Mixing
Proof of the Squeeze theorem for Sequences
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Show that if $x_n le y_n le z_n$ for all $n in mathbb N$, and if $lim(x_n) = lim (z_n) = l$, then $lim (y_n)= l$ as well.
Proof. Since $lim(x_n) = lim (z_n) = l$. Then follows that $|x_n -l| < fracepsilon2$ whenever $n ge N_1$ and similarly $|l-z_n | < fracepsilon2$ whenever $n ge N_2$. Choose $max N_1,N_2$ and by the triangle inequality we get $$|x_n - z_n| le|x_n -l|+ |l-z_n| lt epsilon .$$
since $$|x_n - z_n| lt epsilon $$ it follows that $x_n=z_n$ for sufficiently large $n ge maxN_1, N_2$. Hence it follows that $x_n=y_n= z_n$ (from the fact that $x_n=z_n$ and $x_n le y_n le z_n$). And therefore $lim(x_n) = lim (y_n) = l$.
Question 1: Is my attempt at the proof correct? If not why not? Also if not can you give a correct version or atleast a few hints?
Question 2: I use the property that $|a-b|< epsilon$ is equivalent to $a=b$ but it doesn't sit well with me below that $x_n=z_n$ since $(x_n), (z_n)$ could have different elements and still converge at the same value.) Why am I getting contradictory results here?
Thanks in advance.
I really appreciate this site helping me learn how to write proofs in real analysis :)
real-analysis sequences-and-series proof-verification inequality epsilon-delta
asked Jul 15 at 2:44
Red
1,747733
add a comment |Â
up vote
1
down vote
favorite
Show that if $x_n le y_n le z_n$ for all $n in mathbb N$, and if $lim(x_n) = lim (z_n) = l$, then $lim (y_n)= l$ as well.
Proof. Since $lim(x_n) = lim (z_n) = l$. Then follows that $|x_n -l| < fracepsilon2$ whenever $n ge N_1$ and similarly $|l-z_n | < fracepsilon2$ whenever $n ge N_2$. Choose $max N_1,N_2$ and by the triangle inequality we get $$|x_n - z_n| le|x_n -l|+ |l-z_n| lt epsilon .$$
since $$|x_n - z_n| lt epsilon $$ it follows that $x_n=z_n$ for sufficiently large $n ge maxN_1, N_2$. Hence it follows that $x_n=y_n= z_n$ (from the fact that $x_n=z_n$ and $x_n le y_n le z_n$). And therefore $lim(x_n) = lim (y_n) = l$.
Question 1: Is my attempt at the proof correct? If not why not? Also if not can you give a correct version or atleast a few hints?
Question 2: I use the property that $|a-b|< epsilon$ is equivalent to $a=b$ but it doesn't sit well with me below that $x_n=z_n$ since $(x_n), (z_n)$ could have different elements and still converge at the same value.) Why am I getting contradictory results here?
Thanks in advance.
I really appreciate this site helping me learn how to write proofs in real analysis :)
real-analysis sequences-and-series proof-verification inequality epsilon-delta
asked Jul 15 at 2:44
Red
1,747733
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Show that if $x_n le y_n le z_n$ for all $n in mathbb N$, and if $lim(x_n) = lim (z_n) = l$, then $lim (y_n)= l$ as well.
Proof. Since $lim(x_n) = lim (z_n) = l$. Then follows that $|x_n -l| < fracepsilon2$ whenever $n ge N_1$ and similarly $|l-z_n | < fracepsilon2$ whenever $n ge N_2$. Choose $max N_1,N_2$ and by the triangle inequality we get $$|x_n - z_n| le|x_n -l|+ |l-z_n| lt epsilon .$$
since $$|x_n - z_n| lt epsilon $$ it follows that $x_n=z_n$ for sufficiently large $n ge maxN_1, N_2$. Hence it follows that $x_n=y_n= z_n$ (from the fact that $x_n=z_n$ and $x_n le y_n le z_n$). And therefore $lim(x_n) = lim (y_n) = l$.
Question 1: Is my attempt at the proof correct? If not why not? Also if not can you give a correct version or atleast a few hints?
Question 2: I use the property that $|a-b|< epsilon$ is equivalent to $a=b$ but it doesn't sit well with me below that $x_n=z_n$ since $(x_n), (z_n)$ could have different elements and still converge at the same value.) Why am I getting contradictory results here?
Thanks in advance.
I really appreciate this site helping me learn how to write proofs in real analysis :)
real-analysis sequences-and-series proof-verification inequality epsilon-delta
asked Jul 15 at 2:44
Red
1,747733
Show that if $x_n le y_n le z_n$ for all $n in mathbb N$, and if $lim(x_n) = lim (z_n) = l$, then $lim (y_n)= l$ as well.
Proof. Since $lim(x_n) = lim (z_n) = l$. Then follows that $|x_n -l| < fracepsilon2$ whenever $n ge N_1$ and similarly $|l-z_n | < fracepsilon2$ whenever $n ge N_2$. Choose $max N_1,N_2$ and by the triangle inequality we get $$|x_n - z_n| le|x_n -l|+ |l-z_n| lt epsilon .$$
since $$|x_n - z_n| lt epsilon $$ it follows that $x_n=z_n$ for sufficiently large $n ge maxN_1, N_2$. Hence it follows that $x_n=y_n= z_n$ (from the fact that $x_n=z_n$ and $x_n le y_n le z_n$). And therefore $lim(x_n) = lim (y_n) = l$.
Question 1: Is my attempt at the proof correct? If not why not? Also if not can you give a correct version or atleast a few hints?
Question 2: I use the property that $|a-b|< epsilon$ is equivalent to $a=b$ but it doesn't sit well with me below that $x_n=z_n$ since $(x_n), (z_n)$ could have different elements and still converge at the same value.) Why am I getting contradictory results here?
Thanks in advance.
I really appreciate this site helping me learn how to write proofs in real analysis :)
real-analysis sequences-and-series proof-verification inequality epsilon-delta
asked Jul 15 at 2:44
Red
1,747733
edited Jul 15 at 2:54
asked Jul 15 at 2:44
Red
1,747733
asked Jul 15 at 2:44
Red
1,747733
asked Jul 15 at 2:44
Red
1,747733
1,747733
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
Firstly, $z_n$ does not equal $x_n$ for sufficiently large $n$. For an answer to your second question $frac1n$ and $frac12n$ both converge to zero, but they are never equal for a given $n$. Use the definition of convergence instead. What do you need to show that $lim_nto infty y_n=l$?
answered Jul 15 at 2:54
DeerintheHeadlights
1317
Yes that is correct
– Red
Jul 15 at 2:55
I completely agree with you that $z_n$ does not necessarily equal to $x_n$ but as I have shown the inequality $|x_n - z_n| lt epsilon$ would imply that $x_n = z_n$ doesn't it??
– Red
Jul 15 at 2:59
@Red: why would the inequality imply $z_n=x_n$?
– Paramanand Singh
Jul 15 at 6:06
add a comment |Â
up vote
2
down vote
Most elementary results of convergence for sequences either depend on the triangle inequality (via an $epsilon/2$ trick) or the inequality $0leqfrac1+leq 1$ for all $x$ (to avoid cases where division by $0$ may occur).
The Squeeze Theorem however is probably best proved by a different approach.
Try to show that if $xleq yleq z$ then $|y-l|leq maxleft$ no matter what $l$ may be. This is probably best proved with a picture.
answered Jul 15 at 3:05
Robert Wolfe
5,30722261
Is the aproach I have taken completely wrong or can it be tweaked for a correct proof?
– Red
Jul 15 at 3:11
@Red you have claimed that the sequences $x_n$ and $z_n$ eventually agree. But an easy counter example is $x_n=-1/n$, $z_n=1/n$ and $y_n=0$ for all $n$.
– Robert Wolfe
Jul 15 at 3:12
I completely agree with this, the sequences need not agree eventually.But why is the triangle inequality gving me that result anyway? I feel like I have used it wrong but I cannot figure out where my mistake llies? as I said in th second part of my question
– Red
Jul 15 at 3:16
1
@Red if $a$ and $b$ satisfy $|a-b|<epsilon$ for every $epsilon>0$ then $a=b$. However you have not proved that $|x_n-z_n|<epsilon$ for every $epsilon>0$. You have proved that for every $epsilon>0$ there is an $n$ such that $|x_n-z_n|<epsilon$.
– Robert Wolfe
Jul 15 at 3:27
I see my mistake. Thanks!
– Red
Jul 15 at 3:28
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Firstly, $z_n$ does not equal $x_n$ for sufficiently large $n$. For an answer to your second question $frac1n$ and $frac12n$ both converge to zero, but they are never equal for a given $n$. Use the definition of convergence instead. What do you need to show that $lim_nto infty y_n=l$?
answered Jul 15 at 2:54
DeerintheHeadlights
1317
Yes that is correct
– Red
Jul 15 at 2:55
I completely agree with you that $z_n$ does not necessarily equal to $x_n$ but as I have shown the inequality $|x_n - z_n| lt epsilon$ would imply that $x_n = z_n$ doesn't it??
– Red
Jul 15 at 2:59
@Red: why would the inequality imply $z_n=x_n$?
– Paramanand Singh
Jul 15 at 6:06
add a comment |Â
up vote
2
down vote
Firstly, $z_n$ does not equal $x_n$ for sufficiently large $n$. For an answer to your second question $frac1n$ and $frac12n$ both converge to zero, but they are never equal for a given $n$. Use the definition of convergence instead. What do you need to show that $lim_nto infty y_n=l$?
answered Jul 15 at 2:54
DeerintheHeadlights
1317
Yes that is correct
– Red
Jul 15 at 2:55
I completely agree with you that $z_n$ does not necessarily equal to $x_n$ but as I have shown the inequality $|x_n - z_n| lt epsilon$ would imply that $x_n = z_n$ doesn't it??
– Red
Jul 15 at 2:59
@Red: why would the inequality imply $z_n=x_n$?
– Paramanand Singh
Jul 15 at 6:06
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Firstly, $z_n$ does not equal $x_n$ for sufficiently large $n$. For an answer to your second question $frac1n$ and $frac12n$ both converge to zero, but they are never equal for a given $n$. Use the definition of convergence instead. What do you need to show that $lim_nto infty y_n=l$?
answered Jul 15 at 2:54
DeerintheHeadlights
1317
Firstly, $z_n$ does not equal $x_n$ for sufficiently large $n$. For an answer to your second question $frac1n$ and $frac12n$ both converge to zero, but they are never equal for a given $n$. Use the definition of convergence instead. What do you need to show that $lim_nto infty y_n=l$?
answered Jul 15 at 2:54
DeerintheHeadlights
1317
edited Jul 15 at 2:56
answered Jul 15 at 2:54
DeerintheHeadlights
1317
answered Jul 15 at 2:54
DeerintheHeadlights
1317
answered Jul 15 at 2:54
DeerintheHeadlights
1317
1317
Yes that is correct
– Red
Jul 15 at 2:55
I completely agree with you that $z_n$ does not necessarily equal to $x_n$ but as I have shown the inequality $|x_n - z_n| lt epsilon$ would imply that $x_n = z_n$ doesn't it??
– Red
Jul 15 at 2:59
@Red: why would the inequality imply $z_n=x_n$?
– Paramanand Singh
Jul 15 at 6:06
add a comment |Â
Yes that is correct
– Red
Jul 15 at 2:55
I completely agree with you that $z_n$ does not necessarily equal to $x_n$ but as I have shown the inequality $|x_n - z_n| lt epsilon$ would imply that $x_n = z_n$ doesn't it??
– Red
Jul 15 at 2:59
@Red: why would the inequality imply $z_n=x_n$?
– Paramanand Singh
Jul 15 at 6:06
Yes that is correct
– Red
Jul 15 at 2:55
Yes that is correct
– Red
Jul 15 at 2:55
I completely agree with you that $z_n$ does not necessarily equal to $x_n$ but as I have shown the inequality $|x_n - z_n| lt epsilon$ would imply that $x_n = z_n$ doesn't it??
– Red
Jul 15 at 2:59
I completely agree with you that $z_n$ does not necessarily equal to $x_n$ but as I have shown the inequality $|x_n - z_n| lt epsilon$ would imply that $x_n = z_n$ doesn't it??
– Red
Jul 15 at 2:59
@Red: why would the inequality imply $z_n=x_n$?
– Paramanand Singh
Jul 15 at 6:06
@Red: why would the inequality imply $z_n=x_n$?
– Paramanand Singh
Jul 15 at 6:06
add a comment |Â
up vote
2
down vote
Most elementary results of convergence for sequences either depend on the triangle inequality (via an $epsilon/2$ trick) or the inequality $0leqfrac1+leq 1$ for all $x$ (to avoid cases where division by $0$ may occur).
The Squeeze Theorem however is probably best proved by a different approach.
Try to show that if $xleq yleq z$ then $|y-l|leq maxleft$ no matter what $l$ may be. This is probably best proved with a picture.
answered Jul 15 at 3:05
Robert Wolfe
5,30722261
Is the aproach I have taken completely wrong or can it be tweaked for a correct proof?
– Red
Jul 15 at 3:11
@Red you have claimed that the sequences $x_n$ and $z_n$ eventually agree. But an easy counter example is $x_n=-1/n$, $z_n=1/n$ and $y_n=0$ for all $n$.
– Robert Wolfe
Jul 15 at 3:12
I completely agree with this, the sequences need not agree eventually.But why is the triangle inequality gving me that result anyway? I feel like I have used it wrong but I cannot figure out where my mistake llies? as I said in th second part of my question
– Red
Jul 15 at 3:16
1
@Red if $a$ and $b$ satisfy $|a-b|<epsilon$ for every $epsilon>0$ then $a=b$. However you have not proved that $|x_n-z_n|<epsilon$ for every $epsilon>0$. You have proved that for every $epsilon>0$ there is an $n$ such that $|x_n-z_n|<epsilon$.
– Robert Wolfe
Jul 15 at 3:27
I see my mistake. Thanks!
– Red
Jul 15 at 3:28
add a comment |Â
up vote
2
down vote
Most elementary results of convergence for sequences either depend on the triangle inequality (via an $epsilon/2$ trick) or the inequality $0leqfrac1+leq 1$ for all $x$ (to avoid cases where division by $0$ may occur).
The Squeeze Theorem however is probably best proved by a different approach.
Try to show that if $xleq yleq z$ then $|y-l|leq maxleft$ no matter what $l$ may be. This is probably best proved with a picture.
answered Jul 15 at 3:05
Robert Wolfe
5,30722261
Is the aproach I have taken completely wrong or can it be tweaked for a correct proof?
– Red
Jul 15 at 3:11
@Red you have claimed that the sequences $x_n$ and $z_n$ eventually agree. But an easy counter example is $x_n=-1/n$, $z_n=1/n$ and $y_n=0$ for all $n$.
– Robert Wolfe
Jul 15 at 3:12
I completely agree with this, the sequences need not agree eventually.But why is the triangle inequality gving me that result anyway? I feel like I have used it wrong but I cannot figure out where my mistake llies? as I said in th second part of my question
– Red
Jul 15 at 3:16
1
@Red if $a$ and $b$ satisfy $|a-b|<epsilon$ for every $epsilon>0$ then $a=b$. However you have not proved that $|x_n-z_n|<epsilon$ for every $epsilon>0$. You have proved that for every $epsilon>0$ there is an $n$ such that $|x_n-z_n|<epsilon$.
– Robert Wolfe
Jul 15 at 3:27
I see my mistake. Thanks!
– Red
Jul 15 at 3:28
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Most elementary results of convergence for sequences either depend on the triangle inequality (via an $epsilon/2$ trick) or the inequality $0leqfrac1+leq 1$ for all $x$ (to avoid cases where division by $0$ may occur).
The Squeeze Theorem however is probably best proved by a different approach.
Try to show that if $xleq yleq z$ then $|y-l|leq maxleft$ no matter what $l$ may be. This is probably best proved with a picture.
answered Jul 15 at 3:05
Robert Wolfe
5,30722261
Most elementary results of convergence for sequences either depend on the triangle inequality (via an $epsilon/2$ trick) or the inequality $0leqfrac1+leq 1$ for all $x$ (to avoid cases where division by $0$ may occur).
The Squeeze Theorem however is probably best proved by a different approach.
Try to show that if $xleq yleq z$ then $|y-l|leq maxleft$ no matter what $l$ may be. This is probably best proved with a picture.
answered Jul 15 at 3:05
Robert Wolfe
5,30722261
answered Jul 15 at 3:05
Robert Wolfe
5,30722261
answered Jul 15 at 3:05
Robert Wolfe
5,30722261
answered Jul 15 at 3:05
Robert Wolfe
5,30722261
5,30722261
Is the aproach I have taken completely wrong or can it be tweaked for a correct proof?
– Red
Jul 15 at 3:11
@Red you have claimed that the sequences $x_n$ and $z_n$ eventually agree. But an easy counter example is $x_n=-1/n$, $z_n=1/n$ and $y_n=0$ for all $n$.
– Robert Wolfe
Jul 15 at 3:12
I completely agree with this, the sequences need not agree eventually.But why is the triangle inequality gving me that result anyway? I feel like I have used it wrong but I cannot figure out where my mistake llies? as I said in th second part of my question
– Red
Jul 15 at 3:16
1
@Red if $a$ and $b$ satisfy $|a-b|<epsilon$ for every $epsilon>0$ then $a=b$. However you have not proved that $|x_n-z_n|<epsilon$ for every $epsilon>0$. You have proved that for every $epsilon>0$ there is an $n$ such that $|x_n-z_n|<epsilon$.
– Robert Wolfe
Jul 15 at 3:27
I see my mistake. Thanks!
– Red
Jul 15 at 3:28
add a comment |Â
Is the aproach I have taken completely wrong or can it be tweaked for a correct proof?
– Red
Jul 15 at 3:11
@Red you have claimed that the sequences $x_n$ and $z_n$ eventually agree. But an easy counter example is $x_n=-1/n$, $z_n=1/n$ and $y_n=0$ for all $n$.
– Robert Wolfe
Jul 15 at 3:12
I completely agree with this, the sequences need not agree eventually.But why is the triangle inequality gving me that result anyway? I feel like I have used it wrong but I cannot figure out where my mistake llies? as I said in th second part of my question
– Red
Jul 15 at 3:16
1
@Red if $a$ and $b$ satisfy $|a-b|<epsilon$ for every $epsilon>0$ then $a=b$. However you have not proved that $|x_n-z_n|<epsilon$ for every $epsilon>0$. You have proved that for every $epsilon>0$ there is an $n$ such that $|x_n-z_n|<epsilon$.
– Robert Wolfe
Jul 15 at 3:27
I see my mistake. Thanks!
– Red
Jul 15 at 3:28
Is the aproach I have taken completely wrong or can it be tweaked for a correct proof?
– Red
Jul 15 at 3:11
Is the aproach I have taken completely wrong or can it be tweaked for a correct proof?
– Red
Jul 15 at 3:11
@Red you have claimed that the sequences $x_n$ and $z_n$ eventually agree. But an easy counter example is $x_n=-1/n$, $z_n=1/n$ and $y_n=0$ for all $n$.
– Robert Wolfe
Jul 15 at 3:12
@Red you have claimed that the sequences $x_n$ and $z_n$ eventually agree. But an easy counter example is $x_n=-1/n$, $z_n=1/n$ and $y_n=0$ for all $n$.
– Robert Wolfe
Jul 15 at 3:12
I completely agree with this, the sequences need not agree eventually.But why is the triangle inequality gving me that result anyway? I feel like I have used it wrong but I cannot figure out where my mistake llies? as I said in th second part of my question
– Red
Jul 15 at 3:16
I completely agree with this, the sequences need not agree eventually.But why is the triangle inequality gving me that result anyway? I feel like I have used it wrong but I cannot figure out where my mistake llies? as I said in th second part of my question
– Red
Jul 15 at 3:16
1
1
@Red if $a$ and $b$ satisfy $|a-b|<epsilon$ for every $epsilon>0$ then $a=b$. However you have not proved that $|x_n-z_n|<epsilon$ for every $epsilon>0$. You have proved that for every $epsilon>0$ there is an $n$ such that $|x_n-z_n|<epsilon$.
– Robert Wolfe
Jul 15 at 3:27
@Red if $a$ and $b$ satisfy $|a-b|<epsilon$ for every $epsilon>0$ then $a=b$. However you have not proved that $|x_n-z_n|<epsilon$ for every $epsilon>0$. You have proved that for every $epsilon>0$ there is an $n$ such that $|x_n-z_n|<epsilon$.
– Robert Wolfe
Jul 15 at 3:27
I see my mistake. Thanks!
– Red
Jul 15 at 3:28
I see my mistake. Thanks!
– Red
Jul 15 at 3:28
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2852140%2fproof-of-the-squeeze-theorem-for-sequences%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password