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
edited Jan 12 at 18:31
egreg
186k1486208
asked Jan 12 at 16:45
Marc Ballestero RibóMarc Ballestero Ribó
175
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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.
add a comment |
$begingroup$
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
edited Jan 12 at 18:31
egreg
186k1486208
asked Jan 12 at 16:45
Marc Ballestero RibóMarc Ballestero Ribó
175
$endgroup$
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
add a comment |
$begingroup$
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
edited Jan 12 at 18:31
egreg
186k1486208
asked Jan 12 at 16:45
Marc Ballestero RibóMarc Ballestero Ribó
175
$endgroup$
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
edited Jan 12 at 18:31
egreg
186k1486208
asked Jan 12 at 16:45
Marc Ballestero RibóMarc Ballestero Ribó
175
edited Jan 12 at 18:31
egreg
186k1486208
asked Jan 12 at 16:45
Marc Ballestero RibóMarc Ballestero Ribó
175
edited Jan 12 at 18:31
egreg
186k1486208
edited Jan 12 at 18:31
egreg
186k1486208
edited Jan 12 at 18:31
egreg
186k1486208
186k1486208
asked Jan 12 at 16:45
Marc Ballestero RibóMarc Ballestero Ribó
175
asked Jan 12 at 16:45
Marc Ballestero RibóMarc Ballestero Ribó
175
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
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
active
oldest
votes
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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.
answered Jan 12 at 18:39
egregegreg
186k1486208
$endgroup$
add a comment |
$begingroup$
Defining $sqrt{x+1}=t$ we obtain$$lim_{xto 0}{frac{{left ( left (1+x right )^{frac{1}{2}} -1 -frac{x}{2} right )}^{6m}} {{x^n}}}{=lim_{tto 1}{left(t-1-{t^2-1over 2}right)^{6m}over (t^2-1)^n}\={1over2^{6m}}lim_{tto 1}{(t-1)^{12m}over (t-1)^n(t+1)^n}\={1over2^{6m+n}}lim_{tto 1}{(t-1)^{12m}over (t-1)^n}\=begin{cases}0&,quad12m>n\{1over 2^{6m+n}}&,quad 12m=n\infty&,quad 12m<nquad, quad n text{ is even}\text{undefined}&,quad 12m<nquad, quad n text{ is odd}end{cases}}$$
answered Jan 12 at 17:27
Mostafa AyazMostafa Ayaz
18.1k31040
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add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Jan 12 at 18:39
egregegreg
186k1486208
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 12 at 18:39
egregegreg
186k1486208
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 12 at 18:39
egregegreg
186k1486208
$endgroup$
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.
answered Jan 12 at 18:39
egregegreg
186k1486208
answered Jan 12 at 18:39
egregegreg
186k1486208
answered Jan 12 at 18:39
egregegreg
186k1486208
answered Jan 12 at 18:39
egregegreg
186k1486208
186k1486208
add a comment |
add a comment |
$begingroup$
Defining $sqrt{x+1}=t$ we obtain$$lim_{xto 0}{frac{{left ( left (1+x right )^{frac{1}{2}} -1 -frac{x}{2} right )}^{6m}} {{x^n}}}{=lim_{tto 1}{left(t-1-{t^2-1over 2}right)^{6m}over (t^2-1)^n}\={1over2^{6m}}lim_{tto 1}{(t-1)^{12m}over (t-1)^n(t+1)^n}\={1over2^{6m+n}}lim_{tto 1}{(t-1)^{12m}over (t-1)^n}\=begin{cases}0&,quad12m>n\{1over 2^{6m+n}}&,quad 12m=n\infty&,quad 12m<nquad, quad n text{ is even}\text{undefined}&,quad 12m<nquad, quad n text{ is odd}end{cases}}$$
answered Jan 12 at 17:27
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
$begingroup$
Defining $sqrt{x+1}=t$ we obtain$$lim_{xto 0}{frac{{left ( left (1+x right )^{frac{1}{2}} -1 -frac{x}{2} right )}^{6m}} {{x^n}}}{=lim_{tto 1}{left(t-1-{t^2-1over 2}right)^{6m}over (t^2-1)^n}\={1over2^{6m}}lim_{tto 1}{(t-1)^{12m}over (t-1)^n(t+1)^n}\={1over2^{6m+n}}lim_{tto 1}{(t-1)^{12m}over (t-1)^n}\=begin{cases}0&,quad12m>n\{1over 2^{6m+n}}&,quad 12m=n\infty&,quad 12m<nquad, quad n text{ is even}\text{undefined}&,quad 12m<nquad, quad n text{ is odd}end{cases}}$$
answered Jan 12 at 17:27
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
$begingroup$
Defining $sqrt{x+1}=t$ we obtain$$lim_{xto 0}{frac{{left ( left (1+x right )^{frac{1}{2}} -1 -frac{x}{2} right )}^{6m}} {{x^n}}}{=lim_{tto 1}{left(t-1-{t^2-1over 2}right)^{6m}over (t^2-1)^n}\={1over2^{6m}}lim_{tto 1}{(t-1)^{12m}over (t-1)^n(t+1)^n}\={1over2^{6m+n}}lim_{tto 1}{(t-1)^{12m}over (t-1)^n}\=begin{cases}0&,quad12m>n\{1over 2^{6m+n}}&,quad 12m=n\infty&,quad 12m<nquad, quad n text{ is even}\text{undefined}&,quad 12m<nquad, quad n text{ is odd}end{cases}}$$
answered Jan 12 at 17:27
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
Defining $sqrt{x+1}=t$ we obtain$$lim_{xto 0}{frac{{left ( left (1+x right )^{frac{1}{2}} -1 -frac{x}{2} right )}^{6m}} {{x^n}}}{=lim_{tto 1}{left(t-1-{t^2-1over 2}right)^{6m}over (t^2-1)^n}\={1over2^{6m}}lim_{tto 1}{(t-1)^{12m}over (t-1)^n(t+1)^n}\={1over2^{6m+n}}lim_{tto 1}{(t-1)^{12m}over (t-1)^n}\=begin{cases}0&,quad12m>n\{1over 2^{6m+n}}&,quad 12m=n\infty&,quad 12m<nquad, quad n text{ is even}\text{undefined}&,quad 12m<nquad, quad n text{ is odd}end{cases}}$$
answered Jan 12 at 17:27
Mostafa AyazMostafa Ayaz
18.1k31040
answered Jan 12 at 17:27
Mostafa AyazMostafa Ayaz
18.1k31040
answered Jan 12 at 17:27
Mostafa AyazMostafa Ayaz
18.1k31040
answered Jan 12 at 17:27
Mostafa AyazMostafa Ayaz
18.1k31040
18.1k31040
add a comment |
add a comment |
$begingroup$
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