development after the first step in the first row. What series is it?
      $endgroup$
      – John D
      Dec 28 '18 at 20:00




      1




      1




      $begingroup$
      @JohnD $$sqrt{x^6+3}=sqrt{frac1{t^6}+3}=sqrt{frac1{t^6}(1+3t^6)}=frac1{t^3}(1+3t^6)^{1/2} \ (1+3t^6)^{1/2}=1+frac12(3t^6)-frac1{2cdot4}(3t^6)^2+cdots$$
      $endgroup$
      – TheSimpliFire
      Dec 28 '18 at 20:04






      $begingroup$
      @JohnD $$sqrt{x^6+3}=sqrt{frac1{t^6}+3}=sqrt{frac1{t^6}(1+3t^6)}=frac1{t^3}(1+3t^6)^{1/2} \ (1+3t^6)^{1/2}=1+frac12(3t^6)-frac1{2cdot4}(3t^6)^2+cdots$$
      $endgroup$
      – TheSimpliFire
      Dec 28 '18 at 20:04














      $begingroup$
      Sorry, But I have some a confusion. Did you develop this series with Taylor? what is your point which function is derivative? if we derivative once $f(t)=(1+3t^6)^1/2$ we got $1/2*(1+3t^6)^-1/2 * 18t^5$ So it does not work out according to what development you've done. you don`t need to write again the develop, just if you can explain in words what did you do and what series it it? This would be greatly appreciated!
      $endgroup$
      – John D
      Dec 28 '18 at 20:43




      $begingroup$
      Sorry, But I have some a confusion. Did you develop this series with Taylor? what is your point which function is derivative? if we derivative once $f(t)=(1+3t^6)^1/2$ we got $1/2*(1+3t^6)^-1/2 * 18t^5$ So it does not work out according to what development you've done. you don`t need to write again the develop, just if you can explain in words what did you do and what series it it? This would be greatly appreciated!
      $endgroup$
      – John D
      Dec 28 '18 at 20:43












      $begingroup$
      Please see here and here. The binomial series with $alpha=1/2$ is a particular application of Taylor/Maclaurin series.
      $endgroup$
      – TheSimpliFire
      Dec 28 '18 at 20:58






      $begingroup$
      Please see here and here. The binomial series with $alpha=1/2$ is a particular application of Taylor/Maclaurin series.
      $endgroup$
      – TheSimpliFire
      Dec 28 '18 at 20:58













      1












      $begingroup$

      I think that you don't have to go that far. You know that (for large values of $x$):
      $$
      e^{1/x}=sum_{k=0}^{infty}{frac{1}{k!x^k}}geq1
      $$
      so that:
      $$
      (2x^3-2x^2+x)e^{1/x}-sqrt{x^6+3} geq (2x^3-2x^2+x)-sqrt{x^6+3}geqfrac{x^3}{2}
      $$
      Thus, the limit goes to $infty$.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        The OP specifically wants to find the limit using Taylor series.
        $endgroup$
        – TheSimpliFire
        Dec 28 '18 at 19:20
















      1












      $begingroup$

      I think that you don't have to go that far. You know that (for large values of $x$):
      $$
      e^{1/x}=sum_{k=0}^{infty}{frac{1}{k!x^k}}geq1
      $$
      so that:
      $$
      (2x^3-2x^2+x)e^{1/x}-sqrt{x^6+3} geq (2x^3-2x^2+x)-sqrt{x^6+3}geqfrac{x^3}{2}
      $$
      Thus, the limit goes to $infty$.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        The OP specifically wants to find the limit using Taylor series.
        $endgroup$
        – TheSimpliFire
        Dec 28 '18 at 19:20














      1












      1








      1





      $begingroup$

      I think that you don't have to go that far. You know that (for large values of $x$):
      $$
      e^{1/x}=sum_{k=0}^{infty}{frac{1}{k!x^k}}geq1
      $$
      so that:
      $$
      (2x^3-2x^2+x)e^{1/x}-sqrt{x^6+3} geq (2x^3-2x^2+x)-sqrt{x^6+3}geqfrac{x^3}{2}
      $$
      Thus, the limit goes to $infty$.






      share|cite|improve this answer









      $endgroup$



      I think that you don't have to go that far. You know that (for large values of $x$):
      $$
      e^{1/x}=sum_{k=0}^{infty}{frac{1}{k!x^k}}geq1
      $$
      so that:
      $$
      (2x^3-2x^2+x)e^{1/x}-sqrt{x^6+3} geq (2x^3-2x^2+x)-sqrt{x^6+3}geqfrac{x^3}{2}
      $$
      Thus, the limit goes to $infty$.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 28 '18 at 19:19









      DudeManDudeMan

      1113




      1113












      • $begingroup$
        The OP specifically wants to find the limit using Taylor series.
        $endgroup$
        – TheSimpliFire
        Dec 28 '18 at 19:20


















      • $begingroup$
        The OP specifically wants to find the limit using Taylor series.
        $endgroup$
        – TheSimpliFire
        Dec 28 '18 at 19:20
















      $begingroup$
      The OP specifically wants to find the limit using Taylor series.
      $endgroup$
      – TheSimpliFire
      Dec 28 '18 at 19:20




      $begingroup$
      The OP specifically wants to find the limit using Taylor series.
      $endgroup$
      – TheSimpliFire
      Dec 28 '18 at 19:20











      0












      $begingroup$

      hint



      The square root becomes



      $$frac{sqrt{1+3t^6}}{|t^3|}=$$



      $$frac{1}{|t^3|}Bigl(1+frac{3t^6}{2}-frac 98t^{12}+t^{12}epsilon(t)Bigr)$$






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        hint



        The square root becomes



        $$frac{sqrt{1+3t^6}}{|t^3|}=$$



        $$frac{1}{|t^3|}Bigl(1+frac{3t^6}{2}-frac 98t^{12}+t^{12}epsilon(t)Bigr)$$






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          hint



          The square root becomes



          $$frac{sqrt{1+3t^6}}{|t^3|}=$$



          $$frac{1}{|t^3|}Bigl(1+frac{3t^6}{2}-frac 98t^{12}+t^{12}epsilon(t)Bigr)$$






          share|cite|improve this answer









          $endgroup$



          hint



          The square root becomes



          $$frac{sqrt{1+3t^6}}{|t^3|}=$$



          $$frac{1}{|t^3|}Bigl(1+frac{3t^6}{2}-frac 98t^{12}+t^{12}epsilon(t)Bigr)$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 28 '18 at 19:20









          hamam_Abdallahhamam_Abdallah

          38.1k21634




          38.1k21634























              0












              $begingroup$

              $y = frac 1x$



              then we have



              $lim_limits{yto 0^+} frac {(2 -2y+ y^2)e^y - sqrt {1+3y^6}}{y^3}$



              Now if you want to do a Taylor expansion...



              $lim_limits{yto 0^+} frac {(2 -2y+ y^2)(1+y+frac 12 y^2+cdots) - (1+frac 32 y^6 - cdots )}{y^3}$



              $lim_limits{yto 0^+} frac {1 + O(y)}{y^3} = infty$






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                $y = frac 1x$



                then we have



                $lim_limits{yto 0^+} frac {(2 -2y+ y^2)e^y - sqrt {1+3y^6}}{y^3}$



                Now if you want to do a Taylor expansion...



                $lim_limits{yto 0^+} frac {(2 -2y+ y^2)(1+y+frac 12 y^2+cdots) - (1+frac 32 y^6 - cdots )}{y^3}$



                $lim_limits{yto 0^+} frac {1 + O(y)}{y^3} = infty$






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $y = frac 1x$



                  then we have



                  $lim_limits{yto 0^+} frac {(2 -2y+ y^2)e^y - sqrt {1+3y^6}}{y^3}$



                  Now if you want to do a Taylor expansion...



                  $lim_limits{yto 0^+} frac {(2 -2y+ y^2)(1+y+frac 12 y^2+cdots) - (1+frac 32 y^6 - cdots )}{y^3}$



                  $lim_limits{yto 0^+} frac {1 + O(y)}{y^3} = infty$






                  share|cite|improve this answer











                  $endgroup$



                  $y = frac 1x$



                  then we have



                  $lim_limits{yto 0^+} frac {(2 -2y+ y^2)e^y - sqrt {1+3y^6}}{y^3}$



                  Now if you want to do a Taylor expansion...



                  $lim_limits{yto 0^+} frac {(2 -2y+ y^2)(1+y+frac 12 y^2+cdots) - (1+frac 32 y^6 - cdots )}{y^3}$



                  $lim_limits{yto 0^+} frac {1 + O(y)}{y^3} = infty$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 28 '18 at 19:31

























                  answered Dec 28 '18 at 19:27









                  Doug MDoug M

                  45.2k31854




                  45.2k31854






























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