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Can we cancel the equality mark here?

Limit of function as $x toinfty $ when $f'(x)$ is givenProving that $limlimits_ntoinftyint_1^inftyfrac1nxdx neq 0$how to compute this limits given these conditions.How to evaluate the $limlimits_xto 0frac 2sin(x)-arctan(x)-xcos(x^2)x^5$, using power series?Limit of $x(2pi - 4arctan(x))$Can we still use L'Hospital's rule on limit that does not exists?Calculate the limit: $lim_ntoinftyn^2left(left(1+frac1nright)^8-left(1+frac2nright)^4right)$Limit of $dfrace^-1/x^n$ as $xto 0$Show that $limlimits_n to inftysin n^2$ does not exist.How to calculate the limit $limlimits_ntoinfty dfracx_ny_n$ where $0<x_0<y_0<dfracpi2$ and $x_n+1=sinx_n, y_n+1=siny_n$?Why am I computing $lim limits_x to infty x left(arctan fracx+1x+2 -arctan fracxx+2 right)$ wrong?

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 Apr 22 at 15:23

LutzL

61.2k42157

asked Apr 22 at 10:35

mengdie1982mengdie1982

5,161621

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PeterForeman Sir, $f(x)=-frac1x^2<0$ but $limlimits_x to +inftyf(x)=0.$
  $endgroup$
  – mengdie1982
  Apr 22 at 10:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you asked that the limit is smaller than $1+pi/2$ or than $1+pi/4$?
  $endgroup$
  – kingW3
  Apr 22 at 10:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kingW3 No. I just wonder whether the equality with the inequality may hold or not.
  $endgroup$
  – mengdie1982
  Apr 22 at 10:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry. A typo in the "problem". Corrected. See the new version.
  $endgroup$
  – mengdie1982
  Apr 22 at 10:48
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Limit of function as $x toinfty $ when $f'(x)$ is given
  $endgroup$
  – LutzL
  Apr 22 at 15:26
  

  
  
  
  

 | 
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 Apr 22 at 15:23

LutzL

61.2k42157

asked Apr 22 at 10:35

mengdie1982mengdie1982

5,161621

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PeterForeman Sir, $f(x)=-frac1x^2<0$ but $limlimits_x to +inftyf(x)=0.$
  $endgroup$
  – mengdie1982
  Apr 22 at 10:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you asked that the limit is smaller than $1+pi/2$ or than $1+pi/4$?
  $endgroup$
  – kingW3
  Apr 22 at 10:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kingW3 No. I just wonder whether the equality with the inequality may hold or not.
  $endgroup$
  – mengdie1982
  Apr 22 at 10:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry. A typo in the "problem". Corrected. See the new version.
  $endgroup$
  – mengdie1982
  Apr 22 at 10:48
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Limit of function as $x toinfty $ when $f'(x)$ is given
  $endgroup$
  – LutzL
  Apr 22 at 15:26
  

  
  
  
  

 | 
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 Apr 22 at 15:23

LutzL

61.2k42157

asked Apr 22 at 10:35

mengdie1982mengdie1982

5,161621

$endgroup$

share|cite|improve this question

edited Apr 22 at 15:23

LutzL

61.2k42157

asked Apr 22 at 10:35

mengdie1982mengdie1982

5,161621

share|cite|improve this question

edited Apr 22 at 15:23

LutzL

61.2k42157

asked Apr 22 at 10:35

mengdie1982mengdie1982

5,161621

edited Apr 22 at 15:23

LutzL

61.2k42157

edited Apr 22 at 15:23

LutzL

61.2k42157

asked Apr 22 at 10:35

mengdie1982mengdie1982

5,161621

asked Apr 22 at 10:35

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 Apr 22 at 12:01

answered Apr 22 at 11:04

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
  Apr 22 at 11:43
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  Apr 22 at 11:51
  
  

  
  
  
  

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 Apr 22 at 16:54

Hagen von EitzenHagen von Eitzen

284k23274508

$endgroup$

add a comment | 

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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 Apr 22 at 12:01

answered Apr 22 at 11:04

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
  Apr 22 at 11:43
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  Apr 22 at 11:51
  
  

  
  
  
  

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 Apr 22 at 12:01

answered Apr 22 at 11:04

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
  Apr 22 at 11:43
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  Apr 22 at 11:51
  
  

  
  
  
  

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 Apr 22 at 12:01

answered Apr 22 at 11:04

kingW3kingW3

11.2k72656

$endgroup$

share|cite|improve this answer

edited Apr 22 at 12:01

answered Apr 22 at 11:04

kingW3kingW3

11.2k72656

answered Apr 22 at 11:04

kingW3kingW3

11.2k72656

answered Apr 22 at 11:04

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 Apr 22 at 16:54

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 Apr 22 at 16:54

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 Apr 22 at 16:54

Hagen von EitzenHagen von Eitzen

284k23274508

$endgroup$

share|cite|improve this answer

answered Apr 22 at 16:54

Hagen von EitzenHagen von Eitzen

284k23274508

answered Apr 22 at 16:54

Hagen von EitzenHagen von Eitzen

284k23274508

answered Apr 22 at 16:54

Hagen von EitzenHagen von Eitzen

284k23274508