Mixing
limit of quotient is quotient of limits
Clash Royale CLAN TAG#URR8PPP
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Prove the following:
Let $X$ be a metric space. Let $E subseteq X$. Let $p$ be a limit
point of $E$, and let $f,g: E subseteq X to mathbbK$ be a maps.
Suppose $lim_x to p f(x) = q_1, lim_x to p g(x) = q_2$. Then,
if we define
$$fracfg: Z= x in E mid g(x) neq 0 to mathbbK: x
mapsto f(x)/g(x)$$
we have $lim_x to p left(fracfgright)(x) = fracq_1q_2$
provided that $q_2 neq 0$.
In my proof, I use;
$lim_x to p f(x) = q$ iff for every sequence $(p_n)$ in
$E setminus p$ for which $p_n to p$, we have $f(p_n) to q$
Is my proof correct?
Proof:
First, we check that $p$ is a limit point of $Z$, so that talking about limits makes sense.
Since $lim_x to p g(x) = q_2 neq 0$, there is $delta > 0$ such that $|g(x) - q_2| < |q_2| neq 0$ for all $x in E$ satisfying $0 <d_X(x,p) < delta.$ It follows that $g(x) neq 0$ for $x in (Esetminus p) cap B_X(p, delta)$.
Let $epsilon > 0$. Then we know that we can pick $z in B_X(p, epsilon land delta) cap (E setminus p)$, since $p$ is a limit point of $E$. It follows that $z in B_X(p,epsilon) cap (Z setminus p)$, and $p$ is a limit point of $Z$.
Notice that $lim_x to p, x in Z f(x) = q_1$ and $lim_x to p, x in Z g(x) = q_2$ (this follows trivially from the definition of limit).
Let $(p_n)_n$ be a sequence in $Z setminus p$. Then, we know that $f(p_n) to q_1$ and $g(p_n) to q_2$. Hence, $$fracf(p_n)g(p_n) to fracq_1q_2$$
and the result follows $square$.
calculus limits proof-verification metric-spaces
edited Aug 1 at 11:49
rtybase
8,77221333
asked Aug 1 at 11:44
Math_QED
6,30831344
add a comment |Â
up vote
2
down vote
favorite
Prove the following:
Let $X$ be a metric space. Let $E subseteq X$. Let $p$ be a limit
point of $E$, and let $f,g: E subseteq X to mathbbK$ be a maps.
Suppose $lim_x to p f(x) = q_1, lim_x to p g(x) = q_2$. Then,
if we define
$$fracfg: Z= x in E mid g(x) neq 0 to mathbbK: x
mapsto f(x)/g(x)$$
we have $lim_x to p left(fracfgright)(x) = fracq_1q_2$
provided that $q_2 neq 0$.
In my proof, I use;
$lim_x to p f(x) = q$ iff for every sequence $(p_n)$ in
$E setminus p$ for which $p_n to p$, we have $f(p_n) to q$
Is my proof correct?
Proof:
First, we check that $p$ is a limit point of $Z$, so that talking about limits makes sense.
Since $lim_x to p g(x) = q_2 neq 0$, there is $delta > 0$ such that $|g(x) - q_2| < |q_2| neq 0$ for all $x in E$ satisfying $0 <d_X(x,p) < delta.$ It follows that $g(x) neq 0$ for $x in (Esetminus p) cap B_X(p, delta)$.
Let $epsilon > 0$. Then we know that we can pick $z in B_X(p, epsilon land delta) cap (E setminus p)$, since $p$ is a limit point of $E$. It follows that $z in B_X(p,epsilon) cap (Z setminus p)$, and $p$ is a limit point of $Z$.
Notice that $lim_x to p, x in Z f(x) = q_1$ and $lim_x to p, x in Z g(x) = q_2$ (this follows trivially from the definition of limit).
Let $(p_n)_n$ be a sequence in $Z setminus p$. Then, we know that $f(p_n) to q_1$ and $g(p_n) to q_2$. Hence, $$fracf(p_n)g(p_n) to fracq_1q_2$$
and the result follows $square$.
calculus limits proof-verification metric-spaces
edited Aug 1 at 11:49
rtybase
8,77221333
asked Aug 1 at 11:44
Math_QED
6,30831344
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Prove the following:
Let $X$ be a metric space. Let $E subseteq X$. Let $p$ be a limit
point of $E$, and let $f,g: E subseteq X to mathbbK$ be a maps.
Suppose $lim_x to p f(x) = q_1, lim_x to p g(x) = q_2$. Then,
if we define
$$fracfg: Z= x in E mid g(x) neq 0 to mathbbK: x
mapsto f(x)/g(x)$$
we have $lim_x to p left(fracfgright)(x) = fracq_1q_2$
provided that $q_2 neq 0$.
In my proof, I use;
$lim_x to p f(x) = q$ iff for every sequence $(p_n)$ in
$E setminus p$ for which $p_n to p$, we have $f(p_n) to q$
Is my proof correct?
Proof:
First, we check that $p$ is a limit point of $Z$, so that talking about limits makes sense.
Since $lim_x to p g(x) = q_2 neq 0$, there is $delta > 0$ such that $|g(x) - q_2| < |q_2| neq 0$ for all $x in E$ satisfying $0 <d_X(x,p) < delta.$ It follows that $g(x) neq 0$ for $x in (Esetminus p) cap B_X(p, delta)$.
Let $epsilon > 0$. Then we know that we can pick $z in B_X(p, epsilon land delta) cap (E setminus p)$, since $p$ is a limit point of $E$. It follows that $z in B_X(p,epsilon) cap (Z setminus p)$, and $p$ is a limit point of $Z$.
Notice that $lim_x to p, x in Z f(x) = q_1$ and $lim_x to p, x in Z g(x) = q_2$ (this follows trivially from the definition of limit).
Let $(p_n)_n$ be a sequence in $Z setminus p$. Then, we know that $f(p_n) to q_1$ and $g(p_n) to q_2$. Hence, $$fracf(p_n)g(p_n) to fracq_1q_2$$
and the result follows $square$.
calculus limits proof-verification metric-spaces
edited Aug 1 at 11:49
rtybase
8,77221333
asked Aug 1 at 11:44
Math_QED
6,30831344
Prove the following:
Let $X$ be a metric space. Let $E subseteq X$. Let $p$ be a limit
point of $E$, and let $f,g: E subseteq X to mathbbK$ be a maps.
Suppose $lim_x to p f(x) = q_1, lim_x to p g(x) = q_2$. Then,
if we define
$$fracfg: Z= x in E mid g(x) neq 0 to mathbbK: x
mapsto f(x)/g(x)$$
we have $lim_x to p left(fracfgright)(x) = fracq_1q_2$
provided that $q_2 neq 0$.
In my proof, I use;
$lim_x to p f(x) = q$ iff for every sequence $(p_n)$ in
$E setminus p$ for which $p_n to p$, we have $f(p_n) to q$
Is my proof correct?
Proof:
First, we check that $p$ is a limit point of $Z$, so that talking about limits makes sense.
Since $lim_x to p g(x) = q_2 neq 0$, there is $delta > 0$ such that $|g(x) - q_2| < |q_2| neq 0$ for all $x in E$ satisfying $0 <d_X(x,p) < delta.$ It follows that $g(x) neq 0$ for $x in (Esetminus p) cap B_X(p, delta)$.
Let $epsilon > 0$. Then we know that we can pick $z in B_X(p, epsilon land delta) cap (E setminus p)$, since $p$ is a limit point of $E$. It follows that $z in B_X(p,epsilon) cap (Z setminus p)$, and $p$ is a limit point of $Z$.
Notice that $lim_x to p, x in Z f(x) = q_1$ and $lim_x to p, x in Z g(x) = q_2$ (this follows trivially from the definition of limit).
Let $(p_n)_n$ be a sequence in $Z setminus p$. Then, we know that $f(p_n) to q_1$ and $g(p_n) to q_2$. Hence, $$fracf(p_n)g(p_n) to fracq_1q_2$$
and the result follows $square$.
calculus limits proof-verification metric-spaces
edited Aug 1 at 11:49
rtybase
8,77221333
asked Aug 1 at 11:44
Math_QED
6,30831344
edited Aug 1 at 11:49
rtybase
8,77221333
edited Aug 1 at 11:49
rtybase
8,77221333
edited Aug 1 at 11:49
rtybase
8,77221333
8,77221333
asked Aug 1 at 11:44
Math_QED
6,30831344
asked Aug 1 at 11:44
Math_QED
6,30831344
asked Aug 1 at 11:44
Math_QED
6,30831344
6,30831344
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
It looks pretty solid, but more justification may be needed at the end. What is $Bbb K$? Have you got a previous result to rely on that says, if $(a_n)_n$ and $(b_n)_n$ are sequences in $Bbb K$ with $a_nto ainBbb K$ and $b_nto binBbb K,$ such that $b$ and each $b_n$ are non-zero, then $fraca_nb_ntofracab$? If so, then I'd say you're good to go. If not, then you'll have to prove this, first, either as a Lemma or as part of your proof.
Also, in this line, there's an issue:
Since $lim_x to p g(x) = q_2 neq 0$, there is $delta > 0$ such that $|g(x) - q_2| < |q_2| neq 0$ for all $x in E$ satisfying $0 <d_X(x,p) < delta.$
While this is true, and while it does follow that $g$ is non-zero for $x$ sufficiently close to $p,$ you haven't actually shown it to be true. What specific $epsilon>0$ could you use to find such an appropriate $delta$?
answered Aug 1 at 12:12
Cameron Buie
83.5k771152
Ah sorry, I forgot to mention. $mathbbK$ is the complex numbers or the real numbers. But I think any vector space will do.
– Math_QED
Aug 1 at 12:13
What exactly is the issue? I clearly pick $epsilon = |q_2| > 0$.
– Math_QED
Aug 1 at 12:17
1
Ah! I see. The "$|q_2|neq 0$" part was throwing me. Perhaps it would be clearer to say something like "Since $q_2neq 0,$ then $|q_2|>0,$ so there is $delta>0$ such that $|g(x)-q_2|<|q_2|$ by definition of limit."
– Cameron Buie
Aug 1 at 12:21
You are right. That would have been clearer. Thanks for your time!
– Math_QED
Aug 1 at 12:22
1
As for $Bbb K$ being an arbitrary vector space: division isn't necessarily defined in a given vector space, so I doubt it. It would have to be a field, at least.
– Cameron Buie
Aug 1 at 12:22
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
It looks pretty solid, but more justification may be needed at the end. What is $Bbb K$? Have you got a previous result to rely on that says, if $(a_n)_n$ and $(b_n)_n$ are sequences in $Bbb K$ with $a_nto ainBbb K$ and $b_nto binBbb K,$ such that $b$ and each $b_n$ are non-zero, then $fraca_nb_ntofracab$? If so, then I'd say you're good to go. If not, then you'll have to prove this, first, either as a Lemma or as part of your proof.
Also, in this line, there's an issue:
Since $lim_x to p g(x) = q_2 neq 0$, there is $delta > 0$ such that $|g(x) - q_2| < |q_2| neq 0$ for all $x in E$ satisfying $0 <d_X(x,p) < delta.$
While this is true, and while it does follow that $g$ is non-zero for $x$ sufficiently close to $p,$ you haven't actually shown it to be true. What specific $epsilon>0$ could you use to find such an appropriate $delta$?
answered Aug 1 at 12:12
Cameron Buie
83.5k771152
Ah sorry, I forgot to mention. $mathbbK$ is the complex numbers or the real numbers. But I think any vector space will do.
– Math_QED
Aug 1 at 12:13
What exactly is the issue? I clearly pick $epsilon = |q_2| > 0$.
– Math_QED
Aug 1 at 12:17
1
Ah! I see. The "$|q_2|neq 0$" part was throwing me. Perhaps it would be clearer to say something like "Since $q_2neq 0,$ then $|q_2|>0,$ so there is $delta>0$ such that $|g(x)-q_2|<|q_2|$ by definition of limit."
– Cameron Buie
Aug 1 at 12:21
You are right. That would have been clearer. Thanks for your time!
– Math_QED
Aug 1 at 12:22
1
As for $Bbb K$ being an arbitrary vector space: division isn't necessarily defined in a given vector space, so I doubt it. It would have to be a field, at least.
– Cameron Buie
Aug 1 at 12:22
 |Â
show 1 more comment
up vote
1
down vote
accepted
It looks pretty solid, but more justification may be needed at the end. What is $Bbb K$? Have you got a previous result to rely on that says, if $(a_n)_n$ and $(b_n)_n$ are sequences in $Bbb K$ with $a_nto ainBbb K$ and $b_nto binBbb K,$ such that $b$ and each $b_n$ are non-zero, then $fraca_nb_ntofracab$? If so, then I'd say you're good to go. If not, then you'll have to prove this, first, either as a Lemma or as part of your proof.
Also, in this line, there's an issue:
Since $lim_x to p g(x) = q_2 neq 0$, there is $delta > 0$ such that $|g(x) - q_2| < |q_2| neq 0$ for all $x in E$ satisfying $0 <d_X(x,p) < delta.$
While this is true, and while it does follow that $g$ is non-zero for $x$ sufficiently close to $p,$ you haven't actually shown it to be true. What specific $epsilon>0$ could you use to find such an appropriate $delta$?
answered Aug 1 at 12:12
Cameron Buie
83.5k771152
Ah sorry, I forgot to mention. $mathbbK$ is the complex numbers or the real numbers. But I think any vector space will do.
– Math_QED
Aug 1 at 12:13
What exactly is the issue? I clearly pick $epsilon = |q_2| > 0$.
– Math_QED
Aug 1 at 12:17
1
Ah! I see. The "$|q_2|neq 0$" part was throwing me. Perhaps it would be clearer to say something like "Since $q_2neq 0,$ then $|q_2|>0,$ so there is $delta>0$ such that $|g(x)-q_2|<|q_2|$ by definition of limit."
– Cameron Buie
Aug 1 at 12:21
You are right. That would have been clearer. Thanks for your time!
– Math_QED
Aug 1 at 12:22
1
As for $Bbb K$ being an arbitrary vector space: division isn't necessarily defined in a given vector space, so I doubt it. It would have to be a field, at least.
– Cameron Buie
Aug 1 at 12:22
 |Â
show 1 more comment
up vote
1
down vote
accepted
up vote
1
down vote
accepted
It looks pretty solid, but more justification may be needed at the end. What is $Bbb K$? Have you got a previous result to rely on that says, if $(a_n)_n$ and $(b_n)_n$ are sequences in $Bbb K$ with $a_nto ainBbb K$ and $b_nto binBbb K,$ such that $b$ and each $b_n$ are non-zero, then $fraca_nb_ntofracab$? If so, then I'd say you're good to go. If not, then you'll have to prove this, first, either as a Lemma or as part of your proof.
Also, in this line, there's an issue:
Since $lim_x to p g(x) = q_2 neq 0$, there is $delta > 0$ such that $|g(x) - q_2| < |q_2| neq 0$ for all $x in E$ satisfying $0 <d_X(x,p) < delta.$
While this is true, and while it does follow that $g$ is non-zero for $x$ sufficiently close to $p,$ you haven't actually shown it to be true. What specific $epsilon>0$ could you use to find such an appropriate $delta$?
answered Aug 1 at 12:12
Cameron Buie
83.5k771152
It looks pretty solid, but more justification may be needed at the end. What is $Bbb K$? Have you got a previous result to rely on that says, if $(a_n)_n$ and $(b_n)_n$ are sequences in $Bbb K$ with $a_nto ainBbb K$ and $b_nto binBbb K,$ such that $b$ and each $b_n$ are non-zero, then $fraca_nb_ntofracab$? If so, then I'd say you're good to go. If not, then you'll have to prove this, first, either as a Lemma or as part of your proof.
Also, in this line, there's an issue:
Since $lim_x to p g(x) = q_2 neq 0$, there is $delta > 0$ such that $|g(x) - q_2| < |q_2| neq 0$ for all $x in E$ satisfying $0 <d_X(x,p) < delta.$
While this is true, and while it does follow that $g$ is non-zero for $x$ sufficiently close to $p,$ you haven't actually shown it to be true. What specific $epsilon>0$ could you use to find such an appropriate $delta$?
answered Aug 1 at 12:12
Cameron Buie
83.5k771152
edited Aug 1 at 12:17
answered Aug 1 at 12:12
Cameron Buie
83.5k771152
answered Aug 1 at 12:12
Cameron Buie
83.5k771152
answered Aug 1 at 12:12
Cameron Buie
83.5k771152
83.5k771152
Ah sorry, I forgot to mention. $mathbbK$ is the complex numbers or the real numbers. But I think any vector space will do.
– Math_QED
Aug 1 at 12:13
What exactly is the issue? I clearly pick $epsilon = |q_2| > 0$.
– Math_QED
Aug 1 at 12:17
1
Ah! I see. The "$|q_2|neq 0$" part was throwing me. Perhaps it would be clearer to say something like "Since $q_2neq 0,$ then $|q_2|>0,$ so there is $delta>0$ such that $|g(x)-q_2|<|q_2|$ by definition of limit."
– Cameron Buie
Aug 1 at 12:21
You are right. That would have been clearer. Thanks for your time!
– Math_QED
Aug 1 at 12:22
1
As for $Bbb K$ being an arbitrary vector space: division isn't necessarily defined in a given vector space, so I doubt it. It would have to be a field, at least.
– Cameron Buie
Aug 1 at 12:22
 |Â
show 1 more comment
Ah sorry, I forgot to mention. $mathbbK$ is the complex numbers or the real numbers. But I think any vector space will do.
– Math_QED
Aug 1 at 12:13
What exactly is the issue? I clearly pick $epsilon = |q_2| > 0$.
– Math_QED
Aug 1 at 12:17
1
Ah! I see. The "$|q_2|neq 0$" part was throwing me. Perhaps it would be clearer to say something like "Since $q_2neq 0,$ then $|q_2|>0,$ so there is $delta>0$ such that $|g(x)-q_2|<|q_2|$ by definition of limit."
– Cameron Buie
Aug 1 at 12:21
You are right. That would have been clearer. Thanks for your time!
– Math_QED
Aug 1 at 12:22
1
As for $Bbb K$ being an arbitrary vector space: division isn't necessarily defined in a given vector space, so I doubt it. It would have to be a field, at least.
– Cameron Buie
Aug 1 at 12:22
Ah sorry, I forgot to mention. $mathbbK$ is the complex numbers or the real numbers. But I think any vector space will do.
– Math_QED
Aug 1 at 12:13
Ah sorry, I forgot to mention. $mathbbK$ is the complex numbers or the real numbers. But I think any vector space will do.
– Math_QED
Aug 1 at 12:13
What exactly is the issue? I clearly pick $epsilon = |q_2| > 0$.
– Math_QED
Aug 1 at 12:17
What exactly is the issue? I clearly pick $epsilon = |q_2| > 0$.
– Math_QED
Aug 1 at 12:17
1
1
Ah! I see. The "$|q_2|neq 0$" part was throwing me. Perhaps it would be clearer to say something like "Since $q_2neq 0,$ then $|q_2|>0,$ so there is $delta>0$ such that $|g(x)-q_2|<|q_2|$ by definition of limit."
– Cameron Buie
Aug 1 at 12:21
Ah! I see. The "$|q_2|neq 0$" part was throwing me. Perhaps it would be clearer to say something like "Since $q_2neq 0,$ then $|q_2|>0,$ so there is $delta>0$ such that $|g(x)-q_2|<|q_2|$ by definition of limit."
– Cameron Buie
Aug 1 at 12:21
You are right. That would have been clearer. Thanks for your time!
– Math_QED
Aug 1 at 12:22
You are right. That would have been clearer. Thanks for your time!
– Math_QED
Aug 1 at 12:22
1
1
As for $Bbb K$ being an arbitrary vector space: division isn't necessarily defined in a given vector space, so I doubt it. It would have to be a field, at least.
– Cameron Buie
Aug 1 at 12:22
As for $Bbb K$ being an arbitrary vector space: division isn't necessarily defined in a given vector space, so I doubt it. It would have to be a field, at least.
– Cameron Buie
Aug 1 at 12:22
 |Â
show 1 more comment
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