Mixing
Can one find a $g(x)$ , which is differentiable in $[a,infty)$ and $lim_{xtoinfty}g(x)=infty$,but...
$begingroup$
Can one find a $g(x)$ , which is differentiable in $[a,infty)$ and $$lim_{xtoinfty}|g(x)|=infty$$
but $$lim_{xtoinfty}x^2|g'(x)|neqinfty$$
(Or doesn't exist that limit)
calculus
asked Jan 29 at 3:13
Alexander LauAlexander Lau
1308
$endgroup$
add a comment |
$begingroup$
Can one find a $g(x)$ , which is differentiable in $[a,infty)$ and $$lim_{xtoinfty}|g(x)|=infty$$
but $$lim_{xtoinfty}x^2|g'(x)|neqinfty$$
(Or doesn't exist that limit)
calculus
asked Jan 29 at 3:13
Alexander LauAlexander Lau
1308
$endgroup$
$begingroup$
I don't think there exists such a g， but I don't how to prove.
$endgroup$
– Alexander Lau
Jan 29 at 4:12
add a comment |
$begingroup$
Can one find a $g(x)$ , which is differentiable in $[a,infty)$ and $$lim_{xtoinfty}|g(x)|=infty$$
but $$lim_{xtoinfty}x^2|g'(x)|neqinfty$$
(Or doesn't exist that limit)
calculus
asked Jan 29 at 3:13
Alexander LauAlexander Lau
1308
$endgroup$
Can one find a $g(x)$ , which is differentiable in $[a,infty)$ and $$lim_{xtoinfty}|g(x)|=infty$$
but $$lim_{xtoinfty}x^2|g'(x)|neqinfty$$
(Or doesn't exist that limit)
calculus
calculus
asked Jan 29 at 3:13
Alexander LauAlexander Lau
1308
asked Jan 29 at 3:13
Alexander LauAlexander Lau
1308
asked Jan 29 at 3:13
Alexander LauAlexander Lau
1308
asked Jan 29 at 3:13
Alexander LauAlexander Lau
1308
asked Jan 29 at 3:13
Alexander LauAlexander Lau
1308
1308
$begingroup$
I don't think there exists such a g， but I don't how to prove.
$endgroup$
– Alexander Lau
Jan 29 at 4:12
add a comment |
$begingroup$
I don't think there exists such a g， but I don't how to prove.
$endgroup$
– Alexander Lau
Jan 29 at 4:12
$begingroup$
I don't think there exists such a g， but I don't how to prove.
$endgroup$
– Alexander Lau
Jan 29 at 4:12
$begingroup$
I don't think there exists such a g， but I don't how to prove.
$endgroup$
– Alexander Lau
Jan 29 at 4:12
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Since, if $g$ is at least absolutely continuous and $g'$ integrable,
$$|g(x)| leqslant |g(a)| + int_a^x |g'(t)| , dt$$
and $|g(x)| to infty$, we have
$$int_a^infty |g'(t)| , dt = infty$$
Hence, we can rule out “nice” functions where $lim_{x to infty}x^2 |g'(x)| = L < infty$ which would imply that the integral converges with $|g'(x)| = mathcal{O}(x^{-2})$ as $x to infty$.
The limit may not exist though, for example,
$$g(x) = int_1^x sin^2(t) , dt$$
where $|g(x)| to infty$ but $x^2|g'(x)|= x^2 sin^2 (x) $ does not converge as $x to infty$.
Note that $liminf_{x to infty} x^2 sin^2 (x) = 0$ and $limsup_{x to infty} x^2 sin^2 (x) = infty$.
At this point your question is answered and we don't need to explore more pathological cases.
answered Jan 29 at 4:19
RRLRRL
53.1k52574
$endgroup$
add a comment |
$begingroup$
Let $$g(x)=x+sin x$$
Then clearly, $$lim_{xrightarrow +infty}|g(x)|=+infty$$
but $$lim_{xrightarrow +infty}x^2|g^prime(x)| =lim_{xrightarrow +infty}x^2|1+cos(x)|$$ does not exist.
answered Jan 29 at 4:14
Stefan LafonStefan Lafon
3,005212
$endgroup$
$begingroup$
(+1) All too easy
$endgroup$
– Mark Viola
Jan 29 at 4:17
$begingroup$
Yes, really easy
$endgroup$
– Alexander Lau
Jan 29 at 15:10
add a comment |
$begingroup$
Let
$g(x)=xsin(x)+2x$,
then
$x^2g'(x)=x^2(sin(x)+2)+x^3cos(x)$ .
answered Jan 29 at 4:19
LZYLZY
136
$endgroup$
$begingroup$
原来是你😂😂😂😂😂😂😂😂
$endgroup$
– Alexander Lau
Feb 18 at 2:34
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3091686%2fcan-one-find-a-gx-which-is-differentiable-in-a-infty-and-lim-x-to%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since, if $g$ is at least absolutely continuous and $g'$ integrable,
$$|g(x)| leqslant |g(a)| + int_a^x |g'(t)| , dt$$
and $|g(x)| to infty$, we have
$$int_a^infty |g'(t)| , dt = infty$$
Hence, we can rule out “nice” functions where $lim_{x to infty}x^2 |g'(x)| = L < infty$ which would imply that the integral converges with $|g'(x)| = mathcal{O}(x^{-2})$ as $x to infty$.
The limit may not exist though, for example,
$$g(x) = int_1^x sin^2(t) , dt$$
where $|g(x)| to infty$ but $x^2|g'(x)|= x^2 sin^2 (x) $ does not converge as $x to infty$.
Note that $liminf_{x to infty} x^2 sin^2 (x) = 0$ and $limsup_{x to infty} x^2 sin^2 (x) = infty$.
At this point your question is answered and we don't need to explore more pathological cases.
answered Jan 29 at 4:19
RRLRRL
53.1k52574
$endgroup$
add a comment |
$begingroup$
Since, if $g$ is at least absolutely continuous and $g'$ integrable,
$$|g(x)| leqslant |g(a)| + int_a^x |g'(t)| , dt$$
and $|g(x)| to infty$, we have
$$int_a^infty |g'(t)| , dt = infty$$
Hence, we can rule out “nice” functions where $lim_{x to infty}x^2 |g'(x)| = L < infty$ which would imply that the integral converges with $|g'(x)| = mathcal{O}(x^{-2})$ as $x to infty$.
The limit may not exist though, for example,
$$g(x) = int_1^x sin^2(t) , dt$$
where $|g(x)| to infty$ but $x^2|g'(x)|= x^2 sin^2 (x) $ does not converge as $x to infty$.
Note that $liminf_{x to infty} x^2 sin^2 (x) = 0$ and $limsup_{x to infty} x^2 sin^2 (x) = infty$.
At this point your question is answered and we don't need to explore more pathological cases.
answered Jan 29 at 4:19
RRLRRL
53.1k52574
$endgroup$
add a comment |
$begingroup$
Since, if $g$ is at least absolutely continuous and $g'$ integrable,
$$|g(x)| leqslant |g(a)| + int_a^x |g'(t)| , dt$$
and $|g(x)| to infty$, we have
$$int_a^infty |g'(t)| , dt = infty$$
Hence, we can rule out “nice” functions where $lim_{x to infty}x^2 |g'(x)| = L < infty$ which would imply that the integral converges with $|g'(x)| = mathcal{O}(x^{-2})$ as $x to infty$.
The limit may not exist though, for example,
$$g(x) = int_1^x sin^2(t) , dt$$
where $|g(x)| to infty$ but $x^2|g'(x)|= x^2 sin^2 (x) $ does not converge as $x to infty$.
Note that $liminf_{x to infty} x^2 sin^2 (x) = 0$ and $limsup_{x to infty} x^2 sin^2 (x) = infty$.
At this point your question is answered and we don't need to explore more pathological cases.
answered Jan 29 at 4:19
RRLRRL
53.1k52574
$endgroup$
Since, if $g$ is at least absolutely continuous and $g'$ integrable,
$$|g(x)| leqslant |g(a)| + int_a^x |g'(t)| , dt$$
and $|g(x)| to infty$, we have
$$int_a^infty |g'(t)| , dt = infty$$
Hence, we can rule out “nice” functions where $lim_{x to infty}x^2 |g'(x)| = L < infty$ which would imply that the integral converges with $|g'(x)| = mathcal{O}(x^{-2})$ as $x to infty$.
The limit may not exist though, for example,
$$g(x) = int_1^x sin^2(t) , dt$$
where $|g(x)| to infty$ but $x^2|g'(x)|= x^2 sin^2 (x) $ does not converge as $x to infty$.
Note that $liminf_{x to infty} x^2 sin^2 (x) = 0$ and $limsup_{x to infty} x^2 sin^2 (x) = infty$.
At this point your question is answered and we don't need to explore more pathological cases.
answered Jan 29 at 4:19
RRLRRL
53.1k52574
edited Jan 29 at 7:12
answered Jan 29 at 4:19
RRLRRL
53.1k52574
answered Jan 29 at 4:19
RRLRRL
53.1k52574
answered Jan 29 at 4:19
RRLRRL
53.1k52574
53.1k52574
add a comment |
add a comment |
$begingroup$
Let $$g(x)=x+sin x$$
Then clearly, $$lim_{xrightarrow +infty}|g(x)|=+infty$$
but $$lim_{xrightarrow +infty}x^2|g^prime(x)| =lim_{xrightarrow +infty}x^2|1+cos(x)|$$ does not exist.
answered Jan 29 at 4:14
Stefan LafonStefan Lafon
3,005212
$endgroup$
$begingroup$
(+1) All too easy
$endgroup$
– Mark Viola
Jan 29 at 4:17
$begingroup$
Yes, really easy
$endgroup$
– Alexander Lau
Jan 29 at 15:10
add a comment |
$begingroup$
Let $$g(x)=x+sin x$$
Then clearly, $$lim_{xrightarrow +infty}|g(x)|=+infty$$
but $$lim_{xrightarrow +infty}x^2|g^prime(x)| =lim_{xrightarrow +infty}x^2|1+cos(x)|$$ does not exist.
answered Jan 29 at 4:14
Stefan LafonStefan Lafon
3,005212
$endgroup$
$begingroup$
(+1) All too easy
$endgroup$
– Mark Viola
Jan 29 at 4:17
$begingroup$
Yes, really easy
$endgroup$
– Alexander Lau
Jan 29 at 15:10
add a comment |
$begingroup$
Let $$g(x)=x+sin x$$
Then clearly, $$lim_{xrightarrow +infty}|g(x)|=+infty$$
but $$lim_{xrightarrow +infty}x^2|g^prime(x)| =lim_{xrightarrow +infty}x^2|1+cos(x)|$$ does not exist.
answered Jan 29 at 4:14
Stefan LafonStefan Lafon
3,005212
$endgroup$
Let $$g(x)=x+sin x$$
Then clearly, $$lim_{xrightarrow +infty}|g(x)|=+infty$$
but $$lim_{xrightarrow +infty}x^2|g^prime(x)| =lim_{xrightarrow +infty}x^2|1+cos(x)|$$ does not exist.
answered Jan 29 at 4:14
Stefan LafonStefan Lafon
3,005212
answered Jan 29 at 4:14
Stefan LafonStefan Lafon
3,005212
answered Jan 29 at 4:14
Stefan LafonStefan Lafon
3,005212
answered Jan 29 at 4:14
Stefan LafonStefan Lafon
3,005212
3,005212
$begingroup$
(+1) All too easy
$endgroup$
– Mark Viola
Jan 29 at 4:17
$begingroup$
Yes, really easy
$endgroup$
– Alexander Lau
Jan 29 at 15:10
add a comment |
$begingroup$
(+1) All too easy
$endgroup$
– Mark Viola
Jan 29 at 4:17
$begingroup$
Yes, really easy
$endgroup$
– Alexander Lau
Jan 29 at 15:10
$begingroup$
(+1) All too easy
$endgroup$
– Mark Viola
Jan 29 at 4:17
$begingroup$
(+1) All too easy
$endgroup$
– Mark Viola
Jan 29 at 4:17
$begingroup$
Yes, really easy
$endgroup$
– Alexander Lau
Jan 29 at 15:10
$begingroup$
Yes, really easy
$endgroup$
– Alexander Lau
Jan 29 at 15:10
add a comment |
$begingroup$
Let
$g(x)=xsin(x)+2x$,
then
$x^2g'(x)=x^2(sin(x)+2)+x^3cos(x)$ .
answered Jan 29 at 4:19
LZYLZY
136
$endgroup$
$begingroup$
原来是你😂😂😂😂😂😂😂😂
$endgroup$
– Alexander Lau
Feb 18 at 2:34
add a comment |
$begingroup$
Let
$g(x)=xsin(x)+2x$,
then
$x^2g'(x)=x^2(sin(x)+2)+x^3cos(x)$ .
answered Jan 29 at 4:19
LZYLZY
136
$endgroup$
$begingroup$
原来是你😂😂😂😂😂😂😂😂
$endgroup$
– Alexander Lau
Feb 18 at 2:34
add a comment |
$begingroup$
Let
$g(x)=xsin(x)+2x$,
then
$x^2g'(x)=x^2(sin(x)+2)+x^3cos(x)$ .
answered Jan 29 at 4:19
LZYLZY
136
$endgroup$
Let
$g(x)=xsin(x)+2x$,
then
$x^2g'(x)=x^2(sin(x)+2)+x^3cos(x)$ .
answered Jan 29 at 4:19
LZYLZY
136
edited Feb 18 at 5:31
answered Jan 29 at 4:19
LZYLZY
136
answered Jan 29 at 4:19
LZYLZY
136
answered Jan 29 at 4:19
LZYLZY
136
136
$begingroup$
原来是你😂😂😂😂😂😂😂😂
$endgroup$
– Alexander Lau
Feb 18 at 2:34
add a comment |
$begingroup$
原来是你😂😂😂😂😂😂😂😂
$endgroup$
– Alexander Lau
Feb 18 at 2:34
$begingroup$
原来是你😂😂😂😂😂😂😂😂
$endgroup$
– Alexander Lau
Feb 18 at 2:34
$begingroup$
原来是你😂😂😂😂😂😂😂😂
$endgroup$
– Alexander Lau
Feb 18 at 2:34
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3091686%2fcan-one-find-a-gx-which-is-differentiable-in-a-infty-and-lim-x-to%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
I don't think there exists such a g， but I don't how to prove.
$endgroup$
– Alexander Lau
Jan 29 at 4:12