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Suppose $f$ is defined and differentiable proof.

0

$begingroup$

Suppose $f$ is defined and differentiable for every $x>0$, and $lim_xto infty f'(x)=0$.

Put $g(x)=f(x+1)-f(x)$. Prove that $lim_xto infty g(x)=0$

calculus analysis

share|cite|improve this question

edited Mar 27 at 20:31

Bernard

124k742117

asked Mar 27 at 20:11

BernhardRiemannBernhardRiemann

12

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To get focused answers, it is advisable to indicate effort in your question: what have you tried? Where are you stuck? What are you unsure about? Do you know the definitions of all the concepts involved?
  $endgroup$
  – avs
  Mar 27 at 20:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You keep adding new information in the comments about what your professor said to do or not to do. This information should really be in the question: otherwise, people keep trying to answer with incomplete information.
  $endgroup$
  – avs
  Mar 28 at 5:54
  

  
  
  
  

add a comment | 

0

$begingroup$

Suppose $f$ is defined and differentiable for every $x>0$, and $lim_xto infty f'(x)=0$.

Put $g(x)=f(x+1)-f(x)$. Prove that $lim_xto infty g(x)=0$

calculus analysis

share|cite|improve this question

edited Mar 27 at 20:31

Bernard

124k742117

asked Mar 27 at 20:11

BernhardRiemannBernhardRiemann

12

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To get focused answers, it is advisable to indicate effort in your question: what have you tried? Where are you stuck? What are you unsure about? Do you know the definitions of all the concepts involved?
  $endgroup$
  – avs
  Mar 27 at 20:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You keep adding new information in the comments about what your professor said to do or not to do. This information should really be in the question: otherwise, people keep trying to answer with incomplete information.
  $endgroup$
  – avs
  Mar 28 at 5:54
  

  
  
  
  

add a comment | 

$begingroup$

Suppose $f$ is defined and differentiable for every $x>0$, and $lim_xto infty f'(x)=0$.

Put $g(x)=f(x+1)-f(x)$. Prove that $lim_xto infty g(x)=0$

calculus analysis

share|cite|improve this question

edited Mar 27 at 20:31

Bernard

124k742117

asked Mar 27 at 20:11

BernhardRiemannBernhardRiemann

12

$endgroup$

share|cite|improve this question

edited Mar 27 at 20:31

Bernard

124k742117

asked Mar 27 at 20:11

BernhardRiemannBernhardRiemann

12

share|cite|improve this question

edited Mar 27 at 20:31

Bernard

124k742117

asked Mar 27 at 20:11

BernhardRiemannBernhardRiemann

12

edited Mar 27 at 20:31

Bernard

124k742117

edited Mar 27 at 20:31

Bernard

124k742117

asked Mar 27 at 20:11

BernhardRiemannBernhardRiemann

12

asked Mar 27 at 20:11

BernhardRiemannBernhardRiemann

12

5 Answers
5

active

oldest

votes

2

$begingroup$

Hint:
$$
f(x + 1) - f(x) = int_x^x+1 f'(s) ; ds,
$$
and $f'(s)$ is "getting smaller" as $s$ increases.

share|cite|improve this answer

answered Mar 27 at 20:16

avsavs

4,280515

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i am confused because our professor said the derivative use the definition
  $endgroup$
  – BernhardRiemann
  Mar 27 at 20:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  sorry, not understanding your comment.
  $endgroup$
  – avs
  Mar 27 at 21:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For each positive $epsilon$, there exists an $M_epsilon$ such that $$ |f'(s)| < epsilon quad mbox for s > M_epsilon. $$ Therefore, for $x > M_epsilon$, $$ |f(x+1) - f(x)| = left| int_x^x+1 f'(s) ds right| leq int_x^x+1 |f'(s)| ds leq int_x^x+1 ; epsilon ; ds. $$
  $endgroup$
  – avs
  Mar 28 at 16:19
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Need to show: given any positive epsilon, there's a positive M such that |g(x)| is less than epsilon for x beyond M.

So, take an arbitrary positive epsilon.
By the same limit definition, there exists an "M" such that |f'(x)| is small for x beyond M.

Using the mean-value theorem I think you can show that the same M works for |g(x)| too.

share|cite|improve this answer

answered Mar 27 at 20:47

bluebirdbluebird

214

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes , our professor mentioned about exactly but I did not understand anything
  $endgroup$
  – BernhardRiemann
  Mar 27 at 21:23
  

  
  
  
  

add a comment | 

1

$begingroup$

By the mean value theorem for differentiable functions, for $x>0$ there exists $x< c_x< x+1$ such that $g(x)=f(x+1)-f(x)=f'(c_x)$.

If $xrightarrow infty$, then $c_x rightarrow infty$. Therefore,
$lim_xrightarrow inftyg(x)=lim_xrightarrowinftyf'(c_x)=lim_xrightarrowinftyf'(x)=0$

share|cite|improve this answer

answered Mar 27 at 21:37

folouer of kaklasfolouer of kaklas

338110

$endgroup$

add a comment | 

1

$begingroup$

MVT:

$g(x)=f(x+1)-f(x)= f'(t_x) cdot 1$; where $x <t_x <x+1$.

We have $lim_x rightarrow infty f'(x)=0:$

Let $epsilon >0$ be given.

There is a $M >0$, real, s.t. for $x >M$:

$|f'(x)| < epsilon.$

For $x >M$:

Since $g(x)=f'(t_x)$ , where $x <t_x,$ we have

$|g(x)| =|f'(t_x)| < epsilon,$ i .e

$lim_x rightarrow inftyg(x)=0.$

share|cite|improve this answer

edited Mar 28 at 4:29

answered Mar 27 at 22:27

Peter SzilasPeter Szilas

12k2822

$endgroup$

add a comment | 

0

$begingroup$

The proof is simple:

$lim_x rightarrow infty g(x)=lim_xrightarrow infty f(x+1)-lim_xrightarrow infty f(x)$

For large $x$, $x+1 approx x$, so, by the previous result,

$lim_x rightarrow infty g(x)=0$

share|cite|improve this answer

answered Mar 27 at 21:24

officialnoriaofficialnoria

112

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  One has to also prove that $lim_xrightarrow infty f(x)$ exists!
  $endgroup$
  – folouer of kaklas
  Mar 27 at 21:30
  
  

  
  
  
  

add a comment | 

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2

$begingroup$

Hint:
$$
f(x + 1) - f(x) = int_x^x+1 f'(s) ; ds,
$$
and $f'(s)$ is "getting smaller" as $s$ increases.

share|cite|improve this answer

answered Mar 27 at 20:16

avsavs

4,280515

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i am confused because our professor said the derivative use the definition
  $endgroup$
  – BernhardRiemann
  Mar 27 at 20:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  sorry, not understanding your comment.
  $endgroup$
  – avs
  Mar 27 at 21:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For each positive $epsilon$, there exists an $M_epsilon$ such that $$ |f'(s)| < epsilon quad mbox for s > M_epsilon. $$ Therefore, for $x > M_epsilon$, $$ |f(x+1) - f(x)| = left| int_x^x+1 f'(s) ds right| leq int_x^x+1 |f'(s)| ds leq int_x^x+1 ; epsilon ; ds. $$
  $endgroup$
  – avs
  Mar 28 at 16:19
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Hint:
$$
f(x + 1) - f(x) = int_x^x+1 f'(s) ; ds,
$$
and $f'(s)$ is "getting smaller" as $s$ increases.

share|cite|improve this answer

answered Mar 27 at 20:16

avsavs

4,280515

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i am confused because our professor said the derivative use the definition
  $endgroup$
  – BernhardRiemann
  Mar 27 at 20:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  sorry, not understanding your comment.
  $endgroup$
  – avs
  Mar 27 at 21:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For each positive $epsilon$, there exists an $M_epsilon$ such that $$ |f'(s)| < epsilon quad mbox for s > M_epsilon. $$ Therefore, for $x > M_epsilon$, $$ |f(x+1) - f(x)| = left| int_x^x+1 f'(s) ds right| leq int_x^x+1 |f'(s)| ds leq int_x^x+1 ; epsilon ; ds. $$
  $endgroup$
  – avs
  Mar 28 at 16:19
  
  

  
  
  
  

add a comment | 

$begingroup$

Hint:
$$
f(x + 1) - f(x) = int_x^x+1 f'(s) ; ds,
$$
and $f'(s)$ is "getting smaller" as $s$ increases.

share|cite|improve this answer

answered Mar 27 at 20:16

avsavs

4,280515

$endgroup$

share|cite|improve this answer

answered Mar 27 at 20:16

avsavs

4,280515

answered Mar 27 at 20:16

avsavs

4,280515

answered Mar 27 at 20:16

avsavs

4,280515

1

$begingroup$

Need to show: given any positive epsilon, there's a positive M such that |g(x)| is less than epsilon for x beyond M.

So, take an arbitrary positive epsilon.
By the same limit definition, there exists an "M" such that |f'(x)| is small for x beyond M.

Using the mean-value theorem I think you can show that the same M works for |g(x)| too.

share|cite|improve this answer

answered Mar 27 at 20:47

bluebirdbluebird

214

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes , our professor mentioned about exactly but I did not understand anything
  $endgroup$
  – BernhardRiemann
  Mar 27 at 21:23
  

  
  
  
  

add a comment | 

1

$begingroup$

Need to show: given any positive epsilon, there's a positive M such that |g(x)| is less than epsilon for x beyond M.

So, take an arbitrary positive epsilon.
By the same limit definition, there exists an "M" such that |f'(x)| is small for x beyond M.

Using the mean-value theorem I think you can show that the same M works for |g(x)| too.

share|cite|improve this answer

answered Mar 27 at 20:47

bluebirdbluebird

214

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes , our professor mentioned about exactly but I did not understand anything
  $endgroup$
  – BernhardRiemann
  Mar 27 at 21:23
  

  
  
  
  

add a comment | 

$begingroup$

Need to show: given any positive epsilon, there's a positive M such that |g(x)| is less than epsilon for x beyond M.

So, take an arbitrary positive epsilon.
By the same limit definition, there exists an "M" such that |f'(x)| is small for x beyond M.

Using the mean-value theorem I think you can show that the same M works for |g(x)| too.

share|cite|improve this answer

answered Mar 27 at 20:47

bluebirdbluebird

214

$endgroup$

share|cite|improve this answer

answered Mar 27 at 20:47

bluebirdbluebird

214

answered Mar 27 at 20:47

bluebirdbluebird

214

answered Mar 27 at 20:47

bluebirdbluebird

214

1

$begingroup$

By the mean value theorem for differentiable functions, for $x>0$ there exists $x< c_x< x+1$ such that $g(x)=f(x+1)-f(x)=f'(c_x)$.

If $xrightarrow infty$, then $c_x rightarrow infty$. Therefore,
$lim_xrightarrow inftyg(x)=lim_xrightarrowinftyf'(c_x)=lim_xrightarrowinftyf'(x)=0$

share|cite|improve this answer

answered Mar 27 at 21:37

folouer of kaklasfolouer of kaklas

338110

$endgroup$

add a comment | 

1

$begingroup$

By the mean value theorem for differentiable functions, for $x>0$ there exists $x< c_x< x+1$ such that $g(x)=f(x+1)-f(x)=f'(c_x)$.

If $xrightarrow infty$, then $c_x rightarrow infty$. Therefore,
$lim_xrightarrow inftyg(x)=lim_xrightarrowinftyf'(c_x)=lim_xrightarrowinftyf'(x)=0$

share|cite|improve this answer

answered Mar 27 at 21:37

folouer of kaklasfolouer of kaklas

338110

$endgroup$

add a comment | 

$begingroup$

By the mean value theorem for differentiable functions, for $x>0$ there exists $x< c_x< x+1$ such that $g(x)=f(x+1)-f(x)=f'(c_x)$.

If $xrightarrow infty$, then $c_x rightarrow infty$. Therefore,
$lim_xrightarrow inftyg(x)=lim_xrightarrowinftyf'(c_x)=lim_xrightarrowinftyf'(x)=0$

share|cite|improve this answer

answered Mar 27 at 21:37

folouer of kaklasfolouer of kaklas

338110

$endgroup$

share|cite|improve this answer

answered Mar 27 at 21:37

folouer of kaklasfolouer of kaklas

338110

answered Mar 27 at 21:37

folouer of kaklasfolouer of kaklas

338110

answered Mar 27 at 21:37

folouer of kaklasfolouer of kaklas

338110

1

$begingroup$

MVT:

$g(x)=f(x+1)-f(x)= f'(t_x) cdot 1$; where $x <t_x <x+1$.

We have $lim_x rightarrow infty f'(x)=0:$

Let $epsilon >0$ be given.

There is a $M >0$, real, s.t. for $x >M$:

$|f'(x)| < epsilon.$

For $x >M$:

Since $g(x)=f'(t_x)$ , where $x <t_x,$ we have

$|g(x)| =|f'(t_x)| < epsilon,$ i .e

$lim_x rightarrow inftyg(x)=0.$

share|cite|improve this answer

edited Mar 28 at 4:29

answered Mar 27 at 22:27

Peter SzilasPeter Szilas

12k2822

$endgroup$

add a comment | 

1

$begingroup$

MVT:

$g(x)=f(x+1)-f(x)= f'(t_x) cdot 1$; where $x <t_x <x+1$.

We have $lim_x rightarrow infty f'(x)=0:$

Let $epsilon >0$ be given.

There is a $M >0$, real, s.t. for $x >M$:

$|f'(x)| < epsilon.$

For $x >M$:

Since $g(x)=f'(t_x)$ , where $x <t_x,$ we have

$|g(x)| =|f'(t_x)| < epsilon,$ i .e

$lim_x rightarrow inftyg(x)=0.$

share|cite|improve this answer

edited Mar 28 at 4:29

answered Mar 27 at 22:27

Peter SzilasPeter Szilas

12k2822

$endgroup$

add a comment | 

$begingroup$

MVT:

$g(x)=f(x+1)-f(x)= f'(t_x) cdot 1$; where $x <t_x <x+1$.

We have $lim_x rightarrow infty f'(x)=0:$

Let $epsilon >0$ be given.

There is a $M >0$, real, s.t. for $x >M$:

$|f'(x)| < epsilon.$

For $x >M$:

Since $g(x)=f'(t_x)$ , where $x <t_x,$ we have

$|g(x)| =|f'(t_x)| < epsilon,$ i .e

$lim_x rightarrow inftyg(x)=0.$

share|cite|improve this answer

edited Mar 28 at 4:29

answered Mar 27 at 22:27

Peter SzilasPeter Szilas

12k2822

$endgroup$

share|cite|improve this answer

edited Mar 28 at 4:29

answered Mar 27 at 22:27

Peter SzilasPeter Szilas

12k2822

answered Mar 27 at 22:27

Peter SzilasPeter Szilas

12k2822

answered Mar 27 at 22:27

Peter SzilasPeter Szilas

12k2822

0

$begingroup$

The proof is simple:

$lim_x rightarrow infty g(x)=lim_xrightarrow infty f(x+1)-lim_xrightarrow infty f(x)$

For large $x$, $x+1 approx x$, so, by the previous result,

$lim_x rightarrow infty g(x)=0$

share|cite|improve this answer

answered Mar 27 at 21:24

officialnoriaofficialnoria

112

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  One has to also prove that $lim_xrightarrow infty f(x)$ exists!
  $endgroup$
  – folouer of kaklas
  Mar 27 at 21:30
  
  

  
  
  
  

add a comment | 

0

$begingroup$

The proof is simple:

$lim_x rightarrow infty g(x)=lim_xrightarrow infty f(x+1)-lim_xrightarrow infty f(x)$

For large $x$, $x+1 approx x$, so, by the previous result,

$lim_x rightarrow infty g(x)=0$

share|cite|improve this answer

answered Mar 27 at 21:24

officialnoriaofficialnoria

112

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  One has to also prove that $lim_xrightarrow infty f(x)$ exists!
  $endgroup$
  – folouer of kaklas
  Mar 27 at 21:30
  
  

  
  
  
  

add a comment | 

$begingroup$

The proof is simple:

$lim_x rightarrow infty g(x)=lim_xrightarrow infty f(x+1)-lim_xrightarrow infty f(x)$

For large $x$, $x+1 approx x$, so, by the previous result,

$lim_x rightarrow infty g(x)=0$

share|cite|improve this answer

answered Mar 27 at 21:24

officialnoriaofficialnoria

112

$endgroup$

share|cite|improve this answer

answered Mar 27 at 21:24

officialnoriaofficialnoria

112

answered Mar 27 at 21:24

officialnoriaofficialnoria

112

answered Mar 27 at 21:24

officialnoriaofficialnoria

112