Prove / disprove if $lim((a_n)^2+ (b_n)^2)=0$ then $lim(a_n)=0$ and $lim(b_n)=0$ [closed]

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Prove / disprove:
Let $a_n$ and $b_n$ be real sequences, if $lim((a_n)^2+ (b_n)^2)=0$ then $lim(a_n)=0$ and $lim(b_n)=0$

sequences-and-series limits

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edited Jul 16 at 20:34

asked Jul 16 at 20:29

Oran Sherf

97

closed as off-topic by Math Lover, Alex Francisco, Xander Henderson, Trần Thúc Minh Trí, Parcly Taxel Jul 17 at 3:10

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." – Math Lover, Alex Francisco, Xander Henderson, Trần Thúc Minh Trí, Parcly Taxel

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

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Are $a_n$ and $b_n$ real, or complex?
  – Saucy O'Path
  Jul 16 at 20:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  real. I have tried to disprove it but found nothing yet..
  – Oran Sherf
  Jul 16 at 20:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Separately apply limits to each term and arrive at conclusion that since the sequence is real the only plausible result is the one given to prove.
  – Pi_die_die
  Jul 16 at 20:39
  

  
  
  
  

add a comment | 

up vote
-2
down vote

favorite

Prove / disprove:
Let $a_n$ and $b_n$ be real sequences, if $lim((a_n)^2+ (b_n)^2)=0$ then $lim(a_n)=0$ and $lim(b_n)=0$

sequences-and-series limits

share|cite|improve this question

edited Jul 16 at 20:34

asked Jul 16 at 20:29

Oran Sherf

97

closed as off-topic by Math Lover, Alex Francisco, Xander Henderson, Trần Thúc Minh Trí, Parcly Taxel Jul 17 at 3:10

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." – Math Lover, Alex Francisco, Xander Henderson, Trần Thúc Minh Trí, Parcly Taxel

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

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Are $a_n$ and $b_n$ real, or complex?
  – Saucy O'Path
  Jul 16 at 20:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  real. I have tried to disprove it but found nothing yet..
  – Oran Sherf
  Jul 16 at 20:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Separately apply limits to each term and arrive at conclusion that since the sequence is real the only plausible result is the one given to prove.
  – Pi_die_die
  Jul 16 at 20:39
  

  
  
  
  

add a comment | 

up vote
-2
down vote

favorite

up vote
-2
down vote

favorite

Prove / disprove:
Let $a_n$ and $b_n$ be real sequences, if $lim((a_n)^2+ (b_n)^2)=0$ then $lim(a_n)=0$ and $lim(b_n)=0$

sequences-and-series limits

share|cite|improve this question

edited Jul 16 at 20:34

asked Jul 16 at 20:29

Oran Sherf

97

Prove / disprove:
Let $a_n$ and $b_n$ be real sequences, if $lim((a_n)^2+ (b_n)^2)=0$ then $lim(a_n)=0$ and $lim(b_n)=0$

sequences-and-series limits

share|cite|improve this question

edited Jul 16 at 20:34

asked Jul 16 at 20:29

Oran Sherf

97

share|cite|improve this question

share|cite|improve this question

edited Jul 16 at 20:34

edited Jul 16 at 20:34

edited Jul 16 at 20:34

asked Jul 16 at 20:29

Oran Sherf

97

asked Jul 16 at 20:29

Oran Sherf

97

asked Jul 16 at 20:29

Oran Sherf

97

97

closed as off-topic by Math Lover, Alex Francisco, Xander Henderson, Trần Thúc Minh Trí, Parcly Taxel Jul 17 at 3:10

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." – Math Lover, Alex Francisco, Xander Henderson, Trần Thúc Minh Trí, Parcly Taxel

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

closed as off-topic by Math Lover, Alex Francisco, Xander Henderson, Trần Thúc Minh Trí, Parcly Taxel Jul 17 at 3:10

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." – Math Lover, Alex Francisco, Xander Henderson, Trần Thúc Minh Trí, Parcly Taxel

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

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Are $a_n$ and $b_n$ real, or complex?
  – Saucy O'Path
  Jul 16 at 20:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  real. I have tried to disprove it but found nothing yet..
  – Oran Sherf
  Jul 16 at 20:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Separately apply limits to each term and arrive at conclusion that since the sequence is real the only plausible result is the one given to prove.
  – Pi_die_die
  Jul 16 at 20:39
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Are $a_n$ and $b_n$ real, or complex?
  – Saucy O'Path
  Jul 16 at 20:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  real. I have tried to disprove it but found nothing yet..
  – Oran Sherf
  Jul 16 at 20:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Separately apply limits to each term and arrive at conclusion that since the sequence is real the only plausible result is the one given to prove.
  – Pi_die_die
  Jul 16 at 20:39
  

  
  
  
  

1

1

Are $a_n$ and $b_n$ real, or complex?
– Saucy O'Path
Jul 16 at 20:31

Are $a_n$ and $b_n$ real, or complex?
– Saucy O'Path
Jul 16 at 20:31

real. I have tried to disprove it but found nothing yet..
– Oran Sherf
Jul 16 at 20:33

real. I have tried to disprove it but found nothing yet..
– Oran Sherf
Jul 16 at 20:33

Separately apply limits to each term and arrive at conclusion that since the sequence is real the only plausible result is the one given to prove.
– Pi_die_die
Jul 16 at 20:39

Separately apply limits to each term and arrive at conclusion that since the sequence is real the only plausible result is the one given to prove.
– Pi_die_die
Jul 16 at 20:39

add a comment | 

3 Answers
3

active

oldest

votes

up vote
3
down vote



accepted

Hint:   $0 le a_n^2 le a_n^2+b_n^2,$, so $,a_n^2 to 0,$ by the squeeze theorem.

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answered Jul 16 at 20:40

dxiv

54.3k64797

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0
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Suppose $a_n$ does not go to 0. Then there exists an $epsilon>0$ such that $|a_n|>epsilon$ for infinitely many $n$. This implies $a_n^2>epsilon^2>0$ for infinitely many $n$. Since $a_n^2+b_n^2geq a_n^2$, this implies $lim (a_n^2+b_n^2)geq epsilon^2>0$, which implies that the original limit could not possibly be 0.

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answered Jul 16 at 20:38

Alex R.

23.7k12352

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up vote
0
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Since $$|a_n^2+b_n^2|<epsilon^2$$so is $$|a_n^2|le |a_n^2+b_n^2|<epsilon^2$$

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answered Jul 16 at 20:42

Mostafa Ayaz

8,6023630

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



accepted

Hint:   $0 le a_n^2 le a_n^2+b_n^2,$, so $,a_n^2 to 0,$ by the squeeze theorem.

share|cite|improve this answer

answered Jul 16 at 20:40

dxiv

54.3k64797

add a comment | 

up vote
3
down vote



accepted

Hint:   $0 le a_n^2 le a_n^2+b_n^2,$, so $,a_n^2 to 0,$ by the squeeze theorem.

share|cite|improve this answer

answered Jul 16 at 20:40

dxiv

54.3k64797

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

Hint:   $0 le a_n^2 le a_n^2+b_n^2,$, so $,a_n^2 to 0,$ by the squeeze theorem.

share|cite|improve this answer

answered Jul 16 at 20:40

dxiv

54.3k64797

Hint:   $0 le a_n^2 le a_n^2+b_n^2,$, so $,a_n^2 to 0,$ by the squeeze theorem.

share|cite|improve this answer

answered Jul 16 at 20:40

dxiv

54.3k64797

share|cite|improve this answer

share|cite|improve this answer

answered Jul 16 at 20:40

dxiv

54.3k64797

answered Jul 16 at 20:40

dxiv

54.3k64797

answered Jul 16 at 20:40

dxiv

54.3k64797

54.3k64797

add a comment | 

add a comment | 

up vote
0
down vote

Suppose $a_n$ does not go to 0. Then there exists an $epsilon>0$ such that $|a_n|>epsilon$ for infinitely many $n$. This implies $a_n^2>epsilon^2>0$ for infinitely many $n$. Since $a_n^2+b_n^2geq a_n^2$, this implies $lim (a_n^2+b_n^2)geq epsilon^2>0$, which implies that the original limit could not possibly be 0.

share|cite|improve this answer

answered Jul 16 at 20:38

Alex R.

23.7k12352

add a comment | 

up vote
0
down vote

Suppose $a_n$ does not go to 0. Then there exists an $epsilon>0$ such that $|a_n|>epsilon$ for infinitely many $n$. This implies $a_n^2>epsilon^2>0$ for infinitely many $n$. Since $a_n^2+b_n^2geq a_n^2$, this implies $lim (a_n^2+b_n^2)geq epsilon^2>0$, which implies that the original limit could not possibly be 0.

share|cite|improve this answer

answered Jul 16 at 20:38

Alex R.

23.7k12352

add a comment | 

up vote
0
down vote

up vote
0
down vote

Suppose $a_n$ does not go to 0. Then there exists an $epsilon>0$ such that $|a_n|>epsilon$ for infinitely many $n$. This implies $a_n^2>epsilon^2>0$ for infinitely many $n$. Since $a_n^2+b_n^2geq a_n^2$, this implies $lim (a_n^2+b_n^2)geq epsilon^2>0$, which implies that the original limit could not possibly be 0.

share|cite|improve this answer

answered Jul 16 at 20:38

Alex R.

23.7k12352

Suppose $a_n$ does not go to 0. Then there exists an $epsilon>0$ such that $|a_n|>epsilon$ for infinitely many $n$. This implies $a_n^2>epsilon^2>0$ for infinitely many $n$. Since $a_n^2+b_n^2geq a_n^2$, this implies $lim (a_n^2+b_n^2)geq epsilon^2>0$, which implies that the original limit could not possibly be 0.

share|cite|improve this answer

answered Jul 16 at 20:38

Alex R.

23.7k12352

share|cite|improve this answer

share|cite|improve this answer

answered Jul 16 at 20:38

Alex R.

23.7k12352

answered Jul 16 at 20:38

Alex R.

23.7k12352

answered Jul 16 at 20:38

Alex R.

23.7k12352

23.7k12352

add a comment | 

add a comment | 

up vote
0
down vote

Since $$|a_n^2+b_n^2|<epsilon^2$$so is $$|a_n^2|le |a_n^2+b_n^2|<epsilon^2$$

share|cite|improve this answer

answered Jul 16 at 20:42

Mostafa Ayaz

8,6023630

add a comment | 

up vote
0
down vote

Since $$|a_n^2+b_n^2|<epsilon^2$$so is $$|a_n^2|le |a_n^2+b_n^2|<epsilon^2$$

share|cite|improve this answer

answered Jul 16 at 20:42

Mostafa Ayaz

8,6023630

add a comment | 

up vote
0
down vote

up vote
0
down vote

Since $$|a_n^2+b_n^2|<epsilon^2$$so is $$|a_n^2|le |a_n^2+b_n^2|<epsilon^2$$

share|cite|improve this answer

answered Jul 16 at 20:42

Mostafa Ayaz

8,6023630

Since $$|a_n^2+b_n^2|<epsilon^2$$so is $$|a_n^2|le |a_n^2+b_n^2|<epsilon^2$$

share|cite|improve this answer

answered Jul 16 at 20:42

Mostafa Ayaz

8,6023630

share|cite|improve this answer

share|cite|improve this answer

answered Jul 16 at 20:42

Mostafa Ayaz

8,6023630

answered Jul 16 at 20:42

Mostafa Ayaz

8,6023630

answered Jul 16 at 20:42

Mostafa Ayaz

8,6023630

8,6023630

add a comment | 

add a comment |