Give an example to illustrate that $lim_xto 0 f(x)$ is not always equal to $lim_xto 0 f(2x)$ [closed]

Clash Royale CLAN TAG#URR8PPP

up vote
0
down vote

favorite

It's really hard to come up with such an example. My professor said that the example won't be made up of ILATE functions. It would be something like signum, gif etc.

calculus real-analysis limits

share|cite|improve this question

edited Jul 18 at 12:26

md2perpe

5,93011022

asked Jul 18 at 11:58

aditya gupta

11

closed as off-topic by Alex Francisco, steven gregory, Paramanand Singh, Strants, Daniel Fischer♦ Jul 18 at 20:54

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, steven gregory, Paramanand Singh, Daniel Fischer

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  How could they even be different?
  – md2perpe
  Jul 18 at 12:29
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @adityagupta, if you take the limit at $0$, they will be same.
  – Biswarup Saha
  Jul 18 at 13:28
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Either your professor tricked you, or there is something you don't tell us.
  – Yves Daoust
  Jul 18 at 13:42
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Should $f$ be defined for all $x in mathbb R$? If not, then I can solve it, I think.
  – md2perpe
  Jul 18 at 13:43
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  This is nonsense as stated. Possibly what he asked for is not exactly what you say?
  – David C. Ullrich
  Jul 18 at 13:59
  

  
  
  
  

 | 
show 5 more comments

up vote
0
down vote

favorite

It's really hard to come up with such an example. My professor said that the example won't be made up of ILATE functions. It would be something like signum, gif etc.

calculus real-analysis limits

share|cite|improve this question

edited Jul 18 at 12:26

md2perpe

5,93011022

asked Jul 18 at 11:58

aditya gupta

11

closed as off-topic by Alex Francisco, steven gregory, Paramanand Singh, Strants, Daniel Fischer♦ Jul 18 at 20:54

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, steven gregory, Paramanand Singh, Daniel Fischer

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  How could they even be different?
  – md2perpe
  Jul 18 at 12:29
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @adityagupta, if you take the limit at $0$, they will be same.
  – Biswarup Saha
  Jul 18 at 13:28
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Either your professor tricked you, or there is something you don't tell us.
  – Yves Daoust
  Jul 18 at 13:42
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Should $f$ be defined for all $x in mathbb R$? If not, then I can solve it, I think.
  – md2perpe
  Jul 18 at 13:43
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  This is nonsense as stated. Possibly what he asked for is not exactly what you say?
  – David C. Ullrich
  Jul 18 at 13:59
  

  
  
  
  

 | 
show 5 more comments

up vote
0
down vote

favorite

up vote
0
down vote

favorite

It's really hard to come up with such an example. My professor said that the example won't be made up of ILATE functions. It would be something like signum, gif etc.

calculus real-analysis limits

share|cite|improve this question

edited Jul 18 at 12:26

md2perpe

5,93011022

asked Jul 18 at 11:58

aditya gupta

11

It's really hard to come up with such an example. My professor said that the example won't be made up of ILATE functions. It would be something like signum, gif etc.

calculus real-analysis limits

share|cite|improve this question

edited Jul 18 at 12:26

md2perpe

5,93011022

asked Jul 18 at 11:58

aditya gupta

11

share|cite|improve this question

share|cite|improve this question

edited Jul 18 at 12:26

md2perpe

5,93011022

edited Jul 18 at 12:26

md2perpe

5,93011022

edited Jul 18 at 12:26

md2perpe

5,93011022

5,93011022

asked Jul 18 at 11:58

aditya gupta

11

asked Jul 18 at 11:58

aditya gupta

11

asked Jul 18 at 11:58

aditya gupta

11

11

closed as off-topic by Alex Francisco, steven gregory, Paramanand Singh, Strants, Daniel Fischer♦ Jul 18 at 20:54

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, steven gregory, Paramanand Singh, Daniel Fischer

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Alex Francisco, steven gregory, Paramanand Singh, Strants, Daniel Fischer♦ Jul 18 at 20:54

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, steven gregory, Paramanand Singh, Daniel Fischer

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  How could they even be different?
  – md2perpe
  Jul 18 at 12:29
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @adityagupta, if you take the limit at $0$, they will be same.
  – Biswarup Saha
  Jul 18 at 13:28
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Either your professor tricked you, or there is something you don't tell us.
  – Yves Daoust
  Jul 18 at 13:42
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Should $f$ be defined for all $x in mathbb R$? If not, then I can solve it, I think.
  – md2perpe
  Jul 18 at 13:43
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  This is nonsense as stated. Possibly what he asked for is not exactly what you say?
  – David C. Ullrich
  Jul 18 at 13:59
  

  
  
  
  

 | 
show 5 more comments

• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  How could they even be different?
  – md2perpe
  Jul 18 at 12:29
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @adityagupta, if you take the limit at $0$, they will be same.
  – Biswarup Saha
  Jul 18 at 13:28
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Either your professor tricked you, or there is something you don't tell us.
  – Yves Daoust
  Jul 18 at 13:42
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Should $f$ be defined for all $x in mathbb R$? If not, then I can solve it, I think.
  – md2perpe
  Jul 18 at 13:43
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  This is nonsense as stated. Possibly what he asked for is not exactly what you say?
  – David C. Ullrich
  Jul 18 at 13:59
  

  
  
  
  

6

6

How could they even be different?
– md2perpe
Jul 18 at 12:29

How could they even be different?
– md2perpe
Jul 18 at 12:29

1

1

@adityagupta, if you take the limit at $0$, they will be same.
– Biswarup Saha
Jul 18 at 13:28

@adityagupta, if you take the limit at $0$, they will be same.
– Biswarup Saha
Jul 18 at 13:28

3

3

Either your professor tricked you, or there is something you don't tell us.
– Yves Daoust
Jul 18 at 13:42

Either your professor tricked you, or there is something you don't tell us.
– Yves Daoust
Jul 18 at 13:42

1

1

Should $f$ be defined for all $x in mathbb R$? If not, then I can solve it, I think.
– md2perpe
Jul 18 at 13:43

Should $f$ be defined for all $x in mathbb R$? If not, then I can solve it, I think.
– md2perpe
Jul 18 at 13:43

1

1

This is nonsense as stated. Possibly what he asked for is not exactly what you say?
– David C. Ullrich
Jul 18 at 13:59

This is nonsense as stated. Possibly what he asked for is not exactly what you say?
– David C. Ullrich
Jul 18 at 13:59

 | 
show 5 more comments

3 Answers
3

active

oldest

votes

up vote
3
down vote

Although, I don't know where you get this proposition, my implication shows

If $f$ is a real-valued function on $BbbR$, such that $lim_xto
0 f(x)$ exists then $lim_xto 0 f(2x)$ exists and equal to
$lim_xto 0 f(x)$.

I am going to prove this statement.

Let, $lim_xto 0 f(x)=LinBbbR$ and define $g:BbbRtoBbbR$ by $g(x)=f(2x)quadforall xinBbbR$

To prove $lim_xto 0 f(2x)=L$, it is enough to show that $lim_xto 0 g(x)=L$

Choose $varepsilon >0$

Then $exists delta_varepsilon >0$ such that $|f(x)-L|<varepsilonquadforall xin(-delta_varepsilon,delta_varepsilon)backslash0 $

$implies |f(2x)-L|<varepsilonquad forall xinleft(-fracdelta_varepsilon2,fracdelta_varepsilon2right )backslashlbrace 0 rbrace$

$implies |g(x)-L|<varepsilonquadforall xin left (-delta_varepsilon',delta_varepsilon'right )backslash0$ where $delta_varepsilon' =fracdelta_varepsilon2 >0$

For any $varepsilon >0$, $existsdelta_varepsilon'>0$ such that $|g(x)-L|<varepsilonquadforall xin left (-delta_varepsilon',delta_varepsilon'right )backslash0$
$implies lim_xto 0 g(x)=Limplieslim_xto 0 f(2x)=L=lim_xto 0 f(x)$(Proved)
This is true only when we are taking limit at $0$ i.e. $lim_xto c f(x)$ may not be equal to $lim_xto c f(2x)$ for any $cinBbbR$(Why?)

You can verify this result geometrically also(Hint: To obtain the graph of $xmapsto f(2x)$, shrink the graph of $f$ horizontally by $frac12$).

share|cite|improve this answer

edited Jul 18 at 13:35

answered Jul 18 at 13:26

Biswarup Saha

2318

add a comment | 

up vote
1
down vote

By definition, if $f(xi) to L$ as $xi to 0$, this means that

$$|f(2x) - L| < epsilon quad textwhenever quad |2x| < delta$$

but this just means that $|x| < delta/2$ implies $|f(2x) - L| < epsilon$

so $f(2x) to L$ as $x to 0$.

share|cite|improve this answer

answered Jul 18 at 14:21

JuliusL33t

1,155816

add a comment | 

up vote
0
down vote

Let $f : p/3^n mid p, n in mathbb N to mathbb R$ be defined by $f(p/3^n) = (-1)^p.$ Then $lim_x to 0 f(x)$ is not defined but $lim_x to 0 f(2x)$ is defined and equals $1.$

share|cite|improve this answer

answered Jul 18 at 14:36

md2perpe

5,93011022

• 
  
  
  

  
  

  
  
  
  
  
  

  
  this was kinda the one that i thought, but yours is much more elegant. thanks for the help
  – aditya gupta
  Jul 19 at 14:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Note that one limit exists and the other does not. You can not find a function where both limits exist and are different.
  – md2perpe
  Jul 19 at 19:46
  

  
  
  
  

add a comment | 

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote

Although, I don't know where you get this proposition, my implication shows

If $f$ is a real-valued function on $BbbR$, such that $lim_xto
0 f(x)$ exists then $lim_xto 0 f(2x)$ exists and equal to
$lim_xto 0 f(x)$.

I am going to prove this statement.

Let, $lim_xto 0 f(x)=LinBbbR$ and define $g:BbbRtoBbbR$ by $g(x)=f(2x)quadforall xinBbbR$

To prove $lim_xto 0 f(2x)=L$, it is enough to show that $lim_xto 0 g(x)=L$

Choose $varepsilon >0$

Then $exists delta_varepsilon >0$ such that $|f(x)-L|<varepsilonquadforall xin(-delta_varepsilon,delta_varepsilon)backslash0 $

$implies |f(2x)-L|<varepsilonquad forall xinleft(-fracdelta_varepsilon2,fracdelta_varepsilon2right )backslashlbrace 0 rbrace$

$implies |g(x)-L|<varepsilonquadforall xin left (-delta_varepsilon',delta_varepsilon'right )backslash0$ where $delta_varepsilon' =fracdelta_varepsilon2 >0$

For any $varepsilon >0$, $existsdelta_varepsilon'>0$ such that $|g(x)-L|<varepsilonquadforall xin left (-delta_varepsilon',delta_varepsilon'right )backslash0$
$implies lim_xto 0 g(x)=Limplieslim_xto 0 f(2x)=L=lim_xto 0 f(x)$(Proved)
This is true only when we are taking limit at $0$ i.e. $lim_xto c f(x)$ may not be equal to $lim_xto c f(2x)$ for any $cinBbbR$(Why?)

You can verify this result geometrically also(Hint: To obtain the graph of $xmapsto f(2x)$, shrink the graph of $f$ horizontally by $frac12$).

share|cite|improve this answer

edited Jul 18 at 13:35

answered Jul 18 at 13:26

Biswarup Saha

2318

add a comment | 

up vote
3
down vote

Although, I don't know where you get this proposition, my implication shows

If $f$ is a real-valued function on $BbbR$, such that $lim_xto
0 f(x)$ exists then $lim_xto 0 f(2x)$ exists and equal to
$lim_xto 0 f(x)$.

I am going to prove this statement.

Let, $lim_xto 0 f(x)=LinBbbR$ and define $g:BbbRtoBbbR$ by $g(x)=f(2x)quadforall xinBbbR$

To prove $lim_xto 0 f(2x)=L$, it is enough to show that $lim_xto 0 g(x)=L$

Choose $varepsilon >0$

Then $exists delta_varepsilon >0$ such that $|f(x)-L|<varepsilonquadforall xin(-delta_varepsilon,delta_varepsilon)backslash0 $

$implies |f(2x)-L|<varepsilonquad forall xinleft(-fracdelta_varepsilon2,fracdelta_varepsilon2right )backslashlbrace 0 rbrace$

$implies |g(x)-L|<varepsilonquadforall xin left (-delta_varepsilon',delta_varepsilon'right )backslash0$ where $delta_varepsilon' =fracdelta_varepsilon2 >0$

For any $varepsilon >0$, $existsdelta_varepsilon'>0$ such that $|g(x)-L|<varepsilonquadforall xin left (-delta_varepsilon',delta_varepsilon'right )backslash0$
$implies lim_xto 0 g(x)=Limplieslim_xto 0 f(2x)=L=lim_xto 0 f(x)$(Proved)
This is true only when we are taking limit at $0$ i.e. $lim_xto c f(x)$ may not be equal to $lim_xto c f(2x)$ for any $cinBbbR$(Why?)

You can verify this result geometrically also(Hint: To obtain the graph of $xmapsto f(2x)$, shrink the graph of $f$ horizontally by $frac12$).

share|cite|improve this answer

edited Jul 18 at 13:35

answered Jul 18 at 13:26

Biswarup Saha

2318

add a comment | 

up vote
3
down vote

up vote
3
down vote

Although, I don't know where you get this proposition, my implication shows

If $f$ is a real-valued function on $BbbR$, such that $lim_xto
0 f(x)$ exists then $lim_xto 0 f(2x)$ exists and equal to
$lim_xto 0 f(x)$.

I am going to prove this statement.

Let, $lim_xto 0 f(x)=LinBbbR$ and define $g:BbbRtoBbbR$ by $g(x)=f(2x)quadforall xinBbbR$

To prove $lim_xto 0 f(2x)=L$, it is enough to show that $lim_xto 0 g(x)=L$

Choose $varepsilon >0$

Then $exists delta_varepsilon >0$ such that $|f(x)-L|<varepsilonquadforall xin(-delta_varepsilon,delta_varepsilon)backslash0 $

$implies |f(2x)-L|<varepsilonquad forall xinleft(-fracdelta_varepsilon2,fracdelta_varepsilon2right )backslashlbrace 0 rbrace$

$implies |g(x)-L|<varepsilonquadforall xin left (-delta_varepsilon',delta_varepsilon'right )backslash0$ where $delta_varepsilon' =fracdelta_varepsilon2 >0$

For any $varepsilon >0$, $existsdelta_varepsilon'>0$ such that $|g(x)-L|<varepsilonquadforall xin left (-delta_varepsilon',delta_varepsilon'right )backslash0$
$implies lim_xto 0 g(x)=Limplieslim_xto 0 f(2x)=L=lim_xto 0 f(x)$(Proved)
This is true only when we are taking limit at $0$ i.e. $lim_xto c f(x)$ may not be equal to $lim_xto c f(2x)$ for any $cinBbbR$(Why?)

You can verify this result geometrically also(Hint: To obtain the graph of $xmapsto f(2x)$, shrink the graph of $f$ horizontally by $frac12$).

share|cite|improve this answer

edited Jul 18 at 13:35

answered Jul 18 at 13:26

Biswarup Saha

2318

Although, I don't know where you get this proposition, my implication shows

If $f$ is a real-valued function on $BbbR$, such that $lim_xto
0 f(x)$ exists then $lim_xto 0 f(2x)$ exists and equal to
$lim_xto 0 f(x)$.

I am going to prove this statement.

Let, $lim_xto 0 f(x)=LinBbbR$ and define $g:BbbRtoBbbR$ by $g(x)=f(2x)quadforall xinBbbR$

To prove $lim_xto 0 f(2x)=L$, it is enough to show that $lim_xto 0 g(x)=L$

Choose $varepsilon >0$

Then $exists delta_varepsilon >0$ such that $|f(x)-L|<varepsilonquadforall xin(-delta_varepsilon,delta_varepsilon)backslash0 $

$implies |f(2x)-L|<varepsilonquad forall xinleft(-fracdelta_varepsilon2,fracdelta_varepsilon2right )backslashlbrace 0 rbrace$

$implies |g(x)-L|<varepsilonquadforall xin left (-delta_varepsilon',delta_varepsilon'right )backslash0$ where $delta_varepsilon' =fracdelta_varepsilon2 >0$

For any $varepsilon >0$, $existsdelta_varepsilon'>0$ such that $|g(x)-L|<varepsilonquadforall xin left (-delta_varepsilon',delta_varepsilon'right )backslash0$
$implies lim_xto 0 g(x)=Limplieslim_xto 0 f(2x)=L=lim_xto 0 f(x)$(Proved)
This is true only when we are taking limit at $0$ i.e. $lim_xto c f(x)$ may not be equal to $lim_xto c f(2x)$ for any $cinBbbR$(Why?)

You can verify this result geometrically also(Hint: To obtain the graph of $xmapsto f(2x)$, shrink the graph of $f$ horizontally by $frac12$).

share|cite|improve this answer

edited Jul 18 at 13:35

answered Jul 18 at 13:26

Biswarup Saha

2318

share|cite|improve this answer

share|cite|improve this answer

edited Jul 18 at 13:35

edited Jul 18 at 13:35

edited Jul 18 at 13:35

answered Jul 18 at 13:26

Biswarup Saha

2318

answered Jul 18 at 13:26

Biswarup Saha

2318

answered Jul 18 at 13:26

Biswarup Saha

2318

2318

add a comment | 

add a comment | 

up vote
1
down vote

By definition, if $f(xi) to L$ as $xi to 0$, this means that

$$|f(2x) - L| < epsilon quad textwhenever quad |2x| < delta$$

but this just means that $|x| < delta/2$ implies $|f(2x) - L| < epsilon$

so $f(2x) to L$ as $x to 0$.

share|cite|improve this answer

answered Jul 18 at 14:21

JuliusL33t

1,155816

add a comment | 

up vote
1
down vote

By definition, if $f(xi) to L$ as $xi to 0$, this means that

$$|f(2x) - L| < epsilon quad textwhenever quad |2x| < delta$$

but this just means that $|x| < delta/2$ implies $|f(2x) - L| < epsilon$

so $f(2x) to L$ as $x to 0$.

share|cite|improve this answer

answered Jul 18 at 14:21

JuliusL33t

1,155816

add a comment | 

up vote
1
down vote

up vote
1
down vote

By definition, if $f(xi) to L$ as $xi to 0$, this means that

$$|f(2x) - L| < epsilon quad textwhenever quad |2x| < delta$$

but this just means that $|x| < delta/2$ implies $|f(2x) - L| < epsilon$

so $f(2x) to L$ as $x to 0$.

share|cite|improve this answer

answered Jul 18 at 14:21

JuliusL33t

1,155816

By definition, if $f(xi) to L$ as $xi to 0$, this means that

$$|f(2x) - L| < epsilon quad textwhenever quad |2x| < delta$$

but this just means that $|x| < delta/2$ implies $|f(2x) - L| < epsilon$

so $f(2x) to L$ as $x to 0$.

share|cite|improve this answer

answered Jul 18 at 14:21

JuliusL33t

1,155816

share|cite|improve this answer

share|cite|improve this answer

answered Jul 18 at 14:21

JuliusL33t

1,155816

answered Jul 18 at 14:21

JuliusL33t

1,155816

answered Jul 18 at 14:21

JuliusL33t

1,155816

1,155816

add a comment | 

add a comment | 

up vote
0
down vote

Let $f : p/3^n mid p, n in mathbb N to mathbb R$ be defined by $f(p/3^n) = (-1)^p.$ Then $lim_x to 0 f(x)$ is not defined but $lim_x to 0 f(2x)$ is defined and equals $1.$

share|cite|improve this answer

answered Jul 18 at 14:36

md2perpe

5,93011022

• 
  
  
  

  
  

  
  
  
  
  
  

  
  this was kinda the one that i thought, but yours is much more elegant. thanks for the help
  – aditya gupta
  Jul 19 at 14:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Note that one limit exists and the other does not. You can not find a function where both limits exist and are different.
  – md2perpe
  Jul 19 at 19:46
  

  
  
  
  

add a comment | 

up vote
0
down vote

Let $f : p/3^n mid p, n in mathbb N to mathbb R$ be defined by $f(p/3^n) = (-1)^p.$ Then $lim_x to 0 f(x)$ is not defined but $lim_x to 0 f(2x)$ is defined and equals $1.$

share|cite|improve this answer

answered Jul 18 at 14:36

md2perpe

5,93011022

• 
  
  
  

  
  

  
  
  
  
  
  

  
  this was kinda the one that i thought, but yours is much more elegant. thanks for the help
  – aditya gupta
  Jul 19 at 14:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Note that one limit exists and the other does not. You can not find a function where both limits exist and are different.
  – md2perpe
  Jul 19 at 19:46
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

Let $f : p/3^n mid p, n in mathbb N to mathbb R$ be defined by $f(p/3^n) = (-1)^p.$ Then $lim_x to 0 f(x)$ is not defined but $lim_x to 0 f(2x)$ is defined and equals $1.$

share|cite|improve this answer

answered Jul 18 at 14:36

md2perpe

5,93011022

Let $f : p/3^n mid p, n in mathbb N to mathbb R$ be defined by $f(p/3^n) = (-1)^p.$ Then $lim_x to 0 f(x)$ is not defined but $lim_x to 0 f(2x)$ is defined and equals $1.$

share|cite|improve this answer

answered Jul 18 at 14:36

md2perpe

5,93011022

share|cite|improve this answer

share|cite|improve this answer

answered Jul 18 at 14:36

md2perpe

5,93011022

answered Jul 18 at 14:36

md2perpe

5,93011022

answered Jul 18 at 14:36

md2perpe

5,93011022

5,93011022

• 
  
  
  

  
  

  
  
  
  
  
  

  
  this was kinda the one that i thought, but yours is much more elegant. thanks for the help
  – aditya gupta
  Jul 19 at 14:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Note that one limit exists and the other does not. You can not find a function where both limits exist and are different.
  – md2perpe
  Jul 19 at 19:46
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  this was kinda the one that i thought, but yours is much more elegant. thanks for the help
  – aditya gupta
  Jul 19 at 14:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Note that one limit exists and the other does not. You can not find a function where both limits exist and are different.
  – md2perpe
  Jul 19 at 19:46
  

  
  
  
  

this was kinda the one that i thought, but yours is much more elegant. thanks for the help
– aditya gupta
Jul 19 at 14:34

this was kinda the one that i thought, but yours is much more elegant. thanks for the help
– aditya gupta
Jul 19 at 14:34

Note that one limit exists and the other does not. You can not find a function where both limits exist and are different.
– md2perpe
Jul 19 at 19:46

Note that one limit exists and the other does not. You can not find a function where both limits exist and are different.
– md2perpe
Jul 19 at 19:46

add a comment |