Limits with parameters using a Taylor polynomial [closed]












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We are asked to discuss the existance and values of the following limit with $mathbb{N} ni n,m geq 1$:



$$lim_{xto 0}{frac{{left ( left (1+x right )^{frac{1}{2}} -1 -frac{x}{2} right )}^{6m}} {{x^n}}}$$



We thought about calculating the $k$-th order Taylor polynomial around $0$ of $(1+x)^{frac{1}{2}}$ and discussing the limits with that, but we're not getting anywhere.






algebra-precalculus







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closed as off-topic by RRL, Shailesh, José Carlos Santos, Abcd, Pierre-Guy Plamondon Jan 13 at 15:18


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Shailesh, José Carlos Santos, Abcd, Pierre-Guy Plamondon

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    This is what youneed to do. Focus on $ left(1+x right )^{frac{1}{2}} -1 -frac{x}{2} $ first and then raise to power $6m$ and then divide by $x^n$. What do you get ?
    $endgroup$
    – Claude Leibovici
    Jan 12 at 16:55


















0












$begingroup$


We are asked to discuss the existance and values of the following limit with $mathbb{N} ni n,m geq 1$:



$$lim_{xto 0}{frac{{left ( left (1+x right )^{frac{1}{2}} -1 -frac{x}{2} right )}^{6m}} {{x^n}}}$$



We thought about calculating the $k$-th order Taylor polynomial around $0$ of $(1+x)^{frac{1}{2}}$ and discussing the limits with that, but we're not getting anywhere.






algebra-precalculus







share|cite|improve this question











$endgroup$



closed as off-topic by RRL, Shailesh, José Carlos Santos, Abcd, Pierre-Guy Plamondon Jan 13 at 15:18


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Shailesh, José Carlos Santos, Abcd, Pierre-Guy Plamondon

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    This is what youneed to do. Focus on $ left(1+x right )^{frac{1}{2}} -1 -frac{x}{2} $ first and then raise to power $6m$ and then divide by $x^n$. What do you get ?
    $endgroup$
    – Claude Leibovici
    Jan 12 at 16:55
















0












0








0





$begingroup$


We are asked to discuss the existance and values of the following limit with $mathbb{N} ni n,m geq 1$:



$$lim_{xto 0}{frac{{left ( left (1+x right )^{frac{1}{2}} -1 -frac{x}{2} right )}^{6m}} {{x^n}}}$$



We thought about calculating the $k$-th order Taylor polynomial around $0$ of $(1+x)^{frac{1}{2}}$ and discussing the limits with that, but we're not getting anywhere.






algebra-precalculus







share|cite|improve this question











$endgroup$




We are asked to discuss the existance and values of the following limit with $mathbb{N} ni n,m geq 1$:



$$lim_{xto 0}{frac{{left ( left (1+x right )^{frac{1}{2}} -1 -frac{x}{2} right )}^{6m}} {{x^n}}}$$



We thought about calculating the $k$-th order Taylor polynomial around $0$ of $(1+x)^{frac{1}{2}}$ and discussing the limits with that, but we're not getting anywhere.






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













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edited Jan 12 at 18:31









egreg

184k1486205




184k1486205










asked Jan 12 at 16:45









Marc Ballestero RibóMarc Ballestero Ribó

175




175




closed as off-topic by RRL, Shailesh, José Carlos Santos, Abcd, Pierre-Guy Plamondon Jan 13 at 15:18


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Shailesh, José Carlos Santos, Abcd, Pierre-Guy Plamondon

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by RRL, Shailesh, José Carlos Santos, Abcd, Pierre-Guy Plamondon Jan 13 at 15:18


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Shailesh, José Carlos Santos, Abcd, Pierre-Guy Plamondon

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    This is what youneed to do. Focus on $ left(1+x right )^{frac{1}{2}} -1 -frac{x}{2} $ first and then raise to power $6m$ and then divide by $x^n$. What do you get ?
    $endgroup$
    – Claude Leibovici
    Jan 12 at 16:55




















  • $begingroup$
    This is what youneed to do. Focus on $ left(1+x right )^{frac{1}{2}} -1 -frac{x}{2} $ first and then raise to power $6m$ and then divide by $x^n$. What do you get ?
    $endgroup$
    – Claude Leibovici
    Jan 12 at 16:55


















$begingroup$
This is what youneed to do. Focus on $ left(1+x right )^{frac{1}{2}} -1 -frac{x}{2} $ first and then raise to power $6m$ and then divide by $x^n$. What do you get ?
$endgroup$
– Claude Leibovici
Jan 12 at 16:55






$begingroup$
This is what youneed to do. Focus on $ left(1+x right )^{frac{1}{2}} -1 -frac{x}{2} $ first and then raise to power $6m$ and then divide by $x^n$. What do you get ?
$endgroup$
– Claude Leibovici
Jan 12 at 16:55












2 Answers
2






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The order two Taylor polynomial is
$$
sqrt{1+x}=1+frac{x}{2}-frac{x^2}{8}+o(x^2)
$$
and therefore your expression is
$$
frac{x^{12m}/2^{18m}+o(x^{12m})}{x^n}
$$
You should be able to finish, now.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Defining $sqrt{x+1}=t$ we obtain$$lim_{xto 0}{frac{{left ( left (1+x right )^{frac{1}{2}} -1 -frac{x}{2} right )}^{6m}} {{x^n}}}{=lim_{tto 1}{left(t-1-{t^2-1over 2}right)^{6m}over (t^2-1)^n}\={1over2^{6m}}lim_{tto 1}{(t-1)^{12m}over (t-1)^n(t+1)^n}\={1over2^{6m+n}}lim_{tto 1}{(t-1)^{12m}over (t-1)^n}\=begin{cases}0&,quad12m>n\{1over 2^{6m+n}}&,quad 12m=n\infty&,quad 12m<nquad, quad n text{ is even}\text{undefined}&,quad 12m<nquad, quad n text{ is odd}end{cases}}$$






    share|cite|improve this answer









    $endgroup$




















      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      The order two Taylor polynomial is
      $$
      sqrt{1+x}=1+frac{x}{2}-frac{x^2}{8}+o(x^2)
      $$
      and therefore your expression is
      $$
      frac{x^{12m}/2^{18m}+o(x^{12m})}{x^n}
      $$
      You should be able to finish, now.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        The order two Taylor polynomial is
        $$
        sqrt{1+x}=1+frac{x}{2}-frac{x^2}{8}+o(x^2)
        $$
        and therefore your expression is
        $$
        frac{x^{12m}/2^{18m}+o(x^{12m})}{x^n}
        $$
        You should be able to finish, now.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          The order two Taylor polynomial is
          $$
          sqrt{1+x}=1+frac{x}{2}-frac{x^2}{8}+o(x^2)
          $$
          and therefore your expression is
          $$
          frac{x^{12m}/2^{18m}+o(x^{12m})}{x^n}
          $$
          You should be able to finish, now.






          share|cite|improve this answer









          $endgroup$



          The order two Taylor polynomial is
          $$
          sqrt{1+x}=1+frac{x}{2}-frac{x^2}{8}+o(x^2)
          $$
          and therefore your expression is
          $$
          frac{x^{12m}/2^{18m}+o(x^{12m})}{x^n}
          $$
          You should be able to finish, now.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 12 at 18:39









          egregegreg

          184k1486205




          184k1486205























              0












              $begingroup$

              Defining $sqrt{x+1}=t$ we obtain$$lim_{xto 0}{frac{{left ( left (1+x right )^{frac{1}{2}} -1 -frac{x}{2} right )}^{6m}} {{x^n}}}{=lim_{tto 1}{left(t-1-{t^2-1over 2}right)^{6m}over (t^2-1)^n}\={1over2^{6m}}lim_{tto 1}{(t-1)^{12m}over (t-1)^n(t+1)^n}\={1over2^{6m+n}}lim_{tto 1}{(t-1)^{12m}over (t-1)^n}\=begin{cases}0&,quad12m>n\{1over 2^{6m+n}}&,quad 12m=n\infty&,quad 12m<nquad, quad n text{ is even}\text{undefined}&,quad 12m<nquad, quad n text{ is odd}end{cases}}$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Defining $sqrt{x+1}=t$ we obtain$$lim_{xto 0}{frac{{left ( left (1+x right )^{frac{1}{2}} -1 -frac{x}{2} right )}^{6m}} {{x^n}}}{=lim_{tto 1}{left(t-1-{t^2-1over 2}right)^{6m}over (t^2-1)^n}\={1over2^{6m}}lim_{tto 1}{(t-1)^{12m}over (t-1)^n(t+1)^n}\={1over2^{6m+n}}lim_{tto 1}{(t-1)^{12m}over (t-1)^n}\=begin{cases}0&,quad12m>n\{1over 2^{6m+n}}&,quad 12m=n\infty&,quad 12m<nquad, quad n text{ is even}\text{undefined}&,quad 12m<nquad, quad n text{ is odd}end{cases}}$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Defining $sqrt{x+1}=t$ we obtain$$lim_{xto 0}{frac{{left ( left (1+x right )^{frac{1}{2}} -1 -frac{x}{2} right )}^{6m}} {{x^n}}}{=lim_{tto 1}{left(t-1-{t^2-1over 2}right)^{6m}over (t^2-1)^n}\={1over2^{6m}}lim_{tto 1}{(t-1)^{12m}over (t-1)^n(t+1)^n}\={1over2^{6m+n}}lim_{tto 1}{(t-1)^{12m}over (t-1)^n}\=begin{cases}0&,quad12m>n\{1over 2^{6m+n}}&,quad 12m=n\infty&,quad 12m<nquad, quad n text{ is even}\text{undefined}&,quad 12m<nquad, quad n text{ is odd}end{cases}}$$






                  share|cite|improve this answer









                  $endgroup$



                  Defining $sqrt{x+1}=t$ we obtain$$lim_{xto 0}{frac{{left ( left (1+x right )^{frac{1}{2}} -1 -frac{x}{2} right )}^{6m}} {{x^n}}}{=lim_{tto 1}{left(t-1-{t^2-1over 2}right)^{6m}over (t^2-1)^n}\={1over2^{6m}}lim_{tto 1}{(t-1)^{12m}over (t-1)^n(t+1)^n}\={1over2^{6m+n}}lim_{tto 1}{(t-1)^{12m}over (t-1)^n}\=begin{cases}0&,quad12m>n\{1over 2^{6m+n}}&,quad 12m=n\infty&,quad 12m<nquad, quad n text{ is even}\text{undefined}&,quad 12m<nquad, quad n text{ is odd}end{cases}}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 12 at 17:27









                  Mostafa AyazMostafa Ayaz

                  15.8k3939




                  15.8k3939















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