$lim (a + b);$ when $;lim(b);$ does not exist? [closed]
$begingroup$
Suppose $a$ and $b$ are functions of $x$. Is it guaranteed that
$$
lim_{x to +infty} a + btext{ does not exist}
$$
when
$$
lim_{x to +infty} a = cquadtext{and}quad
lim_{x to +infty} btext{ does not exist ?}
$$
limits
edited Dec 8 '18 at 23:33
user376343
3,9584829
asked May 28 '11 at 23:08
crocscrocs
6714
$endgroup$
closed as off-topic by TheSimpliFire, Namaste, Holo, Did, Xander Henderson Jan 15 at 18:14
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheSimpliFire, Namaste, Holo, Did, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Suppose $a$ and $b$ are functions of $x$. Is it guaranteed that
$$
lim_{x to +infty} a + btext{ does not exist}
$$
when
$$
lim_{x to +infty} a = cquadtext{and}quad
lim_{x to +infty} btext{ does not exist ?}
$$
limits
edited Dec 8 '18 at 23:33
user376343
3,9584829
asked May 28 '11 at 23:08
crocscrocs
6714
$endgroup$
closed as off-topic by TheSimpliFire, Namaste, Holo, Did, Xander Henderson Jan 15 at 18:14
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheSimpliFire, Namaste, Holo, Did, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Suppose $a$ and $b$ are functions of $x$. Is it guaranteed that
$$
lim_{x to +infty} a + btext{ does not exist}
$$
when
$$
lim_{x to +infty} a = cquadtext{and}quad
lim_{x to +infty} btext{ does not exist ?}
$$
limits
edited Dec 8 '18 at 23:33
user376343
3,9584829
asked May 28 '11 at 23:08
crocscrocs
6714
$endgroup$
Suppose $a$ and $b$ are functions of $x$. Is it guaranteed that
$$
lim_{x to +infty} a + btext{ does not exist}
$$
when
$$
lim_{x to +infty} a = cquadtext{and}quad
lim_{x to +infty} btext{ does not exist ?}
$$
limits
limits
edited Dec 8 '18 at 23:33
user376343
3,9584829
asked May 28 '11 at 23:08
crocscrocs
6714
edited Dec 8 '18 at 23:33
user376343
3,9584829
asked May 28 '11 at 23:08
crocscrocs
6714
edited Dec 8 '18 at 23:33
user376343
3,9584829
edited Dec 8 '18 at 23:33
user376343
3,9584829
edited Dec 8 '18 at 23:33
user376343
3,9584829
3,9584829
asked May 28 '11 at 23:08
crocscrocs
6714
asked May 28 '11 at 23:08
crocscrocs
6714
asked May 28 '11 at 23:08
crocscrocs
6714
6714
closed as off-topic by TheSimpliFire, Namaste, Holo, Did, Xander Henderson Jan 15 at 18:14
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheSimpliFire, Namaste, Holo, Did, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by TheSimpliFire, Namaste, Holo, Did, Xander Henderson Jan 15 at 18:14
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheSimpliFire, Namaste, Holo, Did, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Yes. If the limit of $a+b$ existed, it would follow that
$$lim_{x to +infty}b=lim_{x to +infty} [(a + b) - a]=lim_{x to +infty}(a+b)-lim_{x to +infty}a;.$$
answered May 28 '11 at 23:18
jorikijoriki
171k10189349
$endgroup$
add a comment |
$begingroup$
Suppose, to get a contradiction, that our limit exists. That is, suppose $$lim_{xrightarrow infty} a(x)+b(x)=d$$ exists. Then since $$lim_{xrightarrow infty} -a(x)=-c,$$ and as limits are additive, we conclude that $$lim_{xrightarrow infty} a(x)+b(x)-a(x)=d-c$$ which means $$lim_{xrightarrow infty} b(x)=d-c.$$ But this is impossible since we had that $b(x)$ did not tend to a limit.
Hope that helps,
answered May 28 '11 at 23:18
Eric NaslundEric Naslund
60.6k10140241
$endgroup$
add a comment |
$begingroup$
HINT $ $ This follows immediately from the fact that functions whose limit exists at $rm:infty:$ are closed under subtraction, i.e. they comprise an additive subgroup of all functions. Therefore, abstractly, this is essentially the same as the proof that the sum of an integer and non-integer is a non-integer. For further discussion of this complementary view of a subgroup see this post.
edited Apr 13 '17 at 12:21
Community♦
1
answered May 28 '11 at 23:40
Bill DubuqueBill Dubuque
212k29195654
$endgroup$
add a comment |
$begingroup$
Just to complete joriki's answer:
His answer is right only if $c$ is a number, which seems to be the case from the way you wrote the question.
Anyhow if you deal also with infinite limits, it is possible that $lim_{x to infty}b$ does not exist and $lim_{x to infty} a+b$ exists.
Just take $a=x-sin(x)$ and $b=sin(x)$.
answered May 29 '11 at 1:13
N. S.N. S.
104k7114209
$endgroup$
1
$begingroup$
when the limit is $infty$, it does not exist. Writing $limlimits_{xto a}f = infty$ is a way of saying "the limit does not exist" and specifying why it does not exist. Similarly with limits "equal" to $-infty$", and with limits as $xtoinfty$ or as $xto-infty$. Otherwise, limit theorems need to have a whole bunch of exceptions (e.g., it would no longer be true that if the limits of $f$ and $g$ 'exist', then so does the limit of $f+g$ and the latter is equal to their sum).
$endgroup$
– Arturo Magidin
May 29 '11 at 1:17
3
$begingroup$
I think that this depends from book to book and instructor to instructor. And the sum theorem, holds as long as it is not an indeterminate form, I am also pretty sure I saw somewhere that theorem stated for finite limits..
$endgroup$
– N. S.
May 29 '11 at 1:37
2
$begingroup$
@Arturo: One could also argue as follows: In general, it's possible to consider convergence in the extended reals with respect to the order topology; this yields the usual definition of the limits $pminfty$. However, in that case not only the limits but also the function values may be $pminfty$. Then addition wouldn't be defined, so by using addition the OP was implying that the underlying set was just the reals. Allowing infinite limits but only finite values would be like allowing real limits but only rational values, i.e. considering elements outside the set as potential limits.
$endgroup$
– joriki
May 29 '11 at 1:40
1
$begingroup$
BTW: In most books L'H appears in the form: If bla bla and $limfrac{f'(x)}{g'(x)}$ exists then bla bla... Does that cover the case when this limit is $pm infty$? Of course I realize that this is not really a good point since most Calculus books are very poorly written, but I am used to make a distinction between limit exists and limit is finite. This might come from the fact that I learned Analysis the old way, sequences first, and there convergence is already finite, so we always use the distinction "convergent" vs. "has a limit". For functions we replaced convergence with finite lim
$endgroup$
– N. S.
May 29 '11 at 1:45
2
$begingroup$
"The limit exists" is relative to the set you consider. You can consider the reals, the extended reals, or even the one-point compactification of the reals with just one point at infinity. The limit of the sequence $(-1)^nn$ exists in the one-point compactification but not in the extended reals. This shows that it's not just a matter of adding "enough" limit points, but also of choosing a topology. Again, there's an analogy to the rationals, which can be extended either to the reals or to the $p$-adic numbers.
$endgroup$
– joriki
May 29 '11 at 1:56
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes. If the limit of $a+b$ existed, it would follow that
$$lim_{x to +infty}b=lim_{x to +infty} [(a + b) - a]=lim_{x to +infty}(a+b)-lim_{x to +infty}a;.$$
answered May 28 '11 at 23:18
jorikijoriki
171k10189349
$endgroup$
add a comment |
$begingroup$
Yes. If the limit of $a+b$ existed, it would follow that
$$lim_{x to +infty}b=lim_{x to +infty} [(a + b) - a]=lim_{x to +infty}(a+b)-lim_{x to +infty}a;.$$
answered May 28 '11 at 23:18
jorikijoriki
171k10189349
$endgroup$
add a comment |
$begingroup$
Yes. If the limit of $a+b$ existed, it would follow that
$$lim_{x to +infty}b=lim_{x to +infty} [(a + b) - a]=lim_{x to +infty}(a+b)-lim_{x to +infty}a;.$$
answered May 28 '11 at 23:18
jorikijoriki
171k10189349
$endgroup$
Yes. If the limit of $a+b$ existed, it would follow that
$$lim_{x to +infty}b=lim_{x to +infty} [(a + b) - a]=lim_{x to +infty}(a+b)-lim_{x to +infty}a;.$$
answered May 28 '11 at 23:18
jorikijoriki
171k10189349
answered May 28 '11 at 23:18
jorikijoriki
171k10189349
answered May 28 '11 at 23:18
jorikijoriki
171k10189349
answered May 28 '11 at 23:18
jorikijoriki
171k10189349
171k10189349
add a comment |
add a comment |
$begingroup$
Suppose, to get a contradiction, that our limit exists. That is, suppose $$lim_{xrightarrow infty} a(x)+b(x)=d$$ exists. Then since $$lim_{xrightarrow infty} -a(x)=-c,$$ and as limits are additive, we conclude that $$lim_{xrightarrow infty} a(x)+b(x)-a(x)=d-c$$ which means $$lim_{xrightarrow infty} b(x)=d-c.$$ But this is impossible since we had that $b(x)$ did not tend to a limit.
Hope that helps,
answered May 28 '11 at 23:18
Eric NaslundEric Naslund
60.6k10140241
$endgroup$
add a comment |
$begingroup$
Suppose, to get a contradiction, that our limit exists. That is, suppose $$lim_{xrightarrow infty} a(x)+b(x)=d$$ exists. Then since $$lim_{xrightarrow infty} -a(x)=-c,$$ and as limits are additive, we conclude that $$lim_{xrightarrow infty} a(x)+b(x)-a(x)=d-c$$ which means $$lim_{xrightarrow infty} b(x)=d-c.$$ But this is impossible since we had that $b(x)$ did not tend to a limit.
Hope that helps,
answered May 28 '11 at 23:18
Eric NaslundEric Naslund
60.6k10140241
$endgroup$
add a comment |
$begingroup$
Suppose, to get a contradiction, that our limit exists. That is, suppose $$lim_{xrightarrow infty} a(x)+b(x)=d$$ exists. Then since $$lim_{xrightarrow infty} -a(x)=-c,$$ and as limits are additive, we conclude that $$lim_{xrightarrow infty} a(x)+b(x)-a(x)=d-c$$ which means $$lim_{xrightarrow infty} b(x)=d-c.$$ But this is impossible since we had that $b(x)$ did not tend to a limit.
Hope that helps,
answered May 28 '11 at 23:18
Eric NaslundEric Naslund
60.6k10140241
$endgroup$
Suppose, to get a contradiction, that our limit exists. That is, suppose $$lim_{xrightarrow infty} a(x)+b(x)=d$$ exists. Then since $$lim_{xrightarrow infty} -a(x)=-c,$$ and as limits are additive, we conclude that $$lim_{xrightarrow infty} a(x)+b(x)-a(x)=d-c$$ which means $$lim_{xrightarrow infty} b(x)=d-c.$$ But this is impossible since we had that $b(x)$ did not tend to a limit.
Hope that helps,
answered May 28 '11 at 23:18
Eric NaslundEric Naslund
60.6k10140241
answered May 28 '11 at 23:18
Eric NaslundEric Naslund
60.6k10140241
answered May 28 '11 at 23:18
Eric NaslundEric Naslund
60.6k10140241
answered May 28 '11 at 23:18
Eric NaslundEric Naslund
60.6k10140241
60.6k10140241
add a comment |
add a comment |
$begingroup$
HINT $ $ This follows immediately from the fact that functions whose limit exists at $rm:infty:$ are closed under subtraction, i.e. they comprise an additive subgroup of all functions. Therefore, abstractly, this is essentially the same as the proof that the sum of an integer and non-integer is a non-integer. For further discussion of this complementary view of a subgroup see this post.
edited Apr 13 '17 at 12:21
Community♦
1
answered May 28 '11 at 23:40
Bill DubuqueBill Dubuque
212k29195654
$endgroup$
add a comment |
$begingroup$
HINT $ $ This follows immediately from the fact that functions whose limit exists at $rm:infty:$ are closed under subtraction, i.e. they comprise an additive subgroup of all functions. Therefore, abstractly, this is essentially the same as the proof that the sum of an integer and non-integer is a non-integer. For further discussion of this complementary view of a subgroup see this post.
edited Apr 13 '17 at 12:21
Community♦
1
answered May 28 '11 at 23:40
Bill DubuqueBill Dubuque
212k29195654
$endgroup$
add a comment |
$begingroup$
HINT $ $ This follows immediately from the fact that functions whose limit exists at $rm:infty:$ are closed under subtraction, i.e. they comprise an additive subgroup of all functions. Therefore, abstractly, this is essentially the same as the proof that the sum of an integer and non-integer is a non-integer. For further discussion of this complementary view of a subgroup see this post.
edited Apr 13 '17 at 12:21
Community♦
1
answered May 28 '11 at 23:40
Bill DubuqueBill Dubuque
212k29195654
$endgroup$
HINT $ $ This follows immediately from the fact that functions whose limit exists at $rm:infty:$ are closed under subtraction, i.e. they comprise an additive subgroup of all functions. Therefore, abstractly, this is essentially the same as the proof that the sum of an integer and non-integer is a non-integer. For further discussion of this complementary view of a subgroup see this post.
edited Apr 13 '17 at 12:21
Community♦
1
answered May 28 '11 at 23:40
Bill DubuqueBill Dubuque
212k29195654
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
answered May 28 '11 at 23:40
Bill DubuqueBill Dubuque
212k29195654
answered May 28 '11 at 23:40
Bill DubuqueBill Dubuque
212k29195654
answered May 28 '11 at 23:40
Bill DubuqueBill Dubuque
212k29195654
212k29195654
add a comment |
add a comment |
$begingroup$
Just to complete joriki's answer:
His answer is right only if $c$ is a number, which seems to be the case from the way you wrote the question.
Anyhow if you deal also with infinite limits, it is possible that $lim_{x to infty}b$ does not exist and $lim_{x to infty} a+b$ exists.
Just take $a=x-sin(x)$ and $b=sin(x)$.
answered May 29 '11 at 1:13
N. S.N. S.
104k7114209
$endgroup$
1
$begingroup$
when the limit is $infty$, it does not exist. Writing $limlimits_{xto a}f = infty$ is a way of saying "the limit does not exist" and specifying why it does not exist. Similarly with limits "equal" to $-infty$", and with limits as $xtoinfty$ or as $xto-infty$. Otherwise, limit theorems need to have a whole bunch of exceptions (e.g., it would no longer be true that if the limits of $f$ and $g$ 'exist', then so does the limit of $f+g$ and the latter is equal to their sum).
$endgroup$
– Arturo Magidin
May 29 '11 at 1:17
3
$begingroup$
I think that this depends from book to book and instructor to instructor. And the sum theorem, holds as long as it is not an indeterminate form, I am also pretty sure I saw somewhere that theorem stated for finite limits..
$endgroup$
– N. S.
May 29 '11 at 1:37
2
$begingroup$
@Arturo: One could also argue as follows: In general, it's possible to consider convergence in the extended reals with respect to the order topology; this yields the usual definition of the limits $pminfty$. However, in that case not only the limits but also the function values may be $pminfty$. Then addition wouldn't be defined, so by using addition the OP was implying that the underlying set was just the reals. Allowing infinite limits but only finite values would be like allowing real limits but only rational values, i.e. considering elements outside the set as potential limits.
$endgroup$
– joriki
May 29 '11 at 1:40
1
$begingroup$
BTW: In most books L'H appears in the form: If bla bla and $limfrac{f'(x)}{g'(x)}$ exists then bla bla... Does that cover the case when this limit is $pm infty$? Of course I realize that this is not really a good point since most Calculus books are very poorly written, but I am used to make a distinction between limit exists and limit is finite. This might come from the fact that I learned Analysis the old way, sequences first, and there convergence is already finite, so we always use the distinction "convergent" vs. "has a limit". For functions we replaced convergence with finite lim
$endgroup$
– N. S.
May 29 '11 at 1:45
2
$begingroup$
"The limit exists" is relative to the set you consider. You can consider the reals, the extended reals, or even the one-point compactification of the reals with just one point at infinity. The limit of the sequence $(-1)^nn$ exists in the one-point compactification but not in the extended reals. This shows that it's not just a matter of adding "enough" limit points, but also of choosing a topology. Again, there's an analogy to the rationals, which can be extended either to the reals or to the $p$-adic numbers.
$endgroup$
– joriki
May 29 '11 at 1:56
add a comment |
$begingroup$
Just to complete joriki's answer:
His answer is right only if $c$ is a number, which seems to be the case from the way you wrote the question.
Anyhow if you deal also with infinite limits, it is possible that $lim_{x to infty}b$ does not exist and $lim_{x to infty} a+b$ exists.
Just take $a=x-sin(x)$ and $b=sin(x)$.
answered May 29 '11 at 1:13
N. S.N. S.
104k7114209
$endgroup$
1
$begingroup$
when the limit is $infty$, it does not exist. Writing $limlimits_{xto a}f = infty$ is a way of saying "the limit does not exist" and specifying why it does not exist. Similarly with limits "equal" to $-infty$", and with limits as $xtoinfty$ or as $xto-infty$. Otherwise, limit theorems need to have a whole bunch of exceptions (e.g., it would no longer be true that if the limits of $f$ and $g$ 'exist', then so does the limit of $f+g$ and the latter is equal to their sum).
$endgroup$
– Arturo Magidin
May 29 '11 at 1:17
3
$begingroup$
I think that this depends from book to book and instructor to instructor. And the sum theorem, holds as long as it is not an indeterminate form, I am also pretty sure I saw somewhere that theorem stated for finite limits..
$endgroup$
– N. S.
May 29 '11 at 1:37
2
$begingroup$
@Arturo: One could also argue as follows: In general, it's possible to consider convergence in the extended reals with respect to the order topology; this yields the usual definition of the limits $pminfty$. However, in that case not only the limits but also the function values may be $pminfty$. Then addition wouldn't be defined, so by using addition the OP was implying that the underlying set was just the reals. Allowing infinite limits but only finite values would be like allowing real limits but only rational values, i.e. considering elements outside the set as potential limits.
$endgroup$
– joriki
May 29 '11 at 1:40
1
$begingroup$
BTW: In most books L'H appears in the form: If bla bla and $limfrac{f'(x)}{g'(x)}$ exists then bla bla... Does that cover the case when this limit is $pm infty$? Of course I realize that this is not really a good point since most Calculus books are very poorly written, but I am used to make a distinction between limit exists and limit is finite. This might come from the fact that I learned Analysis the old way, sequences first, and there convergence is already finite, so we always use the distinction "convergent" vs. "has a limit". For functions we replaced convergence with finite lim
$endgroup$
– N. S.
May 29 '11 at 1:45
2
$begingroup$
"The limit exists" is relative to the set you consider. You can consider the reals, the extended reals, or even the one-point compactification of the reals with just one point at infinity. The limit of the sequence $(-1)^nn$ exists in the one-point compactification but not in the extended reals. This shows that it's not just a matter of adding "enough" limit points, but also of choosing a topology. Again, there's an analogy to the rationals, which can be extended either to the reals or to the $p$-adic numbers.
$endgroup$
– joriki
May 29 '11 at 1:56
add a comment |
$begingroup$
Just to complete joriki's answer:
His answer is right only if $c$ is a number, which seems to be the case from the way you wrote the question.
Anyhow if you deal also with infinite limits, it is possible that $lim_{x to infty}b$ does not exist and $lim_{x to infty} a+b$ exists.
Just take $a=x-sin(x)$ and $b=sin(x)$.
answered May 29 '11 at 1:13
N. S.N. S.
104k7114209
$endgroup$
Just to complete joriki's answer:
His answer is right only if $c$ is a number, which seems to be the case from the way you wrote the question.
Anyhow if you deal also with infinite limits, it is possible that $lim_{x to infty}b$ does not exist and $lim_{x to infty} a+b$ exists.
Just take $a=x-sin(x)$ and $b=sin(x)$.
answered May 29 '11 at 1:13
N. S.N. S.
104k7114209
answered May 29 '11 at 1:13
N. S.N. S.
104k7114209
answered May 29 '11 at 1:13
N. S.N. S.
104k7114209
answered May 29 '11 at 1:13
N. S.N. S.
104k7114209
104k7114209
1
$begingroup$
when the limit is $infty$, it does not exist. Writing $limlimits_{xto a}f = infty$ is a way of saying "the limit does not exist" and specifying why it does not exist. Similarly with limits "equal" to $-infty$", and with limits as $xtoinfty$ or as $xto-infty$. Otherwise, limit theorems need to have a whole bunch of exceptions (e.g., it would no longer be true that if the limits of $f$ and $g$ 'exist', then so does the limit of $f+g$ and the latter is equal to their sum).
$endgroup$
– Arturo Magidin
May 29 '11 at 1:17
3
$begingroup$
I think that this depends from book to book and instructor to instructor. And the sum theorem, holds as long as it is not an indeterminate form, I am also pretty sure I saw somewhere that theorem stated for finite limits..
$endgroup$
– N. S.
May 29 '11 at 1:37
2
$begingroup$
@Arturo: One could also argue as follows: In general, it's possible to consider convergence in the extended reals with respect to the order topology; this yields the usual definition of the limits $pminfty$. However, in that case not only the limits but also the function values may be $pminfty$. Then addition wouldn't be defined, so by using addition the OP was implying that the underlying set was just the reals. Allowing infinite limits but only finite values would be like allowing real limits but only rational values, i.e. considering elements outside the set as potential limits.
$endgroup$
– joriki
May 29 '11 at 1:40
1
$begingroup$
BTW: In most books L'H appears in the form: If bla bla and $limfrac{f'(x)}{g'(x)}$ exists then bla bla... Does that cover the case when this limit is $pm infty$? Of course I realize that this is not really a good point since most Calculus books are very poorly written, but I am used to make a distinction between limit exists and limit is finite. This might come from the fact that I learned Analysis the old way, sequences first, and there convergence is already finite, so we always use the distinction "convergent" vs. "has a limit". For functions we replaced convergence with finite lim
$endgroup$
– N. S.
May 29 '11 at 1:45
2
$begingroup$
"The limit exists" is relative to the set you consider. You can consider the reals, the extended reals, or even the one-point compactification of the reals with just one point at infinity. The limit of the sequence $(-1)^nn$ exists in the one-point compactification but not in the extended reals. This shows that it's not just a matter of adding "enough" limit points, but also of choosing a topology. Again, there's an analogy to the rationals, which can be extended either to the reals or to the $p$-adic numbers.
$endgroup$
– joriki
May 29 '11 at 1:56
add a comment |
1
$begingroup$
when the limit is $infty$, it does not exist. Writing $limlimits_{xto a}f = infty$ is a way of saying "the limit does not exist" and specifying why it does not exist. Similarly with limits "equal" to $-infty$", and with limits as $xtoinfty$ or as $xto-infty$. Otherwise, limit theorems need to have a whole bunch of exceptions (e.g., it would no longer be true that if the limits of $f$ and $g$ 'exist', then so does the limit of $f+g$ and the latter is equal to their sum).
$endgroup$
– Arturo Magidin
May 29 '11 at 1:17
3
$begingroup$
I think that this depends from book to book and instructor to instructor. And the sum theorem, holds as long as it is not an indeterminate form, I am also pretty sure I saw somewhere that theorem stated for finite limits..
$endgroup$
– N. S.
May 29 '11 at 1:37
2
$begingroup$
@Arturo: One could also argue as follows: In general, it's possible to consider convergence in the extended reals with respect to the order topology; this yields the usual definition of the limits $pminfty$. However, in that case not only the limits but also the function values may be $pminfty$. Then addition wouldn't be defined, so by using addition the OP was implying that the underlying set was just the reals. Allowing infinite limits but only finite values would be like allowing real limits but only rational values, i.e. considering elements outside the set as potential limits.
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– joriki
May 29 '11 at 1:40
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BTW: In most books L'H appears in the form: If bla bla and $limfrac{f'(x)}{g'(x)}$ exists then bla bla... Does that cover the case when this limit is $pm infty$? Of course I realize that this is not really a good point since most Calculus books are very poorly written, but I am used to make a distinction between limit exists and limit is finite. This might come from the fact that I learned Analysis the old way, sequences first, and there convergence is already finite, so we always use the distinction "convergent" vs. "has a limit". For functions we replaced convergence with finite lim
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– N. S.
May 29 '11 at 1:45
2
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"The limit exists" is relative to the set you consider. You can consider the reals, the extended reals, or even the one-point compactification of the reals with just one point at infinity. The limit of the sequence $(-1)^nn$ exists in the one-point compactification but not in the extended reals. This shows that it's not just a matter of adding "enough" limit points, but also of choosing a topology. Again, there's an analogy to the rationals, which can be extended either to the reals or to the $p$-adic numbers.
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– joriki
May 29 '11 at 1:56
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when the limit is $infty$, it does not exist. Writing $limlimits_{xto a}f = infty$ is a way of saying "the limit does not exist" and specifying why it does not exist. Similarly with limits "equal" to $-infty$", and with limits as $xtoinfty$ or as $xto-infty$. Otherwise, limit theorems need to have a whole bunch of exceptions (e.g., it would no longer be true that if the limits of $f$ and $g$ 'exist', then so does the limit of $f+g$ and the latter is equal to their sum).
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– Arturo Magidin
May 29 '11 at 1:17
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when the limit is $infty$, it does not exist. Writing $limlimits_{xto a}f = infty$ is a way of saying "the limit does not exist" and specifying why it does not exist. Similarly with limits "equal" to $-infty$", and with limits as $xtoinfty$ or as $xto-infty$. Otherwise, limit theorems need to have a whole bunch of exceptions (e.g., it would no longer be true that if the limits of $f$ and $g$ 'exist', then so does the limit of $f+g$ and the latter is equal to their sum).
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– Arturo Magidin
May 29 '11 at 1:17
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I think that this depends from book to book and instructor to instructor. And the sum theorem, holds as long as it is not an indeterminate form, I am also pretty sure I saw somewhere that theorem stated for finite limits..
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– N. S.
May 29 '11 at 1:37
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I think that this depends from book to book and instructor to instructor. And the sum theorem, holds as long as it is not an indeterminate form, I am also pretty sure I saw somewhere that theorem stated for finite limits..
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– N. S.
May 29 '11 at 1:37
2
2
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@Arturo: One could also argue as follows: In general, it's possible to consider convergence in the extended reals with respect to the order topology; this yields the usual definition of the limits $pminfty$. However, in that case not only the limits but also the function values may be $pminfty$. Then addition wouldn't be defined, so by using addition the OP was implying that the underlying set was just the reals. Allowing infinite limits but only finite values would be like allowing real limits but only rational values, i.e. considering elements outside the set as potential limits.
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– joriki
May 29 '11 at 1:40
$begingroup$
@Arturo: One could also argue as follows: In general, it's possible to consider convergence in the extended reals with respect to the order topology; this yields the usual definition of the limits $pminfty$. However, in that case not only the limits but also the function values may be $pminfty$. Then addition wouldn't be defined, so by using addition the OP was implying that the underlying set was just the reals. Allowing infinite limits but only finite values would be like allowing real limits but only rational values, i.e. considering elements outside the set as potential limits.
$endgroup$
– joriki
May 29 '11 at 1:40
1
1
$begingroup$
BTW: In most books L'H appears in the form: If bla bla and $limfrac{f'(x)}{g'(x)}$ exists then bla bla... Does that cover the case when this limit is $pm infty$? Of course I realize that this is not really a good point since most Calculus books are very poorly written, but I am used to make a distinction between limit exists and limit is finite. This might come from the fact that I learned Analysis the old way, sequences first, and there convergence is already finite, so we always use the distinction "convergent" vs. "has a limit". For functions we replaced convergence with finite lim
$endgroup$
– N. S.
May 29 '11 at 1:45
$begingroup$
BTW: In most books L'H appears in the form: If bla bla and $limfrac{f'(x)}{g'(x)}$ exists then bla bla... Does that cover the case when this limit is $pm infty$? Of course I realize that this is not really a good point since most Calculus books are very poorly written, but I am used to make a distinction between limit exists and limit is finite. This might come from the fact that I learned Analysis the old way, sequences first, and there convergence is already finite, so we always use the distinction "convergent" vs. "has a limit". For functions we replaced convergence with finite lim
$endgroup$
– N. S.
May 29 '11 at 1:45
2
2
$begingroup$
"The limit exists" is relative to the set you consider. You can consider the reals, the extended reals, or even the one-point compactification of the reals with just one point at infinity. The limit of the sequence $(-1)^nn$ exists in the one-point compactification but not in the extended reals. This shows that it's not just a matter of adding "enough" limit points, but also of choosing a topology. Again, there's an analogy to the rationals, which can be extended either to the reals or to the $p$-adic numbers.
$endgroup$
– joriki
May 29 '11 at 1:56
$begingroup$
"The limit exists" is relative to the set you consider. You can consider the reals, the extended reals, or even the one-point compactification of the reals with just one point at infinity. The limit of the sequence $(-1)^nn$ exists in the one-point compactification but not in the extended reals. This shows that it's not just a matter of adding "enough" limit points, but also of choosing a topology. Again, there's an analogy to the rationals, which can be extended either to the reals or to the $p$-adic numbers.
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– joriki
May 29 '11 at 1:56
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