Can you stand up from being prone using Skirmisher outside of your turn?

What was Apollo 13's "Little Jolt" after MECO?

The art of proof summarizing. Are there known rules, or is it a purely common sense matter?

What is /etc/mtab in Linux?

Is Electric Central Heating worth it if using Solar Panels?

What is the term for a person whose job is to place products on shelves in stores?

A Paper Record is What I Hamper

Can I criticise the more senior developers around me for not writing clean code?

Need of separate security plugins for both root and subfolder sites Wordpress?

Implementing 3DES algorithm in Java: is my code secure?

How to use @AuraEnabled base class method in Lightning Component?

Why does the Cisco show run command not show the full version, while the show version command does?

Could moose/elk survive in the Amazon forest?

Second order approximation of the loss function (Deep learning book, 7.33)

Is a 5 watt UHF/VHF handheld considered QRP?

Passing args from the bash script to the function in the script

Why didn't the Space Shuttle bounce back into space as many times as possible so as to lose a lot of kinetic energy up there?

What is it called when you ride around on your front wheel?

How long after the last departure shall the airport stay open for an emergency return?

Why isn't everyone flabbergasted about Bran's "gift"?

Why is an operator the quantum mechanical analogue of an observable?

Map material from china not allowed to leave the country

How would this chord from "Rocket Man" be analyzed?

Is accepting an invalid credit card number a security issue?

Can we cancel the equality mark here?

4

$begingroup$

Problem

Let $f(x)$ satisfy that $f(1)=1$ and $f'(x)=dfrac1x^2+f^2(x)$. Prove that $limlimits_x to +inftyf(x)$ exists and is less than $1+dfracpi4.$

Proof

Since $f'(x)=dfrac1x^2+f'(x)>0$, $f(x)$ is strictly increasing. Thus, $f(x)>f(1)=1$ holds for all $x>1$, and $limlimits_x to +inftyf(x)$ equals either the positive infinity or some finite value.

Notice that, $forall x>1:$
beginalign*
f(x)-f(1)&=int_1^x f'(t)rm dt=int_1^x frac1t^2+f^2(t)rm dt<int_1^xfrac1t^2+1rm dt=arctan x-fracpi4.
endalign*
Therefore
$$f(x)<arctan x-fracpi4+1<fracpi2-fracpi4+1=1+fracpi4,$$
which implies that $f(x)$ is bounded upward. Thus,$limlimits_x to +inftyf(x)$ exists. Take the limits as $x to +infty$, we have
$limlimits_x to +inftyf(x)leq 1+dfracpi4.$
Can we cancel the equality mark here? In another word, can we obtain $limlimits_x to +inftyf(x)<1+dfracpi4$?

calculus ordinary-differential-equations limits

share|cite|improve this question

edited 2 days ago

LutzL

61.1k42157

asked 2 days ago

mengdie1982mengdie1982

5,161621

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PeterForeman Sir, $f(x)=-frac1x^2<0$ but $limlimits_x to +inftyf(x)=0.$
  $endgroup$
  – mengdie1982
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you asked that the limit is smaller than $1+pi/2$ or than $1+pi/4$?
  $endgroup$
  – kingW3
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kingW3 No. I just wonder whether the equality with the inequality may hold or not.
  $endgroup$
  – mengdie1982
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry. A typo in the "problem". Corrected. See the new version.
  $endgroup$
  – mengdie1982
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Limit of function as $x toinfty $ when $f'(x)$ is given
  $endgroup$
  – LutzL
  2 days ago
  

  
  
  
  

 | 
show 2 more comments

4

$begingroup$

Problem

Let $f(x)$ satisfy that $f(1)=1$ and $f'(x)=dfrac1x^2+f^2(x)$. Prove that $limlimits_x to +inftyf(x)$ exists and is less than $1+dfracpi4.$

Proof

Since $f'(x)=dfrac1x^2+f'(x)>0$, $f(x)$ is strictly increasing. Thus, $f(x)>f(1)=1$ holds for all $x>1$, and $limlimits_x to +inftyf(x)$ equals either the positive infinity or some finite value.

Notice that, $forall x>1:$
beginalign*
f(x)-f(1)&=int_1^x f'(t)rm dt=int_1^x frac1t^2+f^2(t)rm dt<int_1^xfrac1t^2+1rm dt=arctan x-fracpi4.
endalign*
Therefore
$$f(x)<arctan x-fracpi4+1<fracpi2-fracpi4+1=1+fracpi4,$$
which implies that $f(x)$ is bounded upward. Thus,$limlimits_x to +inftyf(x)$ exists. Take the limits as $x to +infty$, we have
$limlimits_x to +inftyf(x)leq 1+dfracpi4.$
Can we cancel the equality mark here? In another word, can we obtain $limlimits_x to +inftyf(x)<1+dfracpi4$?

calculus ordinary-differential-equations limits

share|cite|improve this question

edited 2 days ago

LutzL

61.1k42157

asked 2 days ago

mengdie1982mengdie1982

5,161621

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PeterForeman Sir, $f(x)=-frac1x^2<0$ but $limlimits_x to +inftyf(x)=0.$
  $endgroup$
  – mengdie1982
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you asked that the limit is smaller than $1+pi/2$ or than $1+pi/4$?
  $endgroup$
  – kingW3
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kingW3 No. I just wonder whether the equality with the inequality may hold or not.
  $endgroup$
  – mengdie1982
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry. A typo in the "problem". Corrected. See the new version.
  $endgroup$
  – mengdie1982
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Limit of function as $x toinfty $ when $f'(x)$ is given
  $endgroup$
  – LutzL
  2 days ago
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

Problem

Let $f(x)$ satisfy that $f(1)=1$ and $f'(x)=dfrac1x^2+f^2(x)$. Prove that $limlimits_x to +inftyf(x)$ exists and is less than $1+dfracpi4.$

Proof

Since $f'(x)=dfrac1x^2+f'(x)>0$, $f(x)$ is strictly increasing. Thus, $f(x)>f(1)=1$ holds for all $x>1$, and $limlimits_x to +inftyf(x)$ equals either the positive infinity or some finite value.

Notice that, $forall x>1:$
beginalign*
f(x)-f(1)&=int_1^x f'(t)rm dt=int_1^x frac1t^2+f^2(t)rm dt<int_1^xfrac1t^2+1rm dt=arctan x-fracpi4.
endalign*
Therefore
$$f(x)<arctan x-fracpi4+1<fracpi2-fracpi4+1=1+fracpi4,$$
which implies that $f(x)$ is bounded upward. Thus,$limlimits_x to +inftyf(x)$ exists. Take the limits as $x to +infty$, we have
$limlimits_x to +inftyf(x)leq 1+dfracpi4.$
Can we cancel the equality mark here? In another word, can we obtain $limlimits_x to +inftyf(x)<1+dfracpi4$?

calculus ordinary-differential-equations limits

share|cite|improve this question

edited 2 days ago

LutzL

61.1k42157

asked 2 days ago

mengdie1982mengdie1982

5,161621

$endgroup$

share|cite|improve this question

edited 2 days ago

LutzL

61.1k42157

asked 2 days ago

mengdie1982mengdie1982

5,161621

share|cite|improve this question

edited 2 days ago

LutzL

61.1k42157

asked 2 days ago

mengdie1982mengdie1982

5,161621

edited 2 days ago

LutzL

61.1k42157

edited 2 days ago

LutzL

61.1k42157

asked 2 days ago

mengdie1982mengdie1982

5,161621

asked 2 days ago

mengdie1982mengdie1982

5,161621

2 Answers
2

active

oldest

votes

4

$begingroup$

The function $$g(x)=int_1^xfrac1t^2+1rm dt-int_1^x frac1t^2+f^2(t)rm dt$$
Is strictly increasing and $g(1)=0<g(2)<g(x)$ for $x>2$ hence $lim_xtoinftyg(x)geq g(2)>0$ so$$lim_xtoinftyg(x)=lim_xtoinfty(fracpi4-(f(x)-1))=lim_xtoinfty(fracpi4+1-f(x))>0$$

So $$lim_xtoinftyf(x)<fracpi4+1$$

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

kingW3kingW3

11.2k72656

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  In general, $f(x)>g(x)$ implies $ lim f(x) geq lim g(x)$ not $ lim f(x) > lim g(x)$...
  $endgroup$
  – mengdie1982
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mengdie1982 Yeah but I didn't use that in my proof. I've used that $g$ is strictly increasing and that $g(x)>0$ for $x>1$. Using that you could prove $g(2)<g(x)$ for $x>2$ hence $0<lim_xtoinfty g(2) leq lim_xtoinfty g(x)$.
  $endgroup$
  – kingW3
  2 days ago
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Fix $M>1$ and for $x>M$ break up your estimate as
$$ f(x)-f(M)<arctan x- arctan M$$
and
$$ f(M)-f(1)<arctan M-fracpi 4,$$
so there is a positive constant (i.e., depending only on $M$, but not on $x$) $delta_M:=arctan M-fracpi 4-f(M)+f(1)$. Then for $x>M$,
$$ f(x)-f(1)=f(x)-f(M)+f(M)-f(1)<arctan x-frac pi 4-delta_M$$
and so
$$ lim_xtoinftyf(x)le 1+fracpi 4-delta_M<1+fracpi 4.$$

share|cite|improve this answer

answered 2 days ago

Hagen von EitzenHagen von Eitzen

284k23274508

$endgroup$

add a comment | 

Your Answer

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
);



);

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3196923%2fcan-we-cancel-the-equality-mark-here%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

4

$begingroup$

The function $$g(x)=int_1^xfrac1t^2+1rm dt-int_1^x frac1t^2+f^2(t)rm dt$$
Is strictly increasing and $g(1)=0<g(2)<g(x)$ for $x>2$ hence $lim_xtoinftyg(x)geq g(2)>0$ so$$lim_xtoinftyg(x)=lim_xtoinfty(fracpi4-(f(x)-1))=lim_xtoinfty(fracpi4+1-f(x))>0$$

So $$lim_xtoinftyf(x)<fracpi4+1$$

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

kingW3kingW3

11.2k72656

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  In general, $f(x)>g(x)$ implies $ lim f(x) geq lim g(x)$ not $ lim f(x) > lim g(x)$...
  $endgroup$
  – mengdie1982
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mengdie1982 Yeah but I didn't use that in my proof. I've used that $g$ is strictly increasing and that $g(x)>0$ for $x>1$. Using that you could prove $g(2)<g(x)$ for $x>2$ hence $0<lim_xtoinfty g(2) leq lim_xtoinfty g(x)$.
  $endgroup$
  – kingW3
  2 days ago
  
  

  
  
  
  

add a comment | 

4

$begingroup$

The function $$g(x)=int_1^xfrac1t^2+1rm dt-int_1^x frac1t^2+f^2(t)rm dt$$
Is strictly increasing and $g(1)=0<g(2)<g(x)$ for $x>2$ hence $lim_xtoinftyg(x)geq g(2)>0$ so$$lim_xtoinftyg(x)=lim_xtoinfty(fracpi4-(f(x)-1))=lim_xtoinfty(fracpi4+1-f(x))>0$$

So $$lim_xtoinftyf(x)<fracpi4+1$$

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

kingW3kingW3

11.2k72656

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  In general, $f(x)>g(x)$ implies $ lim f(x) geq lim g(x)$ not $ lim f(x) > lim g(x)$...
  $endgroup$
  – mengdie1982
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mengdie1982 Yeah but I didn't use that in my proof. I've used that $g$ is strictly increasing and that $g(x)>0$ for $x>1$. Using that you could prove $g(2)<g(x)$ for $x>2$ hence $0<lim_xtoinfty g(2) leq lim_xtoinfty g(x)$.
  $endgroup$
  – kingW3
  2 days ago
  
  

  
  
  
  

add a comment | 

$begingroup$

The function $$g(x)=int_1^xfrac1t^2+1rm dt-int_1^x frac1t^2+f^2(t)rm dt$$
Is strictly increasing and $g(1)=0<g(2)<g(x)$ for $x>2$ hence $lim_xtoinftyg(x)geq g(2)>0$ so$$lim_xtoinftyg(x)=lim_xtoinfty(fracpi4-(f(x)-1))=lim_xtoinfty(fracpi4+1-f(x))>0$$

So $$lim_xtoinftyf(x)<fracpi4+1$$

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

kingW3kingW3

11.2k72656

$endgroup$

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

kingW3kingW3

11.2k72656

answered 2 days ago

kingW3kingW3

11.2k72656

answered 2 days ago

kingW3kingW3

11.2k72656

0

$begingroup$

Fix $M>1$ and for $x>M$ break up your estimate as
$$ f(x)-f(M)<arctan x- arctan M$$
and
$$ f(M)-f(1)<arctan M-fracpi 4,$$
so there is a positive constant (i.e., depending only on $M$, but not on $x$) $delta_M:=arctan M-fracpi 4-f(M)+f(1)$. Then for $x>M$,
$$ f(x)-f(1)=f(x)-f(M)+f(M)-f(1)<arctan x-frac pi 4-delta_M$$
and so
$$ lim_xtoinftyf(x)le 1+fracpi 4-delta_M<1+fracpi 4.$$

share|cite|improve this answer

answered 2 days ago

Hagen von EitzenHagen von Eitzen

284k23274508

$endgroup$

add a comment | 

0

$begingroup$

Fix $M>1$ and for $x>M$ break up your estimate as
$$ f(x)-f(M)<arctan x- arctan M$$
and
$$ f(M)-f(1)<arctan M-fracpi 4,$$
so there is a positive constant (i.e., depending only on $M$, but not on $x$) $delta_M:=arctan M-fracpi 4-f(M)+f(1)$. Then for $x>M$,
$$ f(x)-f(1)=f(x)-f(M)+f(M)-f(1)<arctan x-frac pi 4-delta_M$$
and so
$$ lim_xtoinftyf(x)le 1+fracpi 4-delta_M<1+fracpi 4.$$

share|cite|improve this answer

answered 2 days ago

Hagen von EitzenHagen von Eitzen

284k23274508

$endgroup$

add a comment | 

$begingroup$

Fix $M>1$ and for $x>M$ break up your estimate as
$$ f(x)-f(M)<arctan x- arctan M$$
and
$$ f(M)-f(1)<arctan M-fracpi 4,$$
so there is a positive constant (i.e., depending only on $M$, but not on $x$) $delta_M:=arctan M-fracpi 4-f(M)+f(1)$. Then for $x>M$,
$$ f(x)-f(1)=f(x)-f(M)+f(M)-f(1)<arctan x-frac pi 4-delta_M$$
and so
$$ lim_xtoinftyf(x)le 1+fracpi 4-delta_M<1+fracpi 4.$$

share|cite|improve this answer

answered 2 days ago

Hagen von EitzenHagen von Eitzen

284k23274508

$endgroup$

share|cite|improve this answer

answered 2 days ago

Hagen von EitzenHagen von Eitzen

284k23274508

answered 2 days ago

Hagen von EitzenHagen von Eitzen

284k23274508

answered 2 days ago

Hagen von EitzenHagen von Eitzen

284k23274508