Limits and convergence - selecting the right epsilon for proofs

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Show that if limit of sequence $x_n$ goes to $x$ as $n$ goes to $ infty$ and $ x > a$, then $x_n > a$ all but finitely many $ n$.

I have started it like this:
$x_n > x - epsilon$ (as $|x_n- x| < epsilon$ for all $epsilon$)

$x_n > a - epsilon$

From here on, how does one chose the right epsilon to show that $x_n > a$?

real-analysis limits convergence

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edited 3 hours ago

StammeringMathematician

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Aishwarya Deore

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up vote
3
down vote

favorite

1

Show that if limit of sequence $x_n$ goes to $x$ as $n$ goes to $ infty$ and $ x > a$, then $x_n > a$ all but finitely many $ n$.

I have started it like this:
$x_n > x - epsilon$ (as $|x_n- x| < epsilon$ for all $epsilon$)

$x_n > a - epsilon$

From here on, how does one chose the right epsilon to show that $x_n > a$?

real-analysis limits convergence

share|cite|improve this question

edited 3 hours ago

StammeringMathematician

73616

asked 4 hours ago

Aishwarya Deore

212

New contributor

Aishwarya Deore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

up vote
3
down vote

favorite

1

up vote
3
down vote

favorite

1

1

Show that if limit of sequence $x_n$ goes to $x$ as $n$ goes to $ infty$ and $ x > a$, then $x_n > a$ all but finitely many $ n$.

I have started it like this:
$x_n > x - epsilon$ (as $|x_n- x| < epsilon$ for all $epsilon$)

$x_n > a - epsilon$

From here on, how does one chose the right epsilon to show that $x_n > a$?

real-analysis limits convergence

share|cite|improve this question

edited 3 hours ago

StammeringMathematician

73616

asked 4 hours ago

Aishwarya Deore

212

New contributor

Aishwarya Deore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

Show that if limit of sequence $x_n$ goes to $x$ as $n$ goes to $ infty$ and $ x > a$, then $x_n > a$ all but finitely many $ n$.

I have started it like this:
$x_n > x - epsilon$ (as $|x_n- x| < epsilon$ for all $epsilon$)

$x_n > a - epsilon$

From here on, how does one chose the right epsilon to show that $x_n > a$?

real-analysis limits convergence

real-analysis limits convergence

share|cite|improve this question

edited 3 hours ago

StammeringMathematician

73616

asked 4 hours ago

Aishwarya Deore

212

New contributor

Aishwarya Deore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 3 hours ago

StammeringMathematician

73616

asked 4 hours ago

Aishwarya Deore

212

New contributor

Aishwarya Deore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

share|cite|improve this question

edited 3 hours ago

StammeringMathematician

73616

edited 3 hours ago

StammeringMathematician

73616

edited 3 hours ago

StammeringMathematician

73616

73616

asked 4 hours ago

Aishwarya Deore

212

New contributor

Aishwarya Deore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 4 hours ago

Aishwarya Deore

212

asked 4 hours ago

Aishwarya Deore

212

212

New contributor

Aishwarya Deore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

New contributor

Aishwarya Deore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

Aishwarya Deore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

add a comment | 

3 Answers
3

active

oldest

votes

up vote
2
down vote

It is always best to draw a number line (if possible do it mentally rather than on paper) and position $x, a$ on it: $$------a-----x-----$$ Now things are crystal clear. As $ntoinfty$ the values of $x_n$ will come near $x$ and from the figure above you can see that as one moves near $x$ one gets greater than $a$.

The question is "how near should one go to $x$ in order to become greater than $a$?". Since the distance between $x$ and $a$ is $|x-a|=x-a$ one has to move closer to $x$ by anything less than this amount. Choose any positive $epsilon$ less than $x-a$ and you are done. In particular you can choose $epsilon=(x-a) /2$. Note that $epsilon=x-a$ also works because the inequality used in limit definition is $<$ and not $leq $.

share|cite|improve this answer

answered 3 hours ago

Paramanand Singh

45.7k554144

add a comment | 

up vote
1
down vote

If a sequence $x_n$ converges to $x$, by definition it means for every $epsilon >0$ there exist a $N$ such that $$|x_n-x|<epsilontag1$$ for all $n>N$

Now (1) can be rephrased as $x_nin (x-epsilon, x+epsilon)tag2$ for all $n>N$

Take $epsilon=x-a$ (as $x>a, x-a$ is a positive number and hence the choice of taking $x-a=epsilon$ makes sense).

By definition there exist a $N$ such that
$$x_nin (x-epsilon, x+epsilon)$$ for all $n>N$. Now put the value of $epsilon $ you get $$x_nin (x-(x-a),x+x-a)=(a,2x-a)$$

Therefore $x_n>a$ for all $n>N$

share|cite|improve this answer

answered 3 hours ago

StammeringMathematician

73616

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Why the downvote? +1 to compensate.
  – Paramanand Singh
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well the upvote was also to indicate that your answer is correct apart from the compensation.
  – Paramanand Singh
  3 hours ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

Let $epsilon = frac x-a2$

Then $x-epsilon = frac x+a2 >a$

Thus $x_n > x-epsilon implies x_n >a$

share|cite|improve this answer

answered 3 hours ago

Mohammad Riazi-Kermani

31.7k41853

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

It is always best to draw a number line (if possible do it mentally rather than on paper) and position $x, a$ on it: $$------a-----x-----$$ Now things are crystal clear. As $ntoinfty$ the values of $x_n$ will come near $x$ and from the figure above you can see that as one moves near $x$ one gets greater than $a$.

The question is "how near should one go to $x$ in order to become greater than $a$?". Since the distance between $x$ and $a$ is $|x-a|=x-a$ one has to move closer to $x$ by anything less than this amount. Choose any positive $epsilon$ less than $x-a$ and you are done. In particular you can choose $epsilon=(x-a) /2$. Note that $epsilon=x-a$ also works because the inequality used in limit definition is $<$ and not $leq $.

share|cite|improve this answer

answered 3 hours ago

Paramanand Singh

45.7k554144

add a comment | 

up vote
2
down vote

It is always best to draw a number line (if possible do it mentally rather than on paper) and position $x, a$ on it: $$------a-----x-----$$ Now things are crystal clear. As $ntoinfty$ the values of $x_n$ will come near $x$ and from the figure above you can see that as one moves near $x$ one gets greater than $a$.

The question is "how near should one go to $x$ in order to become greater than $a$?". Since the distance between $x$ and $a$ is $|x-a|=x-a$ one has to move closer to $x$ by anything less than this amount. Choose any positive $epsilon$ less than $x-a$ and you are done. In particular you can choose $epsilon=(x-a) /2$. Note that $epsilon=x-a$ also works because the inequality used in limit definition is $<$ and not $leq $.

share|cite|improve this answer

answered 3 hours ago

Paramanand Singh

45.7k554144

add a comment | 

up vote
2
down vote

up vote
2
down vote

It is always best to draw a number line (if possible do it mentally rather than on paper) and position $x, a$ on it: $$------a-----x-----$$ Now things are crystal clear. As $ntoinfty$ the values of $x_n$ will come near $x$ and from the figure above you can see that as one moves near $x$ one gets greater than $a$.

The question is "how near should one go to $x$ in order to become greater than $a$?". Since the distance between $x$ and $a$ is $|x-a|=x-a$ one has to move closer to $x$ by anything less than this amount. Choose any positive $epsilon$ less than $x-a$ and you are done. In particular you can choose $epsilon=(x-a) /2$. Note that $epsilon=x-a$ also works because the inequality used in limit definition is $<$ and not $leq $.

share|cite|improve this answer

answered 3 hours ago

Paramanand Singh

45.7k554144

It is always best to draw a number line (if possible do it mentally rather than on paper) and position $x, a$ on it: $$------a-----x-----$$ Now things are crystal clear. As $ntoinfty$ the values of $x_n$ will come near $x$ and from the figure above you can see that as one moves near $x$ one gets greater than $a$.

The question is "how near should one go to $x$ in order to become greater than $a$?". Since the distance between $x$ and $a$ is $|x-a|=x-a$ one has to move closer to $x$ by anything less than this amount. Choose any positive $epsilon$ less than $x-a$ and you are done. In particular you can choose $epsilon=(x-a) /2$. Note that $epsilon=x-a$ also works because the inequality used in limit definition is $<$ and not $leq $.

share|cite|improve this answer

answered 3 hours ago

Paramanand Singh

45.7k554144

share|cite|improve this answer

share|cite|improve this answer

answered 3 hours ago

Paramanand Singh

45.7k554144

answered 3 hours ago

Paramanand Singh

45.7k554144

answered 3 hours ago

Paramanand Singh

45.7k554144

45.7k554144

add a comment | 

add a comment | 

up vote
1
down vote

If a sequence $x_n$ converges to $x$, by definition it means for every $epsilon >0$ there exist a $N$ such that $$|x_n-x|<epsilontag1$$ for all $n>N$

Now (1) can be rephrased as $x_nin (x-epsilon, x+epsilon)tag2$ for all $n>N$

Take $epsilon=x-a$ (as $x>a, x-a$ is a positive number and hence the choice of taking $x-a=epsilon$ makes sense).

By definition there exist a $N$ such that
$$x_nin (x-epsilon, x+epsilon)$$ for all $n>N$. Now put the value of $epsilon $ you get $$x_nin (x-(x-a),x+x-a)=(a,2x-a)$$

Therefore $x_n>a$ for all $n>N$

share|cite|improve this answer

answered 3 hours ago

StammeringMathematician

73616

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Why the downvote? +1 to compensate.
  – Paramanand Singh
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well the upvote was also to indicate that your answer is correct apart from the compensation.
  – Paramanand Singh
  3 hours ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

If a sequence $x_n$ converges to $x$, by definition it means for every $epsilon >0$ there exist a $N$ such that $$|x_n-x|<epsilontag1$$ for all $n>N$

Now (1) can be rephrased as $x_nin (x-epsilon, x+epsilon)tag2$ for all $n>N$

Take $epsilon=x-a$ (as $x>a, x-a$ is a positive number and hence the choice of taking $x-a=epsilon$ makes sense).

By definition there exist a $N$ such that
$$x_nin (x-epsilon, x+epsilon)$$ for all $n>N$. Now put the value of $epsilon $ you get $$x_nin (x-(x-a),x+x-a)=(a,2x-a)$$

Therefore $x_n>a$ for all $n>N$

share|cite|improve this answer

answered 3 hours ago

StammeringMathematician

73616

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Why the downvote? +1 to compensate.
  – Paramanand Singh
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well the upvote was also to indicate that your answer is correct apart from the compensation.
  – Paramanand Singh
  3 hours ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

If a sequence $x_n$ converges to $x$, by definition it means for every $epsilon >0$ there exist a $N$ such that $$|x_n-x|<epsilontag1$$ for all $n>N$

Now (1) can be rephrased as $x_nin (x-epsilon, x+epsilon)tag2$ for all $n>N$

Take $epsilon=x-a$ (as $x>a, x-a$ is a positive number and hence the choice of taking $x-a=epsilon$ makes sense).

By definition there exist a $N$ such that
$$x_nin (x-epsilon, x+epsilon)$$ for all $n>N$. Now put the value of $epsilon $ you get $$x_nin (x-(x-a),x+x-a)=(a,2x-a)$$

Therefore $x_n>a$ for all $n>N$

share|cite|improve this answer

answered 3 hours ago

StammeringMathematician

73616

If a sequence $x_n$ converges to $x$, by definition it means for every $epsilon >0$ there exist a $N$ such that $$|x_n-x|<epsilontag1$$ for all $n>N$

Now (1) can be rephrased as $x_nin (x-epsilon, x+epsilon)tag2$ for all $n>N$

Take $epsilon=x-a$ (as $x>a, x-a$ is a positive number and hence the choice of taking $x-a=epsilon$ makes sense).

By definition there exist a $N$ such that
$$x_nin (x-epsilon, x+epsilon)$$ for all $n>N$. Now put the value of $epsilon $ you get $$x_nin (x-(x-a),x+x-a)=(a,2x-a)$$

Therefore $x_n>a$ for all $n>N$

share|cite|improve this answer

answered 3 hours ago

StammeringMathematician

73616

share|cite|improve this answer

share|cite|improve this answer

answered 3 hours ago

StammeringMathematician

73616

answered 3 hours ago

StammeringMathematician

73616

answered 3 hours ago

StammeringMathematician

73616

73616

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Why the downvote? +1 to compensate.
  – Paramanand Singh
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well the upvote was also to indicate that your answer is correct apart from the compensation.
  – Paramanand Singh
  3 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Why the downvote? +1 to compensate.
  – Paramanand Singh
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well the upvote was also to indicate that your answer is correct apart from the compensation.
  – Paramanand Singh
  3 hours ago
  

  
  
  
  

1

1

Why the downvote? +1 to compensate.
– Paramanand Singh
3 hours ago

Why the downvote? +1 to compensate.
– Paramanand Singh
3 hours ago

Well the upvote was also to indicate that your answer is correct apart from the compensation.
– Paramanand Singh
3 hours ago

Well the upvote was also to indicate that your answer is correct apart from the compensation.
– Paramanand Singh
3 hours ago

add a comment | 

up vote
1
down vote

Let $epsilon = frac x-a2$

Then $x-epsilon = frac x+a2 >a$

Thus $x_n > x-epsilon implies x_n >a$

share|cite|improve this answer

answered 3 hours ago

Mohammad Riazi-Kermani

31.7k41853

add a comment | 

up vote
1
down vote

Let $epsilon = frac x-a2$

Then $x-epsilon = frac x+a2 >a$

Thus $x_n > x-epsilon implies x_n >a$

share|cite|improve this answer

answered 3 hours ago

Mohammad Riazi-Kermani

31.7k41853

add a comment | 

up vote
1
down vote

up vote
1
down vote

Let $epsilon = frac x-a2$

Then $x-epsilon = frac x+a2 >a$

Thus $x_n > x-epsilon implies x_n >a$

share|cite|improve this answer

answered 3 hours ago

Mohammad Riazi-Kermani

31.7k41853

Let $epsilon = frac x-a2$

Then $x-epsilon = frac x+a2 >a$

Thus $x_n > x-epsilon implies x_n >a$

share|cite|improve this answer

answered 3 hours ago

Mohammad Riazi-Kermani

31.7k41853

share|cite|improve this answer

share|cite|improve this answer

answered 3 hours ago

Mohammad Riazi-Kermani

31.7k41853

answered 3 hours ago

Mohammad Riazi-Kermani

31.7k41853

answered 3 hours ago

Mohammad Riazi-Kermani

31.7k41853

31.7k41853

add a comment | 

add a comment | 

Aishwarya Deore is a new contributor. Be nice, and check out our Code of Conduct.

 

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