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Limit of sequence (by definiton)

Prove:$limlimits_nto infty frac(-1)^nn = 0$How to find the limit of this rational sequence?2 Sub-limits of sequence converge epsilon proofLimit of a function involving a sequence.Understanding how to use $epsilon-delta$ definition of a limitFind the limit of a sequence defined recursivelyCubic root of sequence translates into cubic root of limit?Prove that $a_n=fracnn+1$ is convergentProving Limit of sequence using Epsilon-N definitionProof of limit of quotient sequence property

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty
margin-bottom:0;

.everyonelovesstackoverflowposition:absolute;height:1px;width:1px;opacity:0;top:0;left:0;pointer-events:none;

5

$begingroup$

I have this assignment:

$$lim_nto infty frac(-1)^n-3n^2=0$$

I have to prove that limit by definition, but I am stuck with

$$|a_n - 0|< epsilon$$

Dont know how to deal with that absolute value.

The solution is

$$n_0=left[frac2sqrtepsilonright]+1$$

calculus limits

share|cite|improve this question

edited 9 hours ago

Don Thousand

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asked 10 hours ago

naruto25naruto25

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New contributor

naruto25 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Have you tried to see what the numerator is for the first few $n$? That might give you an idea of how to deal with the absolute value. Note that the denominator is always positive.
  $endgroup$
  – Theoretical Economist
  10 hours ago
  

  
  
  
  

add a comment
 | 

5

$begingroup$

I have this assignment:

$$lim_nto infty frac(-1)^n-3n^2=0$$

I have to prove that limit by definition, but I am stuck with

$$|a_n - 0|< epsilon$$

Dont know how to deal with that absolute value.

The solution is

$$n_0=left[frac2sqrtepsilonright]+1$$

calculus limits

share|cite|improve this question

edited 9 hours ago

Don Thousand

5,04411 gold badge1111 silver badges3535 bronze badges

asked 10 hours ago

naruto25naruto25

5766 bronze badges

New contributor

naruto25 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Have you tried to see what the numerator is for the first few $n$? That might give you an idea of how to deal with the absolute value. Note that the denominator is always positive.
  $endgroup$
  – Theoretical Economist
  10 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

I have this assignment:

$$lim_nto infty frac(-1)^n-3n^2=0$$

I have to prove that limit by definition, but I am stuck with

$$|a_n - 0|< epsilon$$

Dont know how to deal with that absolute value.

The solution is

$$n_0=left[frac2sqrtepsilonright]+1$$

calculus limits

share|cite|improve this question

edited 9 hours ago

Don Thousand

5,04411 gold badge1111 silver badges3535 bronze badges

asked 10 hours ago

naruto25naruto25

5766 bronze badges

New contributor

naruto25 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 9 hours ago

Don Thousand

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asked 10 hours ago

naruto25naruto25

5766 bronze badges

New contributor

naruto25 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 9 hours ago

Don Thousand

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asked 10 hours ago

naruto25naruto25

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naruto25 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 9 hours ago

Don Thousand

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edited 9 hours ago

Don Thousand

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asked 10 hours ago

naruto25naruto25

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naruto25 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 10 hours ago

naruto25naruto25

5766 bronze badges

3 Answers
3

active

oldest

votes

3

$begingroup$

We have that

$$frac-2n^2lefrac(-1)^n-3n^2le frac-4n^2 $$

and then

$$frac2n^2leleft|frac(-1)^n-3n^2right|le frac4n^2 $$

therefore

$$left|frac(-1)^n-3n^2right|le frac4n^2 <epsilon implies n^2 > frac4epsilon implies n>frac2sqrtepsilon$$

therefore it suffices to take

$$n_0=overbracecolorblueleft[frac2sqrtepsilonright]^colorredtextinteger part+1$$

share|cite|improve this answer

edited 10 hours ago

answered 10 hours ago

gimusigimusi

97.4k99 gold badges4747 silver badges9696 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Since we need $n>frac2sqrtepsilon$ we take the integer part of that and add +1.
  $endgroup$
  – gimusi
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @naruto25 By definition of limit we are looking for $n_0 in mathbb N$ but $frac2sqrtepsilon in mathbb R$ therefore we need to extract the integer part and then add $1$. Try with a numerica example: suppose we obtain $n> 100.45$ then we take $n_0=100+1$.
  $endgroup$
  – gimusi
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @naruto25 We are taking an upper bound that is $4/n^2$, we don't need the exact bound but a simple upper which works. For example we also could take $6/n^2$ and the proof works at the same way.
  $endgroup$
  – gimusi
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @naruto25 We are thinking $epsilon$ as a small number, for example $epsilon=.002 implies sqrt epsilon =0.04472...$ and $frac2sqrtepsilon>44.72...$ then we can take $n_0=45$.
  $endgroup$
  – gimusi
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @naruto25 Exactly!
  $endgroup$
  – gimusi
  9 hours ago
  

  
  
  
  

 | 
show 5 more comments

4

$begingroup$

Given $epsilon>0 $, we should find $N$ such that if $nge N$ then $$|frac(-1)^n-3n^2|<epsilon$$

but

$$|frac(-1)^n-3n^2|le |frac(-1)^nn^2|+|frac3n^2|le frac4n^2$$

thus it is sufficient to find one $N$ such that,

$$nge N ;; implies ;; frac4n^2<epsilon$$

or

$$nge N ;; implies ;; n^2 > frac4epsilon$$

or
$$nge N ;; implies ;; n > frac2sqrtepsilon$$

So, each $N$ satisfying $N>frac2sqrtepsilon$ will work.

the smallest of these $N$ is

$$lfloor frac2sqrtepsilon rfloor +1$$

share|cite|improve this answer

answered 10 hours ago

hamam_Abdallahhamam_Abdallah

40k22 gold badges1717 silver badges3535 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So if I had epsilon=1000000 and then it was 0. I do not have null member of that sequence and thus I have to add that 1?
  $endgroup$
  – naruto25
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $epsilon=1000000$ then $N=1$
  $endgroup$
  – hamam_Abdallah
  10 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In general, $epsilon$ Must be Small. Say if $epsilon =0.000001$ then $N =?$
  $endgroup$
  – hamam_Abdallah
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am still curious why do we have to add that one to that solution. If I had epsilon=1, then the first member of sequence(n0) from which that limit starts should be 2. And every next member should fit in distance epsilon.
  $endgroup$
  – naruto25
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @naruto25 If it is true for small epsilons, it will be also true for greater epsilons.
  $endgroup$
  – hamam_Abdallah
  10 hours ago
  

  
  
  
  

add a comment
 | 

1

$begingroup$

Notice, that:

$$frac(-1)^n-3n^2 geq frac-4n^2$$

And is always negative. So for every $epsilon$ you can find $N$ ($=frac2sqrtepsilon)$, that for every $n>N$ the following holds: $$|frac(-1)^n-3n^2| leq epsilon$$

share|cite|improve this answer

answered 10 hours ago

AndronicusAndronicus

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$endgroup$

add a comment
 | 

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3

$begingroup$

We have that

$$frac-2n^2lefrac(-1)^n-3n^2le frac-4n^2 $$

and then

$$frac2n^2leleft|frac(-1)^n-3n^2right|le frac4n^2 $$

therefore

$$left|frac(-1)^n-3n^2right|le frac4n^2 <epsilon implies n^2 > frac4epsilon implies n>frac2sqrtepsilon$$

therefore it suffices to take

$$n_0=overbracecolorblueleft[frac2sqrtepsilonright]^colorredtextinteger part+1$$

share|cite|improve this answer

edited 10 hours ago

answered 10 hours ago

gimusigimusi

97.4k99 gold badges4747 silver badges9696 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Since we need $n>frac2sqrtepsilon$ we take the integer part of that and add +1.
  $endgroup$
  – gimusi
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @naruto25 By definition of limit we are looking for $n_0 in mathbb N$ but $frac2sqrtepsilon in mathbb R$ therefore we need to extract the integer part and then add $1$. Try with a numerica example: suppose we obtain $n> 100.45$ then we take $n_0=100+1$.
  $endgroup$
  – gimusi
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @naruto25 We are taking an upper bound that is $4/n^2$, we don't need the exact bound but a simple upper which works. For example we also could take $6/n^2$ and the proof works at the same way.
  $endgroup$
  – gimusi
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @naruto25 We are thinking $epsilon$ as a small number, for example $epsilon=.002 implies sqrt epsilon =0.04472...$ and $frac2sqrtepsilon>44.72...$ then we can take $n_0=45$.
  $endgroup$
  – gimusi
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @naruto25 Exactly!
  $endgroup$
  – gimusi
  9 hours ago
  

  
  
  
  

 | 
show 5 more comments

3

$begingroup$

We have that

$$frac-2n^2lefrac(-1)^n-3n^2le frac-4n^2 $$

and then

$$frac2n^2leleft|frac(-1)^n-3n^2right|le frac4n^2 $$

therefore

$$left|frac(-1)^n-3n^2right|le frac4n^2 <epsilon implies n^2 > frac4epsilon implies n>frac2sqrtepsilon$$

therefore it suffices to take

$$n_0=overbracecolorblueleft[frac2sqrtepsilonright]^colorredtextinteger part+1$$

share|cite|improve this answer

edited 10 hours ago

answered 10 hours ago

gimusigimusi

97.4k99 gold badges4747 silver badges9696 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Since we need $n>frac2sqrtepsilon$ we take the integer part of that and add +1.
  $endgroup$
  – gimusi
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @naruto25 By definition of limit we are looking for $n_0 in mathbb N$ but $frac2sqrtepsilon in mathbb R$ therefore we need to extract the integer part and then add $1$. Try with a numerica example: suppose we obtain $n> 100.45$ then we take $n_0=100+1$.
  $endgroup$
  – gimusi
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @naruto25 We are taking an upper bound that is $4/n^2$, we don't need the exact bound but a simple upper which works. For example we also could take $6/n^2$ and the proof works at the same way.
  $endgroup$
  – gimusi
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @naruto25 We are thinking $epsilon$ as a small number, for example $epsilon=.002 implies sqrt epsilon =0.04472...$ and $frac2sqrtepsilon>44.72...$ then we can take $n_0=45$.
  $endgroup$
  – gimusi
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @naruto25 Exactly!
  $endgroup$
  – gimusi
  9 hours ago
  

  
  
  
  

 | 
show 5 more comments

$begingroup$

We have that

$$frac-2n^2lefrac(-1)^n-3n^2le frac-4n^2 $$

and then

$$frac2n^2leleft|frac(-1)^n-3n^2right|le frac4n^2 $$

therefore

$$left|frac(-1)^n-3n^2right|le frac4n^2 <epsilon implies n^2 > frac4epsilon implies n>frac2sqrtepsilon$$

therefore it suffices to take

$$n_0=overbracecolorblueleft[frac2sqrtepsilonright]^colorredtextinteger part+1$$

share|cite|improve this answer

edited 10 hours ago

answered 10 hours ago

gimusigimusi

97.4k99 gold badges4747 silver badges9696 bronze badges

$endgroup$

share|cite|improve this answer

edited 10 hours ago

answered 10 hours ago

gimusigimusi

97.4k99 gold badges4747 silver badges9696 bronze badges

answered 10 hours ago

gimusigimusi

97.4k99 gold badges4747 silver badges9696 bronze badges

answered 10 hours ago

gimusigimusi

97.4k99 gold badges4747 silver badges9696 bronze badges

4

$begingroup$

Given $epsilon>0 $, we should find $N$ such that if $nge N$ then $$|frac(-1)^n-3n^2|<epsilon$$

but

$$|frac(-1)^n-3n^2|le |frac(-1)^nn^2|+|frac3n^2|le frac4n^2$$

thus it is sufficient to find one $N$ such that,

$$nge N ;; implies ;; frac4n^2<epsilon$$

or

$$nge N ;; implies ;; n^2 > frac4epsilon$$

or
$$nge N ;; implies ;; n > frac2sqrtepsilon$$

So, each $N$ satisfying $N>frac2sqrtepsilon$ will work.

the smallest of these $N$ is

$$lfloor frac2sqrtepsilon rfloor +1$$

share|cite|improve this answer

answered 10 hours ago

hamam_Abdallahhamam_Abdallah

40k22 gold badges1717 silver badges3535 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So if I had epsilon=1000000 and then it was 0. I do not have null member of that sequence and thus I have to add that 1?
  $endgroup$
  – naruto25
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $epsilon=1000000$ then $N=1$
  $endgroup$
  – hamam_Abdallah
  10 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In general, $epsilon$ Must be Small. Say if $epsilon =0.000001$ then $N =?$
  $endgroup$
  – hamam_Abdallah
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am still curious why do we have to add that one to that solution. If I had epsilon=1, then the first member of sequence(n0) from which that limit starts should be 2. And every next member should fit in distance epsilon.
  $endgroup$
  – naruto25
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @naruto25 If it is true for small epsilons, it will be also true for greater epsilons.
  $endgroup$
  – hamam_Abdallah
  10 hours ago
  

  
  
  
  

add a comment
 | 

4

$begingroup$

Given $epsilon>0 $, we should find $N$ such that if $nge N$ then $$|frac(-1)^n-3n^2|<epsilon$$

but

$$|frac(-1)^n-3n^2|le |frac(-1)^nn^2|+|frac3n^2|le frac4n^2$$

thus it is sufficient to find one $N$ such that,

$$nge N ;; implies ;; frac4n^2<epsilon$$

or

$$nge N ;; implies ;; n^2 > frac4epsilon$$

or
$$nge N ;; implies ;; n > frac2sqrtepsilon$$

So, each $N$ satisfying $N>frac2sqrtepsilon$ will work.

the smallest of these $N$ is

$$lfloor frac2sqrtepsilon rfloor +1$$

share|cite|improve this answer

answered 10 hours ago

hamam_Abdallahhamam_Abdallah

40k22 gold badges1717 silver badges3535 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So if I had epsilon=1000000 and then it was 0. I do not have null member of that sequence and thus I have to add that 1?
  $endgroup$
  – naruto25
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $epsilon=1000000$ then $N=1$
  $endgroup$
  – hamam_Abdallah
  10 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In general, $epsilon$ Must be Small. Say if $epsilon =0.000001$ then $N =?$
  $endgroup$
  – hamam_Abdallah
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am still curious why do we have to add that one to that solution. If I had epsilon=1, then the first member of sequence(n0) from which that limit starts should be 2. And every next member should fit in distance epsilon.
  $endgroup$
  – naruto25
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @naruto25 If it is true for small epsilons, it will be also true for greater epsilons.
  $endgroup$
  – hamam_Abdallah
  10 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

Given $epsilon>0 $, we should find $N$ such that if $nge N$ then $$|frac(-1)^n-3n^2|<epsilon$$

but

$$|frac(-1)^n-3n^2|le |frac(-1)^nn^2|+|frac3n^2|le frac4n^2$$

thus it is sufficient to find one $N$ such that,

$$nge N ;; implies ;; frac4n^2<epsilon$$

or

$$nge N ;; implies ;; n^2 > frac4epsilon$$

or
$$nge N ;; implies ;; n > frac2sqrtepsilon$$

So, each $N$ satisfying $N>frac2sqrtepsilon$ will work.

the smallest of these $N$ is

$$lfloor frac2sqrtepsilon rfloor +1$$

share|cite|improve this answer

answered 10 hours ago

hamam_Abdallahhamam_Abdallah

40k22 gold badges1717 silver badges3535 bronze badges

$endgroup$

share|cite|improve this answer

answered 10 hours ago

hamam_Abdallahhamam_Abdallah

40k22 gold badges1717 silver badges3535 bronze badges

answered 10 hours ago

hamam_Abdallahhamam_Abdallah

40k22 gold badges1717 silver badges3535 bronze badges

answered 10 hours ago

hamam_Abdallahhamam_Abdallah

40k22 gold badges1717 silver badges3535 bronze badges

1

$begingroup$

Notice, that:

$$frac(-1)^n-3n^2 geq frac-4n^2$$

And is always negative. So for every $epsilon$ you can find $N$ ($=frac2sqrtepsilon)$, that for every $n>N$ the following holds: $$|frac(-1)^n-3n^2| leq epsilon$$

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answered 10 hours ago

AndronicusAndronicus

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$begingroup$

Notice, that:

$$frac(-1)^n-3n^2 geq frac-4n^2$$

And is always negative. So for every $epsilon$ you can find $N$ ($=frac2sqrtepsilon)$, that for every $n>N$ the following holds: $$|frac(-1)^n-3n^2| leq epsilon$$

share|cite|improve this answer

answered 10 hours ago

AndronicusAndronicus

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$endgroup$

add a comment
 | 

$begingroup$

Notice, that:

$$frac(-1)^n-3n^2 geq frac-4n^2$$

And is always negative. So for every $epsilon$ you can find $N$ ($=frac2sqrtepsilon)$, that for every $n>N$ the following holds: $$|frac(-1)^n-3n^2| leq epsilon$$

share|cite|improve this answer

answered 10 hours ago

AndronicusAndronicus

79811 gold badge22 silver badges1414 bronze badges

$endgroup$

share|cite|improve this answer

answered 10 hours ago

AndronicusAndronicus

79811 gold badge22 silver badges1414 bronze badges

answered 10 hours ago

AndronicusAndronicus

79811 gold badge22 silver badges1414 bronze badges

answered 10 hours ago

AndronicusAndronicus

79811 gold badge22 silver badges1414 bronze badges