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Proving $lim_xto c f(x) le lim_x to c g(x)$
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Here's what I need to prove:
Suppose that $f(x)le g(x)$ on some deleted neighborhood of the point
$c$ and that $lim_xto c f(x)$ and $lim_xto c g(x)$ exist. Then
$lim_xto c f(x) le lim_xto c g(x)$.
The authors ask to prove it directly using the definition and not to use the fact that if $L > 0$, there is a $delta >0$ such that $f(x)>0$ for $0<|x-c|<delta$.
Here's my attempt:
If $f(x)le g(x)$ on some deleted neighborhood of $c$ then there's a $delta_1 > 0$ such that $(c-delta_1 , c+ delta_1) setminus c $ is contained in that neighborhood. Let $L_1 := lim_x to c f(x)$ and Let $L_2 := lim_x to c g(x)$. Assume that $L_1 > L_2$. Now, there are $delta_2 > 0$ and $delta_3>0$ such that $|f(x)-L_1| < L_1-L_2$ i.e. $L_2 < f(x)$ for $0<|x-c|< delta_2$ and similarly $g(x) < L_1 $ for $0<|x-c|< delta_3$. Now for $0<|x-c|< min delta_1 , delta_2 , delta_3 $ we have $L_2 < f(x) le g(x) < L_1 $ which contradicts our assumption.
Is my proof fine?
real-analysis limits
asked Jul 31 at 17:51
Ashish K
437312
 |Â
show 4 more comments
up vote
2
down vote
favorite
Here's what I need to prove:
Suppose that $f(x)le g(x)$ on some deleted neighborhood of the point
$c$ and that $lim_xto c f(x)$ and $lim_xto c g(x)$ exist. Then
$lim_xto c f(x) le lim_xto c g(x)$.
The authors ask to prove it directly using the definition and not to use the fact that if $L > 0$, there is a $delta >0$ such that $f(x)>0$ for $0<|x-c|<delta$.
Here's my attempt:
If $f(x)le g(x)$ on some deleted neighborhood of $c$ then there's a $delta_1 > 0$ such that $(c-delta_1 , c+ delta_1) setminus c $ is contained in that neighborhood. Let $L_1 := lim_x to c f(x)$ and Let $L_2 := lim_x to c g(x)$. Assume that $L_1 > L_2$. Now, there are $delta_2 > 0$ and $delta_3>0$ such that $|f(x)-L_1| < L_1-L_2$ i.e. $L_2 < f(x)$ for $0<|x-c|< delta_2$ and similarly $g(x) < L_1 $ for $0<|x-c|< delta_3$. Now for $0<|x-c|< min delta_1 , delta_2 , delta_3 $ we have $L_2 < f(x) le g(x) < L_1 $ which contradicts our assumption.
Is my proof fine?
real-analysis limits
asked Jul 31 at 17:51
Ashish K
437312
I don't understand your contradiction. You assumed that $L_1>L_2$ and that's exactly what you got in the end.
– Mark
Jul 31 at 17:56
If authors ask for a direct proof, then you can't prove it by contradiction, can you
– Rumpelstiltskin
Jul 31 at 18:00
Adam, direct proof means to use the definition of the limit. It doesn't mean he can't prove by contradiction.
– Mark
Jul 31 at 18:02
@Mark oh, thank you. I didn't knew that
– Rumpelstiltskin
Jul 31 at 18:02
2
Think that way: if we assume $L_1>L_2$ then you can find a small enough $epsilon$ for which $L_1-epsilon>L_2+epsilon$. Now, for values of x that are very close to c you must have $L_1-epsilon<f(x)$ and $g(x)<L_2+epsilon$. Can you continue from here?
– Mark
Jul 31 at 18:15
 |Â
show 4 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Here's what I need to prove:
Suppose that $f(x)le g(x)$ on some deleted neighborhood of the point
$c$ and that $lim_xto c f(x)$ and $lim_xto c g(x)$ exist. Then
$lim_xto c f(x) le lim_xto c g(x)$.
The authors ask to prove it directly using the definition and not to use the fact that if $L > 0$, there is a $delta >0$ such that $f(x)>0$ for $0<|x-c|<delta$.
Here's my attempt:
If $f(x)le g(x)$ on some deleted neighborhood of $c$ then there's a $delta_1 > 0$ such that $(c-delta_1 , c+ delta_1) setminus c $ is contained in that neighborhood. Let $L_1 := lim_x to c f(x)$ and Let $L_2 := lim_x to c g(x)$. Assume that $L_1 > L_2$. Now, there are $delta_2 > 0$ and $delta_3>0$ such that $|f(x)-L_1| < L_1-L_2$ i.e. $L_2 < f(x)$ for $0<|x-c|< delta_2$ and similarly $g(x) < L_1 $ for $0<|x-c|< delta_3$. Now for $0<|x-c|< min delta_1 , delta_2 , delta_3 $ we have $L_2 < f(x) le g(x) < L_1 $ which contradicts our assumption.
Is my proof fine?
real-analysis limits
asked Jul 31 at 17:51
Ashish K
437312
Here's what I need to prove:
Suppose that $f(x)le g(x)$ on some deleted neighborhood of the point
$c$ and that $lim_xto c f(x)$ and $lim_xto c g(x)$ exist. Then
$lim_xto c f(x) le lim_xto c g(x)$.
The authors ask to prove it directly using the definition and not to use the fact that if $L > 0$, there is a $delta >0$ such that $f(x)>0$ for $0<|x-c|<delta$.
Here's my attempt:
If $f(x)le g(x)$ on some deleted neighborhood of $c$ then there's a $delta_1 > 0$ such that $(c-delta_1 , c+ delta_1) setminus c $ is contained in that neighborhood. Let $L_1 := lim_x to c f(x)$ and Let $L_2 := lim_x to c g(x)$. Assume that $L_1 > L_2$. Now, there are $delta_2 > 0$ and $delta_3>0$ such that $|f(x)-L_1| < L_1-L_2$ i.e. $L_2 < f(x)$ for $0<|x-c|< delta_2$ and similarly $g(x) < L_1 $ for $0<|x-c|< delta_3$. Now for $0<|x-c|< min delta_1 , delta_2 , delta_3 $ we have $L_2 < f(x) le g(x) < L_1 $ which contradicts our assumption.
Is my proof fine?
real-analysis limits
asked Jul 31 at 17:51
Ashish K
437312
asked Jul 31 at 17:51
Ashish K
437312
asked Jul 31 at 17:51
Ashish K
437312
asked Jul 31 at 17:51
Ashish K
437312
437312
I don't understand your contradiction. You assumed that $L_1>L_2$ and that's exactly what you got in the end.
– Mark
Jul 31 at 17:56
If authors ask for a direct proof, then you can't prove it by contradiction, can you
– Rumpelstiltskin
Jul 31 at 18:00
Adam, direct proof means to use the definition of the limit. It doesn't mean he can't prove by contradiction.
– Mark
Jul 31 at 18:02
@Mark oh, thank you. I didn't knew that
– Rumpelstiltskin
Jul 31 at 18:02
2
Think that way: if we assume $L_1>L_2$ then you can find a small enough $epsilon$ for which $L_1-epsilon>L_2+epsilon$. Now, for values of x that are very close to c you must have $L_1-epsilon<f(x)$ and $g(x)<L_2+epsilon$. Can you continue from here?
– Mark
Jul 31 at 18:15
 |Â
show 4 more comments
I don't understand your contradiction. You assumed that $L_1>L_2$ and that's exactly what you got in the end.
– Mark
Jul 31 at 17:56
If authors ask for a direct proof, then you can't prove it by contradiction, can you
– Rumpelstiltskin
Jul 31 at 18:00
Adam, direct proof means to use the definition of the limit. It doesn't mean he can't prove by contradiction.
– Mark
Jul 31 at 18:02
@Mark oh, thank you. I didn't knew that
– Rumpelstiltskin
Jul 31 at 18:02
2
Think that way: if we assume $L_1>L_2$ then you can find a small enough $epsilon$ for which $L_1-epsilon>L_2+epsilon$. Now, for values of x that are very close to c you must have $L_1-epsilon<f(x)$ and $g(x)<L_2+epsilon$. Can you continue from here?
– Mark
Jul 31 at 18:15
I don't understand your contradiction. You assumed that $L_1>L_2$ and that's exactly what you got in the end.
– Mark
Jul 31 at 17:56
I don't understand your contradiction. You assumed that $L_1>L_2$ and that's exactly what you got in the end.
– Mark
Jul 31 at 17:56
If authors ask for a direct proof, then you can't prove it by contradiction, can you
– Rumpelstiltskin
Jul 31 at 18:00
If authors ask for a direct proof, then you can't prove it by contradiction, can you
– Rumpelstiltskin
Jul 31 at 18:00
Adam, direct proof means to use the definition of the limit. It doesn't mean he can't prove by contradiction.
– Mark
Jul 31 at 18:02
Adam, direct proof means to use the definition of the limit. It doesn't mean he can't prove by contradiction.
– Mark
Jul 31 at 18:02
@Mark oh, thank you. I didn't knew that
– Rumpelstiltskin
Jul 31 at 18:02
@Mark oh, thank you. I didn't knew that
– Rumpelstiltskin
Jul 31 at 18:02
2
2
Think that way: if we assume $L_1>L_2$ then you can find a small enough $epsilon$ for which $L_1-epsilon>L_2+epsilon$. Now, for values of x that are very close to c you must have $L_1-epsilon<f(x)$ and $g(x)<L_2+epsilon$. Can you continue from here?
– Mark
Jul 31 at 18:15
Think that way: if we assume $L_1>L_2$ then you can find a small enough $epsilon$ for which $L_1-epsilon>L_2+epsilon$. Now, for values of x that are very close to c you must have $L_1-epsilon<f(x)$ and $g(x)<L_2+epsilon$. Can you continue from here?
– Mark
Jul 31 at 18:15
 |Â
show 4 more comments
1 Answer
1
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Let $L_1, L_2$ be the limits of $f,g$ respectively $implies -epsilon < f -L_1< epsilon , -epsilon <L_2-g < epsilonimplies -2epsilon < (f-g) + (L_2-L_1) < 2epsilonimplies L_2-L_1> -2epsilon +(g-f)ge -2epsilon$. Since this is true for any $epsilon > 0$, we have: $L_2 - L_1 ge 0implies L_1 le L_2$.
answered Jul 31 at 18:34
DeepSea
68.8k54284
(+1) All too easy.
– Mark Viola
Jul 31 at 19:33
Thank you Mark for the upvote. Are you running for Moderator position? you should.
– DeepSea
Jul 31 at 22:04
You're welcome. I'm not running for a moderator position, but thank you for asking. I'm flattered.
– Mark Viola
Jul 31 at 23:09
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Let $L_1, L_2$ be the limits of $f,g$ respectively $implies -epsilon < f -L_1< epsilon , -epsilon <L_2-g < epsilonimplies -2epsilon < (f-g) + (L_2-L_1) < 2epsilonimplies L_2-L_1> -2epsilon +(g-f)ge -2epsilon$. Since this is true for any $epsilon > 0$, we have: $L_2 - L_1 ge 0implies L_1 le L_2$.
answered Jul 31 at 18:34
DeepSea
68.8k54284
(+1) All too easy.
– Mark Viola
Jul 31 at 19:33
Thank you Mark for the upvote. Are you running for Moderator position? you should.
– DeepSea
Jul 31 at 22:04
You're welcome. I'm not running for a moderator position, but thank you for asking. I'm flattered.
– Mark Viola
Jul 31 at 23:09
add a comment |Â
up vote
1
down vote
Let $L_1, L_2$ be the limits of $f,g$ respectively $implies -epsilon < f -L_1< epsilon , -epsilon <L_2-g < epsilonimplies -2epsilon < (f-g) + (L_2-L_1) < 2epsilonimplies L_2-L_1> -2epsilon +(g-f)ge -2epsilon$. Since this is true for any $epsilon > 0$, we have: $L_2 - L_1 ge 0implies L_1 le L_2$.
answered Jul 31 at 18:34
DeepSea
68.8k54284
(+1) All too easy.
– Mark Viola
Jul 31 at 19:33
Thank you Mark for the upvote. Are you running for Moderator position? you should.
– DeepSea
Jul 31 at 22:04
You're welcome. I'm not running for a moderator position, but thank you for asking. I'm flattered.
– Mark Viola
Jul 31 at 23:09
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let $L_1, L_2$ be the limits of $f,g$ respectively $implies -epsilon < f -L_1< epsilon , -epsilon <L_2-g < epsilonimplies -2epsilon < (f-g) + (L_2-L_1) < 2epsilonimplies L_2-L_1> -2epsilon +(g-f)ge -2epsilon$. Since this is true for any $epsilon > 0$, we have: $L_2 - L_1 ge 0implies L_1 le L_2$.
answered Jul 31 at 18:34
DeepSea
68.8k54284
Let $L_1, L_2$ be the limits of $f,g$ respectively $implies -epsilon < f -L_1< epsilon , -epsilon <L_2-g < epsilonimplies -2epsilon < (f-g) + (L_2-L_1) < 2epsilonimplies L_2-L_1> -2epsilon +(g-f)ge -2epsilon$. Since this is true for any $epsilon > 0$, we have: $L_2 - L_1 ge 0implies L_1 le L_2$.
answered Jul 31 at 18:34
DeepSea
68.8k54284
answered Jul 31 at 18:34
DeepSea
68.8k54284
answered Jul 31 at 18:34
DeepSea
68.8k54284
answered Jul 31 at 18:34
DeepSea
68.8k54284
68.8k54284
(+1) All too easy.
– Mark Viola
Jul 31 at 19:33
Thank you Mark for the upvote. Are you running for Moderator position? you should.
– DeepSea
Jul 31 at 22:04
You're welcome. I'm not running for a moderator position, but thank you for asking. I'm flattered.
– Mark Viola
Jul 31 at 23:09
add a comment |Â
(+1) All too easy.
– Mark Viola
Jul 31 at 19:33
Thank you Mark for the upvote. Are you running for Moderator position? you should.
– DeepSea
Jul 31 at 22:04
You're welcome. I'm not running for a moderator position, but thank you for asking. I'm flattered.
– Mark Viola
Jul 31 at 23:09
(+1) All too easy.
– Mark Viola
Jul 31 at 19:33
(+1) All too easy.
– Mark Viola
Jul 31 at 19:33
Thank you Mark for the upvote. Are you running for Moderator position? you should.
– DeepSea
Jul 31 at 22:04
Thank you Mark for the upvote. Are you running for Moderator position? you should.
– DeepSea
Jul 31 at 22:04
You're welcome. I'm not running for a moderator position, but thank you for asking. I'm flattered.
– Mark Viola
Jul 31 at 23:09
You're welcome. I'm not running for a moderator position, but thank you for asking. I'm flattered.
– Mark Viola
Jul 31 at 23:09
add a comment |Â
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I don't understand your contradiction. You assumed that $L_1>L_2$ and that's exactly what you got in the end.
– Mark
Jul 31 at 17:56
If authors ask for a direct proof, then you can't prove it by contradiction, can you
– Rumpelstiltskin
Jul 31 at 18:00
Adam, direct proof means to use the definition of the limit. It doesn't mean he can't prove by contradiction.
– Mark
Jul 31 at 18:02
@Mark oh, thank you. I didn't knew that
– Rumpelstiltskin
Jul 31 at 18:02
2
Think that way: if we assume $L_1>L_2$ then you can find a small enough $epsilon$ for which $L_1-epsilon>L_2+epsilon$. Now, for values of x that are very close to c you must have $L_1-epsilon<f(x)$ and $g(x)<L_2+epsilon$. Can you continue from here?
– Mark
Jul 31 at 18:15