Proof verification of $x_n = frac{2-cospi n}{2+cos pi n}$ does not have a limit using $varepsilon$ definition
up vote
2
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Let:
$$
begin{cases}
x_n = frac{2-cospi n}{2+cos pi n} \
a = 3
end{cases}
$$
Prove that $a$ is not the limit of ${x_n}$ using $varepsilon$ definition.
Start with the definition of limit:
$$
lim_{nto infty}x_n = a stackrel{text{def}}{iff} {forall varepsilon > 0, exists N in mathbb N, forall n>N:|x_n - a| < varepsilon }
$$
Now negate that definition. For the sequence to not have a limit we have the following to be true:
$$
{exists varepsilon >0, forall N in mathbb N, exists n > N: |x_n - a| ge varepsilon}
$$
Try to find such epsilon. Since $n in mathbb N$ we have that for $kinmathbb N$:
$$
begin{equation}
|x_n| =
begin{cases}
3, & n=2k -1, \
{1over 3}, & n = 2k
end{cases}
end{equation}
$$
Now take for instance $varepsilon = 1$. With that $varepsilon$ for any $N$ we may find infinitely many $n$ such that $n > N text{and} |x_n -3| ge 1$. Thus the sequence does not have a limit.
I'm kindly asking to verify two things:
- The negation of limit definition
- The proof itself
Btw is it true that any periodic sequence doesn't have a limit?
Thank you!
calculus limits proof-verification epsilon-delta
asked Nov 22 at 15:49
roman
1,46111119
add a comment |
up vote
2
down vote
favorite
Let:
$$
begin{cases}
x_n = frac{2-cospi n}{2+cos pi n} \
a = 3
end{cases}
$$
Prove that $a$ is not the limit of ${x_n}$ using $varepsilon$ definition.
Start with the definition of limit:
$$
lim_{nto infty}x_n = a stackrel{text{def}}{iff} {forall varepsilon > 0, exists N in mathbb N, forall n>N:|x_n - a| < varepsilon }
$$
Now negate that definition. For the sequence to not have a limit we have the following to be true:
$$
{exists varepsilon >0, forall N in mathbb N, exists n > N: |x_n - a| ge varepsilon}
$$
Try to find such epsilon. Since $n in mathbb N$ we have that for $kinmathbb N$:
$$
begin{equation}
|x_n| =
begin{cases}
3, & n=2k -1, \
{1over 3}, & n = 2k
end{cases}
end{equation}
$$
Now take for instance $varepsilon = 1$. With that $varepsilon$ for any $N$ we may find infinitely many $n$ such that $n > N text{and} |x_n -3| ge 1$. Thus the sequence does not have a limit.
I'm kindly asking to verify two things:
- The negation of limit definition
- The proof itself
Btw is it true that any periodic sequence doesn't have a limit?
Thank you!
calculus limits proof-verification epsilon-delta
asked Nov 22 at 15:49
roman
1,46111119
$x_n=frac{1}{3}$ if n=even , and if n=odd $x_n=3$
– Dadrahm
Nov 22 at 15:54
@Dadrahm right, i've spotted that as well. It should be $3$ instead of $2$
– roman
Nov 22 at 15:55
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let:
$$
begin{cases}
x_n = frac{2-cospi n}{2+cos pi n} \
a = 3
end{cases}
$$
Prove that $a$ is not the limit of ${x_n}$ using $varepsilon$ definition.
Start with the definition of limit:
$$
lim_{nto infty}x_n = a stackrel{text{def}}{iff} {forall varepsilon > 0, exists N in mathbb N, forall n>N:|x_n - a| < varepsilon }
$$
Now negate that definition. For the sequence to not have a limit we have the following to be true:
$$
{exists varepsilon >0, forall N in mathbb N, exists n > N: |x_n - a| ge varepsilon}
$$
Try to find such epsilon. Since $n in mathbb N$ we have that for $kinmathbb N$:
$$
begin{equation}
|x_n| =
begin{cases}
3, & n=2k -1, \
{1over 3}, & n = 2k
end{cases}
end{equation}
$$
Now take for instance $varepsilon = 1$. With that $varepsilon$ for any $N$ we may find infinitely many $n$ such that $n > N text{and} |x_n -3| ge 1$. Thus the sequence does not have a limit.
I'm kindly asking to verify two things:
- The negation of limit definition
- The proof itself
Btw is it true that any periodic sequence doesn't have a limit?
Thank you!
calculus limits proof-verification epsilon-delta
asked Nov 22 at 15:49
roman
1,46111119
Let:
$$
begin{cases}
x_n = frac{2-cospi n}{2+cos pi n} \
a = 3
end{cases}
$$
Prove that $a$ is not the limit of ${x_n}$ using $varepsilon$ definition.
Start with the definition of limit:
$$
lim_{nto infty}x_n = a stackrel{text{def}}{iff} {forall varepsilon > 0, exists N in mathbb N, forall n>N:|x_n - a| < varepsilon }
$$
Now negate that definition. For the sequence to not have a limit we have the following to be true:
$$
{exists varepsilon >0, forall N in mathbb N, exists n > N: |x_n - a| ge varepsilon}
$$
Try to find such epsilon. Since $n in mathbb N$ we have that for $kinmathbb N$:
$$
begin{equation}
|x_n| =
begin{cases}
3, & n=2k -1, \
{1over 3}, & n = 2k
end{cases}
end{equation}
$$
Now take for instance $varepsilon = 1$. With that $varepsilon$ for any $N$ we may find infinitely many $n$ such that $n > N text{and} |x_n -3| ge 1$. Thus the sequence does not have a limit.
I'm kindly asking to verify two things:
- The negation of limit definition
- The proof itself
Btw is it true that any periodic sequence doesn't have a limit?
Thank you!
calculus limits proof-verification epsilon-delta
calculus limits proof-verification epsilon-delta
asked Nov 22 at 15:49
roman
1,46111119
asked Nov 22 at 15:49
roman
1,46111119
asked Nov 22 at 15:49
roman
1,46111119
asked Nov 22 at 15:49
roman
1,46111119
asked Nov 22 at 15:49
roman
1,46111119
1,46111119
$x_n=frac{1}{3}$ if n=even , and if n=odd $x_n=3$
– Dadrahm
Nov 22 at 15:54
@Dadrahm right, i've spotted that as well. It should be $3$ instead of $2$
– roman
Nov 22 at 15:55
add a comment |
$x_n=frac{1}{3}$ if n=even , and if n=odd $x_n=3$
– Dadrahm
Nov 22 at 15:54
@Dadrahm right, i've spotted that as well. It should be $3$ instead of $2$
– roman
Nov 22 at 15:55
$x_n=frac{1}{3}$ if n=even , and if n=odd $x_n=3$
– Dadrahm
Nov 22 at 15:54
$x_n=frac{1}{3}$ if n=even , and if n=odd $x_n=3$
– Dadrahm
Nov 22 at 15:54
@Dadrahm right, i've spotted that as well. It should be $3$ instead of $2$
– roman
Nov 22 at 15:55
@Dadrahm right, i've spotted that as well. It should be $3$ instead of $2$
– roman
Nov 22 at 15:55
add a comment |
1 Answer
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votes
up vote
1
down vote
accepted
The definition of the limit relies on finding an arbitrary $varepsilon >0$ but as small as you would like to, essentialy "reaching" zero thus $x_n to a$. Since you found a $varepsilon >0$ such that the sequence does not converge (meaning that $|x_n - a | geq epsilon$) then indeed the sequence does not have a limit.
answered Nov 22 at 15:56
Rebellos
13.9k31243
Did OP really prove that, no matter what $a$ is chosen, there is an $x_n$ further than $1$ from $a$? [I think that would be straightforward from the formulas for $a_n$ depending on parity of $n$ that he put in post.]
– coffeemath
Nov 22 at 16:03
I don't see how $frac{sin(n)}{2^n}$ is periodic. What would be it's period?
– roman
Nov 22 at 16:04
1
@coffeemath but $a$ is given to equal $3$, do we really need to consider the case of "any" $a$?
– roman
Nov 22 at 16:10
@roman Sorry, I thought you meant a sequence involving periodic terms. As for the convergence part, your proof is correct. Since you are only interested for $a=3$ it is enough what you've done. The case of any $a$ determines just a wider form of limit.
– Rebellos
Nov 22 at 16:17
@coffeemath Since the OP is interested for the value of a supposed limit of $a=3$ he/she does not have to check for all the cases of limits $a in mathbb R$.
– Rebellos
Nov 22 at 16:18
|
show 2 more comments
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The definition of the limit relies on finding an arbitrary $varepsilon >0$ but as small as you would like to, essentialy "reaching" zero thus $x_n to a$. Since you found a $varepsilon >0$ such that the sequence does not converge (meaning that $|x_n - a | geq epsilon$) then indeed the sequence does not have a limit.
answered Nov 22 at 15:56
Rebellos
13.9k31243
Did OP really prove that, no matter what $a$ is chosen, there is an $x_n$ further than $1$ from $a$? [I think that would be straightforward from the formulas for $a_n$ depending on parity of $n$ that he put in post.]
– coffeemath
Nov 22 at 16:03
I don't see how $frac{sin(n)}{2^n}$ is periodic. What would be it's period?
– roman
Nov 22 at 16:04
1
@coffeemath but $a$ is given to equal $3$, do we really need to consider the case of "any" $a$?
– roman
Nov 22 at 16:10
@roman Sorry, I thought you meant a sequence involving periodic terms. As for the convergence part, your proof is correct. Since you are only interested for $a=3$ it is enough what you've done. The case of any $a$ determines just a wider form of limit.
– Rebellos
Nov 22 at 16:17
@coffeemath Since the OP is interested for the value of a supposed limit of $a=3$ he/she does not have to check for all the cases of limits $a in mathbb R$.
– Rebellos
Nov 22 at 16:18
|
show 2 more comments
up vote
1
down vote
accepted
The definition of the limit relies on finding an arbitrary $varepsilon >0$ but as small as you would like to, essentialy "reaching" zero thus $x_n to a$. Since you found a $varepsilon >0$ such that the sequence does not converge (meaning that $|x_n - a | geq epsilon$) then indeed the sequence does not have a limit.
answered Nov 22 at 15:56
Rebellos
13.9k31243
Did OP really prove that, no matter what $a$ is chosen, there is an $x_n$ further than $1$ from $a$? [I think that would be straightforward from the formulas for $a_n$ depending on parity of $n$ that he put in post.]
– coffeemath
Nov 22 at 16:03
I don't see how $frac{sin(n)}{2^n}$ is periodic. What would be it's period?
– roman
Nov 22 at 16:04
1
@coffeemath but $a$ is given to equal $3$, do we really need to consider the case of "any" $a$?
– roman
Nov 22 at 16:10
@roman Sorry, I thought you meant a sequence involving periodic terms. As for the convergence part, your proof is correct. Since you are only interested for $a=3$ it is enough what you've done. The case of any $a$ determines just a wider form of limit.
– Rebellos
Nov 22 at 16:17
@coffeemath Since the OP is interested for the value of a supposed limit of $a=3$ he/she does not have to check for all the cases of limits $a in mathbb R$.
– Rebellos
Nov 22 at 16:18
|
show 2 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The definition of the limit relies on finding an arbitrary $varepsilon >0$ but as small as you would like to, essentialy "reaching" zero thus $x_n to a$. Since you found a $varepsilon >0$ such that the sequence does not converge (meaning that $|x_n - a | geq epsilon$) then indeed the sequence does not have a limit.
answered Nov 22 at 15:56
Rebellos
13.9k31243
The definition of the limit relies on finding an arbitrary $varepsilon >0$ but as small as you would like to, essentialy "reaching" zero thus $x_n to a$. Since you found a $varepsilon >0$ such that the sequence does not converge (meaning that $|x_n - a | geq epsilon$) then indeed the sequence does not have a limit.
answered Nov 22 at 15:56
Rebellos
13.9k31243
edited Nov 22 at 16:16
answered Nov 22 at 15:56
Rebellos
13.9k31243
answered Nov 22 at 15:56
Rebellos
13.9k31243
answered Nov 22 at 15:56
Rebellos
13.9k31243
13.9k31243
Did OP really prove that, no matter what $a$ is chosen, there is an $x_n$ further than $1$ from $a$? [I think that would be straightforward from the formulas for $a_n$ depending on parity of $n$ that he put in post.]
– coffeemath
Nov 22 at 16:03
I don't see how $frac{sin(n)}{2^n}$ is periodic. What would be it's period?
– roman
Nov 22 at 16:04
1
@coffeemath but $a$ is given to equal $3$, do we really need to consider the case of "any" $a$?
– roman
Nov 22 at 16:10
@roman Sorry, I thought you meant a sequence involving periodic terms. As for the convergence part, your proof is correct. Since you are only interested for $a=3$ it is enough what you've done. The case of any $a$ determines just a wider form of limit.
– Rebellos
Nov 22 at 16:17
@coffeemath Since the OP is interested for the value of a supposed limit of $a=3$ he/she does not have to check for all the cases of limits $a in mathbb R$.
– Rebellos
Nov 22 at 16:18
|
show 2 more comments
Did OP really prove that, no matter what $a$ is chosen, there is an $x_n$ further than $1$ from $a$? [I think that would be straightforward from the formulas for $a_n$ depending on parity of $n$ that he put in post.]
– coffeemath
Nov 22 at 16:03
I don't see how $frac{sin(n)}{2^n}$ is periodic. What would be it's period?
– roman
Nov 22 at 16:04
1
@coffeemath but $a$ is given to equal $3$, do we really need to consider the case of "any" $a$?
– roman
Nov 22 at 16:10
@roman Sorry, I thought you meant a sequence involving periodic terms. As for the convergence part, your proof is correct. Since you are only interested for $a=3$ it is enough what you've done. The case of any $a$ determines just a wider form of limit.
– Rebellos
Nov 22 at 16:17
@coffeemath Since the OP is interested for the value of a supposed limit of $a=3$ he/she does not have to check for all the cases of limits $a in mathbb R$.
– Rebellos
Nov 22 at 16:18
Did OP really prove that, no matter what $a$ is chosen, there is an $x_n$ further than $1$ from $a$? [I think that would be straightforward from the formulas for $a_n$ depending on parity of $n$ that he put in post.]
– coffeemath
Nov 22 at 16:03
Did OP really prove that, no matter what $a$ is chosen, there is an $x_n$ further than $1$ from $a$? [I think that would be straightforward from the formulas for $a_n$ depending on parity of $n$ that he put in post.]
– coffeemath
Nov 22 at 16:03
I don't see how $frac{sin(n)}{2^n}$ is periodic. What would be it's period?
– roman
Nov 22 at 16:04
I don't see how $frac{sin(n)}{2^n}$ is periodic. What would be it's period?
– roman
Nov 22 at 16:04
1
1
@coffeemath but $a$ is given to equal $3$, do we really need to consider the case of "any" $a$?
– roman
Nov 22 at 16:10
@coffeemath but $a$ is given to equal $3$, do we really need to consider the case of "any" $a$?
– roman
Nov 22 at 16:10
@roman Sorry, I thought you meant a sequence involving periodic terms. As for the convergence part, your proof is correct. Since you are only interested for $a=3$ it is enough what you've done. The case of any $a$ determines just a wider form of limit.
– Rebellos
Nov 22 at 16:17
@roman Sorry, I thought you meant a sequence involving periodic terms. As for the convergence part, your proof is correct. Since you are only interested for $a=3$ it is enough what you've done. The case of any $a$ determines just a wider form of limit.
– Rebellos
Nov 22 at 16:17
@coffeemath Since the OP is interested for the value of a supposed limit of $a=3$ he/she does not have to check for all the cases of limits $a in mathbb R$.
– Rebellos
Nov 22 at 16:18
@coffeemath Since the OP is interested for the value of a supposed limit of $a=3$ he/she does not have to check for all the cases of limits $a in mathbb R$.
– Rebellos
Nov 22 at 16:18
|
show 2 more comments
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$x_n=frac{1}{3}$ if n=even , and if n=odd $x_n=3$
– Dadrahm
Nov 22 at 15:54
@Dadrahm right, i've spotted that as well. It should be $3$ instead of $2$
– roman
Nov 22 at 15:55