Mixing
Show that if $x$ is non-negative and $x_n$ is a non-negative sequence converging to $x$ then...
$begingroup$
So I've let $|x_n-x| < epsilon$ but am not sure where to go now.
I know
$sqrt{x_n}-sqrt{x}=frac{x_n-x}{sqrt{x_n}+sqrt{x}}$, but I'm not sure how to use that.
analysis
edited Jan 27 at 17:40
J.G.
31.9k23250
asked Jan 27 at 17:36
AnoUser1AnoUser1
836
$endgroup$
add a comment |
$begingroup$
So I've let $|x_n-x| < epsilon$ but am not sure where to go now.
I know
$sqrt{x_n}-sqrt{x}=frac{x_n-x}{sqrt{x_n}+sqrt{x}}$, but I'm not sure how to use that.
analysis
edited Jan 27 at 17:40
J.G.
31.9k23250
asked Jan 27 at 17:36
AnoUser1AnoUser1
836
$endgroup$
$begingroup$
ayy lmao keith ball?
$endgroup$
– Dis-integrating
Jan 28 at 4:25
add a comment |
$begingroup$
So I've let $|x_n-x| < epsilon$ but am not sure where to go now.
I know
$sqrt{x_n}-sqrt{x}=frac{x_n-x}{sqrt{x_n}+sqrt{x}}$, but I'm not sure how to use that.
analysis
edited Jan 27 at 17:40
J.G.
31.9k23250
asked Jan 27 at 17:36
AnoUser1AnoUser1
836
$endgroup$
So I've let $|x_n-x| < epsilon$ but am not sure where to go now.
I know
$sqrt{x_n}-sqrt{x}=frac{x_n-x}{sqrt{x_n}+sqrt{x}}$, but I'm not sure how to use that.
analysis
analysis
edited Jan 27 at 17:40
J.G.
31.9k23250
asked Jan 27 at 17:36
AnoUser1AnoUser1
836
edited Jan 27 at 17:40
J.G.
31.9k23250
asked Jan 27 at 17:36
AnoUser1AnoUser1
836
edited Jan 27 at 17:40
J.G.
31.9k23250
edited Jan 27 at 17:40
J.G.
31.9k23250
edited Jan 27 at 17:40
J.G.
31.9k23250
31.9k23250
asked Jan 27 at 17:36
AnoUser1AnoUser1
836
asked Jan 27 at 17:36
AnoUser1AnoUser1
836
asked Jan 27 at 17:36
AnoUser1AnoUser1
836
836
$begingroup$
ayy lmao keith ball?
$endgroup$
– Dis-integrating
Jan 28 at 4:25
add a comment |
$begingroup$
ayy lmao keith ball?
$endgroup$
– Dis-integrating
Jan 28 at 4:25
$begingroup$
ayy lmao keith ball?
$endgroup$
– Dis-integrating
Jan 28 at 4:25
$begingroup$
ayy lmao keith ball?
$endgroup$
– Dis-integrating
Jan 28 at 4:25
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Hint: For $x = 0$ this is easy. So let $x neq 0$. Then $$leftvert frac{x_n - x}{sqrt{x_n} + sqrt{x}} rightvertleq frac{1}{sqrt{x}} left vert x_n - xrightvert$$ and $frac {1}{sqrt x}$ is a constant.
answered Jan 27 at 17:47
YaddleYaddle
3,136829
$endgroup$
$begingroup$
I know why it's true, but is there a reason you've chose to make ≤ 1/x^0.5, or is it simply to make it nice and you know x is non-negative, so must be true?
$endgroup$
– AnoUser1
Jan 27 at 17:59
1
$begingroup$
@AnoUser1 It's using $sqrt{x_n}ge 0impliesfrac{1}{sqrt{x_n}+sqrt{x}}lefrac{1}{sqrt{x}}$.
$endgroup$
– J.G.
Jan 27 at 18:25
add a comment |
$begingroup$
We want to force $|sqrt{x_n}-sqrt{x}|<delta$, with arbitrarily small $delta>0$. If $x=0$ take $epsilon=delta^2$. If $epsilonle x$ so that $xne 0$, we have $$|sqrt{x_n}-sqrt{x}|=frac{|x_n-x|}{sqrt{x_n}+sqrt{x}}lefrac{epsilon}{sqrt{x}},$$so it suffices to take $epsilon<deltasqrt{x}$.
answered Jan 27 at 17:48
J.G.J.G.
31.9k23250
$endgroup$
$begingroup$
You say We want to force |√xn −√x|<δ, is this allowed? It looks to me like assuming what you want to prove is true
$endgroup$
– AnoUser1
Jan 27 at 18:13
$begingroup$
@AnoUser1 "We want to force" means "we want to prove"; that statement was a way to introduce the symbol $delta$. I've not assumed the inequality we wished to force; I've pointed out how to choose $epsilon$ so as to prove the desired result.
$endgroup$
– J.G.
Jan 27 at 18:16
$begingroup$
Got it. One last bit, how does ϵ<δ√x help us? I get that |√xn−√x| needs to be less than a constant for all x, and δ√x is a constant but why epsilon less than this?
$endgroup$
– AnoUser1
Jan 27 at 18:21
$begingroup$
@AnoUser1 The point of the exercise is to prove for any $delta>0$ there exists some $epsilon>0$ such that $|x_n-x|<epsilonimplies|sqrt{x_n}-sqrt{x}|<delta$. You can choose any $epsilon$ that works; it turns out that one does.
$endgroup$
– J.G.
Jan 27 at 18:24
add a comment |
$begingroup$
Suppose $x>0$. Then, since $frac{1}{sqrt{x_n} + sqrt{x}}$ is bounded, you have that $sqrt{x_n} - sqrt{x} rightarrow 0$ as $n rightarrow infty$. The case $x=0$ is easier.
answered Jan 27 at 17:49
lzralbulzralbu
640512
$endgroup$
add a comment |
$begingroup$
Suppose $x > 0$. Since $x_n rightarrow x$ as $n rightarrow infty$ and also $displaystylefrac{1}{sqrt{x_n}+sqrt{x}}$ is finite, then $$sqrt{x_n}-sqrt{x}=frac{x_n-x}{sqrt{x_n}+sqrt{x}} rightarrow 0$$
$$implies sqrt{x_n}-sqrt{x} = 0 text{ as } n rightarrow infty$$
answered Jan 27 at 17:49
Exp ikxExp ikx
4489
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: For $x = 0$ this is easy. So let $x neq 0$. Then $$leftvert frac{x_n - x}{sqrt{x_n} + sqrt{x}} rightvertleq frac{1}{sqrt{x}} left vert x_n - xrightvert$$ and $frac {1}{sqrt x}$ is a constant.
answered Jan 27 at 17:47
YaddleYaddle
3,136829
$endgroup$
$begingroup$
I know why it's true, but is there a reason you've chose to make ≤ 1/x^0.5, or is it simply to make it nice and you know x is non-negative, so must be true?
$endgroup$
– AnoUser1
Jan 27 at 17:59
1
$begingroup$
@AnoUser1 It's using $sqrt{x_n}ge 0impliesfrac{1}{sqrt{x_n}+sqrt{x}}lefrac{1}{sqrt{x}}$.
$endgroup$
– J.G.
Jan 27 at 18:25
add a comment |
$begingroup$
Hint: For $x = 0$ this is easy. So let $x neq 0$. Then $$leftvert frac{x_n - x}{sqrt{x_n} + sqrt{x}} rightvertleq frac{1}{sqrt{x}} left vert x_n - xrightvert$$ and $frac {1}{sqrt x}$ is a constant.
answered Jan 27 at 17:47
YaddleYaddle
3,136829
$endgroup$
$begingroup$
I know why it's true, but is there a reason you've chose to make ≤ 1/x^0.5, or is it simply to make it nice and you know x is non-negative, so must be true?
$endgroup$
– AnoUser1
Jan 27 at 17:59
1
$begingroup$
@AnoUser1 It's using $sqrt{x_n}ge 0impliesfrac{1}{sqrt{x_n}+sqrt{x}}lefrac{1}{sqrt{x}}$.
$endgroup$
– J.G.
Jan 27 at 18:25
add a comment |
$begingroup$
Hint: For $x = 0$ this is easy. So let $x neq 0$. Then $$leftvert frac{x_n - x}{sqrt{x_n} + sqrt{x}} rightvertleq frac{1}{sqrt{x}} left vert x_n - xrightvert$$ and $frac {1}{sqrt x}$ is a constant.
answered Jan 27 at 17:47
YaddleYaddle
3,136829
$endgroup$
Hint: For $x = 0$ this is easy. So let $x neq 0$. Then $$leftvert frac{x_n - x}{sqrt{x_n} + sqrt{x}} rightvertleq frac{1}{sqrt{x}} left vert x_n - xrightvert$$ and $frac {1}{sqrt x}$ is a constant.
answered Jan 27 at 17:47
YaddleYaddle
3,136829
answered Jan 27 at 17:47
YaddleYaddle
3,136829
answered Jan 27 at 17:47
YaddleYaddle
3,136829
answered Jan 27 at 17:47
YaddleYaddle
3,136829
3,136829
$begingroup$
I know why it's true, but is there a reason you've chose to make ≤ 1/x^0.5, or is it simply to make it nice and you know x is non-negative, so must be true?
$endgroup$
– AnoUser1
Jan 27 at 17:59
1
$begingroup$
@AnoUser1 It's using $sqrt{x_n}ge 0impliesfrac{1}{sqrt{x_n}+sqrt{x}}lefrac{1}{sqrt{x}}$.
$endgroup$
– J.G.
Jan 27 at 18:25
add a comment |
$begingroup$
I know why it's true, but is there a reason you've chose to make ≤ 1/x^0.5, or is it simply to make it nice and you know x is non-negative, so must be true?
$endgroup$
– AnoUser1
Jan 27 at 17:59
1
$begingroup$
@AnoUser1 It's using $sqrt{x_n}ge 0impliesfrac{1}{sqrt{x_n}+sqrt{x}}lefrac{1}{sqrt{x}}$.
$endgroup$
– J.G.
Jan 27 at 18:25
$begingroup$
I know why it's true, but is there a reason you've chose to make ≤ 1/x^0.5, or is it simply to make it nice and you know x is non-negative, so must be true?
$endgroup$
– AnoUser1
Jan 27 at 17:59
$begingroup$
I know why it's true, but is there a reason you've chose to make ≤ 1/x^0.5, or is it simply to make it nice and you know x is non-negative, so must be true?
$endgroup$
– AnoUser1
Jan 27 at 17:59
1
1
$begingroup$
@AnoUser1 It's using $sqrt{x_n}ge 0impliesfrac{1}{sqrt{x_n}+sqrt{x}}lefrac{1}{sqrt{x}}$.
$endgroup$
– J.G.
Jan 27 at 18:25
$begingroup$
@AnoUser1 It's using $sqrt{x_n}ge 0impliesfrac{1}{sqrt{x_n}+sqrt{x}}lefrac{1}{sqrt{x}}$.
$endgroup$
– J.G.
Jan 27 at 18:25
add a comment |
$begingroup$
We want to force $|sqrt{x_n}-sqrt{x}|<delta$, with arbitrarily small $delta>0$. If $x=0$ take $epsilon=delta^2$. If $epsilonle x$ so that $xne 0$, we have $$|sqrt{x_n}-sqrt{x}|=frac{|x_n-x|}{sqrt{x_n}+sqrt{x}}lefrac{epsilon}{sqrt{x}},$$so it suffices to take $epsilon<deltasqrt{x}$.
answered Jan 27 at 17:48
J.G.J.G.
31.9k23250
$endgroup$
$begingroup$
You say We want to force |√xn −√x|<δ, is this allowed? It looks to me like assuming what you want to prove is true
$endgroup$
– AnoUser1
Jan 27 at 18:13
$begingroup$
@AnoUser1 "We want to force" means "we want to prove"; that statement was a way to introduce the symbol $delta$. I've not assumed the inequality we wished to force; I've pointed out how to choose $epsilon$ so as to prove the desired result.
$endgroup$
– J.G.
Jan 27 at 18:16
$begingroup$
Got it. One last bit, how does ϵ<δ√x help us? I get that |√xn−√x| needs to be less than a constant for all x, and δ√x is a constant but why epsilon less than this?
$endgroup$
– AnoUser1
Jan 27 at 18:21
$begingroup$
@AnoUser1 The point of the exercise is to prove for any $delta>0$ there exists some $epsilon>0$ such that $|x_n-x|<epsilonimplies|sqrt{x_n}-sqrt{x}|<delta$. You can choose any $epsilon$ that works; it turns out that one does.
$endgroup$
– J.G.
Jan 27 at 18:24
add a comment |
$begingroup$
We want to force $|sqrt{x_n}-sqrt{x}|<delta$, with arbitrarily small $delta>0$. If $x=0$ take $epsilon=delta^2$. If $epsilonle x$ so that $xne 0$, we have $$|sqrt{x_n}-sqrt{x}|=frac{|x_n-x|}{sqrt{x_n}+sqrt{x}}lefrac{epsilon}{sqrt{x}},$$so it suffices to take $epsilon<deltasqrt{x}$.
answered Jan 27 at 17:48
J.G.J.G.
31.9k23250
$endgroup$
$begingroup$
You say We want to force |√xn −√x|<δ, is this allowed? It looks to me like assuming what you want to prove is true
$endgroup$
– AnoUser1
Jan 27 at 18:13
$begingroup$
@AnoUser1 "We want to force" means "we want to prove"; that statement was a way to introduce the symbol $delta$. I've not assumed the inequality we wished to force; I've pointed out how to choose $epsilon$ so as to prove the desired result.
$endgroup$
– J.G.
Jan 27 at 18:16
$begingroup$
Got it. One last bit, how does ϵ<δ√x help us? I get that |√xn−√x| needs to be less than a constant for all x, and δ√x is a constant but why epsilon less than this?
$endgroup$
– AnoUser1
Jan 27 at 18:21
$begingroup$
@AnoUser1 The point of the exercise is to prove for any $delta>0$ there exists some $epsilon>0$ such that $|x_n-x|<epsilonimplies|sqrt{x_n}-sqrt{x}|<delta$. You can choose any $epsilon$ that works; it turns out that one does.
$endgroup$
– J.G.
Jan 27 at 18:24
add a comment |
$begingroup$
We want to force $|sqrt{x_n}-sqrt{x}|<delta$, with arbitrarily small $delta>0$. If $x=0$ take $epsilon=delta^2$. If $epsilonle x$ so that $xne 0$, we have $$|sqrt{x_n}-sqrt{x}|=frac{|x_n-x|}{sqrt{x_n}+sqrt{x}}lefrac{epsilon}{sqrt{x}},$$so it suffices to take $epsilon<deltasqrt{x}$.
answered Jan 27 at 17:48
J.G.J.G.
31.9k23250
$endgroup$
We want to force $|sqrt{x_n}-sqrt{x}|<delta$, with arbitrarily small $delta>0$. If $x=0$ take $epsilon=delta^2$. If $epsilonle x$ so that $xne 0$, we have $$|sqrt{x_n}-sqrt{x}|=frac{|x_n-x|}{sqrt{x_n}+sqrt{x}}lefrac{epsilon}{sqrt{x}},$$so it suffices to take $epsilon<deltasqrt{x}$.
answered Jan 27 at 17:48
J.G.J.G.
31.9k23250
answered Jan 27 at 17:48
J.G.J.G.
31.9k23250
answered Jan 27 at 17:48
J.G.J.G.
31.9k23250
answered Jan 27 at 17:48
J.G.J.G.
31.9k23250
31.9k23250
$begingroup$
You say We want to force |√xn −√x|<δ, is this allowed? It looks to me like assuming what you want to prove is true
$endgroup$
– AnoUser1
Jan 27 at 18:13
$begingroup$
@AnoUser1 "We want to force" means "we want to prove"; that statement was a way to introduce the symbol $delta$. I've not assumed the inequality we wished to force; I've pointed out how to choose $epsilon$ so as to prove the desired result.
$endgroup$
– J.G.
Jan 27 at 18:16
$begingroup$
Got it. One last bit, how does ϵ<δ√x help us? I get that |√xn−√x| needs to be less than a constant for all x, and δ√x is a constant but why epsilon less than this?
$endgroup$
– AnoUser1
Jan 27 at 18:21
$begingroup$
@AnoUser1 The point of the exercise is to prove for any $delta>0$ there exists some $epsilon>0$ such that $|x_n-x|<epsilonimplies|sqrt{x_n}-sqrt{x}|<delta$. You can choose any $epsilon$ that works; it turns out that one does.
$endgroup$
– J.G.
Jan 27 at 18:24
add a comment |
$begingroup$
You say We want to force |√xn −√x|<δ, is this allowed? It looks to me like assuming what you want to prove is true
$endgroup$
– AnoUser1
Jan 27 at 18:13
$begingroup$
@AnoUser1 "We want to force" means "we want to prove"; that statement was a way to introduce the symbol $delta$. I've not assumed the inequality we wished to force; I've pointed out how to choose $epsilon$ so as to prove the desired result.
$endgroup$
– J.G.
Jan 27 at 18:16
$begingroup$
Got it. One last bit, how does ϵ<δ√x help us? I get that |√xn−√x| needs to be less than a constant for all x, and δ√x is a constant but why epsilon less than this?
$endgroup$
– AnoUser1
Jan 27 at 18:21
$begingroup$
@AnoUser1 The point of the exercise is to prove for any $delta>0$ there exists some $epsilon>0$ such that $|x_n-x|<epsilonimplies|sqrt{x_n}-sqrt{x}|<delta$. You can choose any $epsilon$ that works; it turns out that one does.
$endgroup$
– J.G.
Jan 27 at 18:24
$begingroup$
You say We want to force |√xn −√x|<δ, is this allowed? It looks to me like assuming what you want to prove is true
$endgroup$
– AnoUser1
Jan 27 at 18:13
$begingroup$
You say We want to force |√xn −√x|<δ, is this allowed? It looks to me like assuming what you want to prove is true
$endgroup$
– AnoUser1
Jan 27 at 18:13
$begingroup$
@AnoUser1 "We want to force" means "we want to prove"; that statement was a way to introduce the symbol $delta$. I've not assumed the inequality we wished to force; I've pointed out how to choose $epsilon$ so as to prove the desired result.
$endgroup$
– J.G.
Jan 27 at 18:16
$begingroup$
@AnoUser1 "We want to force" means "we want to prove"; that statement was a way to introduce the symbol $delta$. I've not assumed the inequality we wished to force; I've pointed out how to choose $epsilon$ so as to prove the desired result.
$endgroup$
– J.G.
Jan 27 at 18:16
$begingroup$
Got it. One last bit, how does ϵ<δ√x help us? I get that |√xn−√x| needs to be less than a constant for all x, and δ√x is a constant but why epsilon less than this?
$endgroup$
– AnoUser1
Jan 27 at 18:21
$begingroup$
Got it. One last bit, how does ϵ<δ√x help us? I get that |√xn−√x| needs to be less than a constant for all x, and δ√x is a constant but why epsilon less than this?
$endgroup$
– AnoUser1
Jan 27 at 18:21
$begingroup$
@AnoUser1 The point of the exercise is to prove for any $delta>0$ there exists some $epsilon>0$ such that $|x_n-x|<epsilonimplies|sqrt{x_n}-sqrt{x}|<delta$. You can choose any $epsilon$ that works; it turns out that one does.
$endgroup$
– J.G.
Jan 27 at 18:24
$begingroup$
@AnoUser1 The point of the exercise is to prove for any $delta>0$ there exists some $epsilon>0$ such that $|x_n-x|<epsilonimplies|sqrt{x_n}-sqrt{x}|<delta$. You can choose any $epsilon$ that works; it turns out that one does.
$endgroup$
– J.G.
Jan 27 at 18:24
add a comment |
$begingroup$
Suppose $x>0$. Then, since $frac{1}{sqrt{x_n} + sqrt{x}}$ is bounded, you have that $sqrt{x_n} - sqrt{x} rightarrow 0$ as $n rightarrow infty$. The case $x=0$ is easier.
answered Jan 27 at 17:49
lzralbulzralbu
640512
$endgroup$
add a comment |
$begingroup$
Suppose $x>0$. Then, since $frac{1}{sqrt{x_n} + sqrt{x}}$ is bounded, you have that $sqrt{x_n} - sqrt{x} rightarrow 0$ as $n rightarrow infty$. The case $x=0$ is easier.
answered Jan 27 at 17:49
lzralbulzralbu
640512
$endgroup$
add a comment |
$begingroup$
Suppose $x>0$. Then, since $frac{1}{sqrt{x_n} + sqrt{x}}$ is bounded, you have that $sqrt{x_n} - sqrt{x} rightarrow 0$ as $n rightarrow infty$. The case $x=0$ is easier.
answered Jan 27 at 17:49
lzralbulzralbu
640512
$endgroup$
Suppose $x>0$. Then, since $frac{1}{sqrt{x_n} + sqrt{x}}$ is bounded, you have that $sqrt{x_n} - sqrt{x} rightarrow 0$ as $n rightarrow infty$. The case $x=0$ is easier.
answered Jan 27 at 17:49
lzralbulzralbu
640512
answered Jan 27 at 17:49
lzralbulzralbu
640512
answered Jan 27 at 17:49
lzralbulzralbu
640512
answered Jan 27 at 17:49
lzralbulzralbu
640512
640512
add a comment |
add a comment |
$begingroup$
Suppose $x > 0$. Since $x_n rightarrow x$ as $n rightarrow infty$ and also $displaystylefrac{1}{sqrt{x_n}+sqrt{x}}$ is finite, then $$sqrt{x_n}-sqrt{x}=frac{x_n-x}{sqrt{x_n}+sqrt{x}} rightarrow 0$$
$$implies sqrt{x_n}-sqrt{x} = 0 text{ as } n rightarrow infty$$
answered Jan 27 at 17:49
Exp ikxExp ikx
4489
$endgroup$
add a comment |
$begingroup$
Suppose $x > 0$. Since $x_n rightarrow x$ as $n rightarrow infty$ and also $displaystylefrac{1}{sqrt{x_n}+sqrt{x}}$ is finite, then $$sqrt{x_n}-sqrt{x}=frac{x_n-x}{sqrt{x_n}+sqrt{x}} rightarrow 0$$
$$implies sqrt{x_n}-sqrt{x} = 0 text{ as } n rightarrow infty$$
answered Jan 27 at 17:49
Exp ikxExp ikx
4489
$endgroup$
add a comment |
$begingroup$
Suppose $x > 0$. Since $x_n rightarrow x$ as $n rightarrow infty$ and also $displaystylefrac{1}{sqrt{x_n}+sqrt{x}}$ is finite, then $$sqrt{x_n}-sqrt{x}=frac{x_n-x}{sqrt{x_n}+sqrt{x}} rightarrow 0$$
$$implies sqrt{x_n}-sqrt{x} = 0 text{ as } n rightarrow infty$$
answered Jan 27 at 17:49
Exp ikxExp ikx
4489
$endgroup$
Suppose $x > 0$. Since $x_n rightarrow x$ as $n rightarrow infty$ and also $displaystylefrac{1}{sqrt{x_n}+sqrt{x}}$ is finite, then $$sqrt{x_n}-sqrt{x}=frac{x_n-x}{sqrt{x_n}+sqrt{x}} rightarrow 0$$
$$implies sqrt{x_n}-sqrt{x} = 0 text{ as } n rightarrow infty$$
answered Jan 27 at 17:49
Exp ikxExp ikx
4489
answered Jan 27 at 17:49
Exp ikxExp ikx
4489
answered Jan 27 at 17:49
Exp ikxExp ikx
4489
answered Jan 27 at 17:49
Exp ikxExp ikx
4489
4489
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$begingroup$
ayy lmao keith ball?
$endgroup$
– Dis-integrating
Jan 28 at 4:25