Why Can L'Hôpital's Rule Not be Applied to the Sum or Difference of Limits?
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5
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Consider $$lim_{xtoinfty}frac{f(x)}{g(x)} + lim_{xtoinfty}frac{h(x)}{i(x)}$$
I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits as
$$lim_{xtoinfty}frac{f(x)}{g(x)} +frac{h(x)}{i(x)}$$
Why is this incorrect?
calculus limits
edited Nov 29 at 4:27
snorepion
31
asked Nov 29 at 0:16
Danielle
1819
add a comment |
up vote
5
down vote
favorite
Consider $$lim_{xtoinfty}frac{f(x)}{g(x)} + lim_{xtoinfty}frac{h(x)}{i(x)}$$
I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits as
$$lim_{xtoinfty}frac{f(x)}{g(x)} +frac{h(x)}{i(x)}$$
Why is this incorrect?
calculus limits
edited Nov 29 at 4:27
snorepion
31
asked Nov 29 at 0:16
Danielle
1819
1
Limit operator is not a linear transformation, that is why.
– Bertrand Wittgenstein's Ghost
Nov 29 at 0:20
1
I think it only goes wrong when you will end up with something of the form $+infty - infty$. So, if you avoid these cases you can apply it if I am correct.
– Stan Tendijck
Nov 29 at 0:54
add a comment |
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Consider $$lim_{xtoinfty}frac{f(x)}{g(x)} + lim_{xtoinfty}frac{h(x)}{i(x)}$$
I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits as
$$lim_{xtoinfty}frac{f(x)}{g(x)} +frac{h(x)}{i(x)}$$
Why is this incorrect?
calculus limits
edited Nov 29 at 4:27
snorepion
31
asked Nov 29 at 0:16
Danielle
1819
Consider $$lim_{xtoinfty}frac{f(x)}{g(x)} + lim_{xtoinfty}frac{h(x)}{i(x)}$$
I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits as
$$lim_{xtoinfty}frac{f(x)}{g(x)} +frac{h(x)}{i(x)}$$
Why is this incorrect?
calculus limits
calculus limits
edited Nov 29 at 4:27
snorepion
31
asked Nov 29 at 0:16
Danielle
1819
edited Nov 29 at 4:27
snorepion
31
asked Nov 29 at 0:16
Danielle
1819
edited Nov 29 at 4:27
snorepion
31
edited Nov 29 at 4:27
snorepion
31
edited Nov 29 at 4:27
snorepion
31
31
asked Nov 29 at 0:16
Danielle
1819
asked Nov 29 at 0:16
Danielle
1819
asked Nov 29 at 0:16
Danielle
1819
1819
1
Limit operator is not a linear transformation, that is why.
– Bertrand Wittgenstein's Ghost
Nov 29 at 0:20
1
I think it only goes wrong when you will end up with something of the form $+infty - infty$. So, if you avoid these cases you can apply it if I am correct.
– Stan Tendijck
Nov 29 at 0:54
add a comment |
1
Limit operator is not a linear transformation, that is why.
– Bertrand Wittgenstein's Ghost
Nov 29 at 0:20
1
I think it only goes wrong when you will end up with something of the form $+infty - infty$. So, if you avoid these cases you can apply it if I am correct.
– Stan Tendijck
Nov 29 at 0:54
1
1
Limit operator is not a linear transformation, that is why.
– Bertrand Wittgenstein's Ghost
Nov 29 at 0:20
Limit operator is not a linear transformation, that is why.
– Bertrand Wittgenstein's Ghost
Nov 29 at 0:20
1
1
I think it only goes wrong when you will end up with something of the form $+infty - infty$. So, if you avoid these cases you can apply it if I am correct.
– Stan Tendijck
Nov 29 at 0:54
I think it only goes wrong when you will end up with something of the form $+infty - infty$. So, if you avoid these cases you can apply it if I am correct.
– Stan Tendijck
Nov 29 at 0:54
add a comment |
4 Answers
4
active
oldest
votes
up vote
13
down vote
accepted
I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits as ...
This has nothing to do with the L'Hôpital's rule itself.
The rule that you cannot use is:
$limlimits_{xto...}f(x)+limlimits_{xto...}g(x)=limlimits_{xto...}(f(x)+g(x))$
And you can see from JDMan4444's answer that there are situations where this rule does not work.
However, if you are sure that $limlimits_{xto...}f(x)$ and $limlimits_{xto...}g(x)$ exist (and are not $pminfty$), you can apply that rule. (And of course you can apply L'Hôpital's rule to the sum in this case.)
This is important because in some cases it is possible to prove that both limits exist but calculating the limits directly is very difficult or even impossible. In such cases calculating the limit of the sum may be easier.
answered Nov 29 at 7:09
Martin Rosenau
1,077128
The OP is related to THAT and the main doubt is on the application of HR.
– gimusi
Nov 29 at 7:42
add a comment |
up vote
10
down vote
Consider the following example:
$$
lim_{xrightarrowinfty}frac{x^2}{x}+lim_{xrightarrowinfty}frac{-x^2}{x} = infty - infty quad text{(which is undefined)}
$$
$$
lim_{xrightarrowinfty}frac{x^2}{x}+frac{-x^2}{x} = 0
$$
edited Nov 29 at 5:08
Tanner Swett
3,9561638
answered Nov 29 at 0:18
JDMan4444
23514
New contributor
JDMan4444 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ypu example is fine and correct but note that the main doubt was in the application of l'HR and is related to this other OP.
– gimusi
Nov 29 at 7:44
1
Your second line "$lim_{xrightarrowinfty}frac{x^2}{x}+frac{-x^2}{x} = 0$" is wrong. Addition has lower precedence than the limit operator, by convention.
– user21820
Nov 29 at 13:51
1
@user21820 indeed; round brackets are needed around both fractions together, otherwise it's +∞ because the limit applies only to the left fraction, the right remains -x^2/x. But that was obvious from the context. More importantly, this applies to the original question as much as here, so comment should be there.
– user3445853
Nov 29 at 15:17
@user21820 It could be convention but since the expression $frac{x^2}{x}+frac{-x^2}{x}$ is identically equal to zero for all $xneq 0$ I think that it is a noce assumption read that as $$lim_{xrightarrowinfty} left(frac{x^2}{x}+frac{-x^2}{x}right) =lim_{xrightarrowinfty} 0= 0$$
– gimusi
Nov 29 at 15:29
@user3445853: No. Teachers making errors are one big reason why students don't learn. I've nothing more to say here.
– user21820
Nov 29 at 16:47
|
show 7 more comments
up vote
3
down vote
The following identity
$$lim_{xto infty}left(frac{f(x)}{g(x)} +frac{h(x)}{i(x)}right)=lim_{xto infty}frac{f(x)}{g(x)} + lim_{xto infty}frac{h(x)}{i(x)}$$
doesn't hold in general and to solve the LHS limit by l'Hopital, if necessary, we need to put it in the form
$$lim_{xto infty}left(frac{f(x)}{g(x)} +frac{h(x)}{i(x)}right)=lim_{xto infty}frac{f(x)i(x)+h(x)g(x)}{g(x)i(x)} $$
the reason is that the case you are referring to is not among the cases considered by l'Hopital theorem.
answered Nov 29 at 0:23
gimusi
89k74495
add a comment |
up vote
1
down vote
It seems to me you got some bad, or incomplete, advice. It's not incorrect if both of those limits exist and are finite. This doesn't have much to do with L'Hopital. The $infty - infty$ or $-infty + infty$ cases are problematic whether you're contemplating L'Hopital or not.
answered Nov 29 at 7:28
zhw.
70.6k43075
The advice was given HERE for that specific example and it was a completely different case to the one you are referring to.
– gimusi
Nov 29 at 7:41
@gimusi I responded to the OP's question here. She wrote "I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits" and then asked why this is incorrect. I answered that. Why are you sending me to some other question?
– zhw.
Nov 29 at 19:19
My aim was solely give you a better context for the question posed.
– gimusi
Nov 29 at 19:23
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
13
down vote
accepted
I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits as ...
This has nothing to do with the L'Hôpital's rule itself.
The rule that you cannot use is:
$limlimits_{xto...}f(x)+limlimits_{xto...}g(x)=limlimits_{xto...}(f(x)+g(x))$
And you can see from JDMan4444's answer that there are situations where this rule does not work.
However, if you are sure that $limlimits_{xto...}f(x)$ and $limlimits_{xto...}g(x)$ exist (and are not $pminfty$), you can apply that rule. (And of course you can apply L'Hôpital's rule to the sum in this case.)
This is important because in some cases it is possible to prove that both limits exist but calculating the limits directly is very difficult or even impossible. In such cases calculating the limit of the sum may be easier.
answered Nov 29 at 7:09
Martin Rosenau
1,077128
The OP is related to THAT and the main doubt is on the application of HR.
– gimusi
Nov 29 at 7:42
add a comment |
up vote
13
down vote
accepted
I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits as ...
This has nothing to do with the L'Hôpital's rule itself.
The rule that you cannot use is:
$limlimits_{xto...}f(x)+limlimits_{xto...}g(x)=limlimits_{xto...}(f(x)+g(x))$
And you can see from JDMan4444's answer that there are situations where this rule does not work.
However, if you are sure that $limlimits_{xto...}f(x)$ and $limlimits_{xto...}g(x)$ exist (and are not $pminfty$), you can apply that rule. (And of course you can apply L'Hôpital's rule to the sum in this case.)
This is important because in some cases it is possible to prove that both limits exist but calculating the limits directly is very difficult or even impossible. In such cases calculating the limit of the sum may be easier.
answered Nov 29 at 7:09
Martin Rosenau
1,077128
The OP is related to THAT and the main doubt is on the application of HR.
– gimusi
Nov 29 at 7:42
add a comment |
up vote
13
down vote
accepted
up vote
13
down vote
accepted
I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits as ...
This has nothing to do with the L'Hôpital's rule itself.
The rule that you cannot use is:
$limlimits_{xto...}f(x)+limlimits_{xto...}g(x)=limlimits_{xto...}(f(x)+g(x))$
And you can see from JDMan4444's answer that there are situations where this rule does not work.
However, if you are sure that $limlimits_{xto...}f(x)$ and $limlimits_{xto...}g(x)$ exist (and are not $pminfty$), you can apply that rule. (And of course you can apply L'Hôpital's rule to the sum in this case.)
This is important because in some cases it is possible to prove that both limits exist but calculating the limits directly is very difficult or even impossible. In such cases calculating the limit of the sum may be easier.
answered Nov 29 at 7:09
Martin Rosenau
1,077128
I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits as ...
This has nothing to do with the L'Hôpital's rule itself.
The rule that you cannot use is:
$limlimits_{xto...}f(x)+limlimits_{xto...}g(x)=limlimits_{xto...}(f(x)+g(x))$
And you can see from JDMan4444's answer that there are situations where this rule does not work.
However, if you are sure that $limlimits_{xto...}f(x)$ and $limlimits_{xto...}g(x)$ exist (and are not $pminfty$), you can apply that rule. (And of course you can apply L'Hôpital's rule to the sum in this case.)
This is important because in some cases it is possible to prove that both limits exist but calculating the limits directly is very difficult or even impossible. In such cases calculating the limit of the sum may be easier.
answered Nov 29 at 7:09
Martin Rosenau
1,077128
answered Nov 29 at 7:09
Martin Rosenau
1,077128
answered Nov 29 at 7:09
Martin Rosenau
1,077128
answered Nov 29 at 7:09
Martin Rosenau
1,077128
1,077128
The OP is related to THAT and the main doubt is on the application of HR.
– gimusi
Nov 29 at 7:42
add a comment |
The OP is related to THAT and the main doubt is on the application of HR.
– gimusi
Nov 29 at 7:42
The OP is related to THAT and the main doubt is on the application of HR.
– gimusi
Nov 29 at 7:42
The OP is related to THAT and the main doubt is on the application of HR.
– gimusi
Nov 29 at 7:42
add a comment |
up vote
10
down vote
Consider the following example:
$$
lim_{xrightarrowinfty}frac{x^2}{x}+lim_{xrightarrowinfty}frac{-x^2}{x} = infty - infty quad text{(which is undefined)}
$$
$$
lim_{xrightarrowinfty}frac{x^2}{x}+frac{-x^2}{x} = 0
$$
edited Nov 29 at 5:08
Tanner Swett
3,9561638
answered Nov 29 at 0:18
JDMan4444
23514
New contributor
JDMan4444 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ypu example is fine and correct but note that the main doubt was in the application of l'HR and is related to this other OP.
– gimusi
Nov 29 at 7:44
1
Your second line "$lim_{xrightarrowinfty}frac{x^2}{x}+frac{-x^2}{x} = 0$" is wrong. Addition has lower precedence than the limit operator, by convention.
– user21820
Nov 29 at 13:51
1
@user21820 indeed; round brackets are needed around both fractions together, otherwise it's +∞ because the limit applies only to the left fraction, the right remains -x^2/x. But that was obvious from the context. More importantly, this applies to the original question as much as here, so comment should be there.
– user3445853
Nov 29 at 15:17
@user21820 It could be convention but since the expression $frac{x^2}{x}+frac{-x^2}{x}$ is identically equal to zero for all $xneq 0$ I think that it is a noce assumption read that as $$lim_{xrightarrowinfty} left(frac{x^2}{x}+frac{-x^2}{x}right) =lim_{xrightarrowinfty} 0= 0$$
– gimusi
Nov 29 at 15:29
@user3445853: No. Teachers making errors are one big reason why students don't learn. I've nothing more to say here.
– user21820
Nov 29 at 16:47
|
show 7 more comments
up vote
10
down vote
Consider the following example:
$$
lim_{xrightarrowinfty}frac{x^2}{x}+lim_{xrightarrowinfty}frac{-x^2}{x} = infty - infty quad text{(which is undefined)}
$$
$$
lim_{xrightarrowinfty}frac{x^2}{x}+frac{-x^2}{x} = 0
$$
edited Nov 29 at 5:08
Tanner Swett
3,9561638
answered Nov 29 at 0:18
JDMan4444
23514
New contributor
JDMan4444 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ypu example is fine and correct but note that the main doubt was in the application of l'HR and is related to this other OP.
– gimusi
Nov 29 at 7:44
1
Your second line "$lim_{xrightarrowinfty}frac{x^2}{x}+frac{-x^2}{x} = 0$" is wrong. Addition has lower precedence than the limit operator, by convention.
– user21820
Nov 29 at 13:51
1
@user21820 indeed; round brackets are needed around both fractions together, otherwise it's +∞ because the limit applies only to the left fraction, the right remains -x^2/x. But that was obvious from the context. More importantly, this applies to the original question as much as here, so comment should be there.
– user3445853
Nov 29 at 15:17
@user21820 It could be convention but since the expression $frac{x^2}{x}+frac{-x^2}{x}$ is identically equal to zero for all $xneq 0$ I think that it is a noce assumption read that as $$lim_{xrightarrowinfty} left(frac{x^2}{x}+frac{-x^2}{x}right) =lim_{xrightarrowinfty} 0= 0$$
– gimusi
Nov 29 at 15:29
@user3445853: No. Teachers making errors are one big reason why students don't learn. I've nothing more to say here.
– user21820
Nov 29 at 16:47
|
show 7 more comments
up vote
10
down vote
up vote
10
down vote
Consider the following example:
$$
lim_{xrightarrowinfty}frac{x^2}{x}+lim_{xrightarrowinfty}frac{-x^2}{x} = infty - infty quad text{(which is undefined)}
$$
$$
lim_{xrightarrowinfty}frac{x^2}{x}+frac{-x^2}{x} = 0
$$
edited Nov 29 at 5:08
Tanner Swett
3,9561638
answered Nov 29 at 0:18
JDMan4444
23514
New contributor
JDMan4444 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Consider the following example:
$$
lim_{xrightarrowinfty}frac{x^2}{x}+lim_{xrightarrowinfty}frac{-x^2}{x} = infty - infty quad text{(which is undefined)}
$$
$$
lim_{xrightarrowinfty}frac{x^2}{x}+frac{-x^2}{x} = 0
$$
edited Nov 29 at 5:08
Tanner Swett
3,9561638
answered Nov 29 at 0:18
JDMan4444
23514
New contributor
JDMan4444 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 29 at 5:08
Tanner Swett
3,9561638
edited Nov 29 at 5:08
Tanner Swett
3,9561638
edited Nov 29 at 5:08
Tanner Swett
3,9561638
3,9561638
answered Nov 29 at 0:18
JDMan4444
23514
New contributor
JDMan4444 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Nov 29 at 0:18
JDMan4444
23514
answered Nov 29 at 0:18
JDMan4444
23514
23514
New contributor
JDMan4444 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
JDMan4444 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
JDMan4444 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ypu example is fine and correct but note that the main doubt was in the application of l'HR and is related to this other OP.
– gimusi
Nov 29 at 7:44
1
Your second line "$lim_{xrightarrowinfty}frac{x^2}{x}+frac{-x^2}{x} = 0$" is wrong. Addition has lower precedence than the limit operator, by convention.
– user21820
Nov 29 at 13:51
1
@user21820 indeed; round brackets are needed around both fractions together, otherwise it's +∞ because the limit applies only to the left fraction, the right remains -x^2/x. But that was obvious from the context. More importantly, this applies to the original question as much as here, so comment should be there.
– user3445853
Nov 29 at 15:17
@user21820 It could be convention but since the expression $frac{x^2}{x}+frac{-x^2}{x}$ is identically equal to zero for all $xneq 0$ I think that it is a noce assumption read that as $$lim_{xrightarrowinfty} left(frac{x^2}{x}+frac{-x^2}{x}right) =lim_{xrightarrowinfty} 0= 0$$
– gimusi
Nov 29 at 15:29
@user3445853: No. Teachers making errors are one big reason why students don't learn. I've nothing more to say here.
– user21820
Nov 29 at 16:47
|
show 7 more comments
Ypu example is fine and correct but note that the main doubt was in the application of l'HR and is related to this other OP.
– gimusi
Nov 29 at 7:44
1
Your second line "$lim_{xrightarrowinfty}frac{x^2}{x}+frac{-x^2}{x} = 0$" is wrong. Addition has lower precedence than the limit operator, by convention.
– user21820
Nov 29 at 13:51
1
@user21820 indeed; round brackets are needed around both fractions together, otherwise it's +∞ because the limit applies only to the left fraction, the right remains -x^2/x. But that was obvious from the context. More importantly, this applies to the original question as much as here, so comment should be there.
– user3445853
Nov 29 at 15:17
@user21820 It could be convention but since the expression $frac{x^2}{x}+frac{-x^2}{x}$ is identically equal to zero for all $xneq 0$ I think that it is a noce assumption read that as $$lim_{xrightarrowinfty} left(frac{x^2}{x}+frac{-x^2}{x}right) =lim_{xrightarrowinfty} 0= 0$$
– gimusi
Nov 29 at 15:29
@user3445853: No. Teachers making errors are one big reason why students don't learn. I've nothing more to say here.
– user21820
Nov 29 at 16:47
Ypu example is fine and correct but note that the main doubt was in the application of l'HR and is related to this other OP.
– gimusi
Nov 29 at 7:44
Ypu example is fine and correct but note that the main doubt was in the application of l'HR and is related to this other OP.
– gimusi
Nov 29 at 7:44
1
1
Your second line "$lim_{xrightarrowinfty}frac{x^2}{x}+frac{-x^2}{x} = 0$" is wrong. Addition has lower precedence than the limit operator, by convention.
– user21820
Nov 29 at 13:51
Your second line "$lim_{xrightarrowinfty}frac{x^2}{x}+frac{-x^2}{x} = 0$" is wrong. Addition has lower precedence than the limit operator, by convention.
– user21820
Nov 29 at 13:51
1
1
@user21820 indeed; round brackets are needed around both fractions together, otherwise it's +∞ because the limit applies only to the left fraction, the right remains -x^2/x. But that was obvious from the context. More importantly, this applies to the original question as much as here, so comment should be there.
– user3445853
Nov 29 at 15:17
@user21820 indeed; round brackets are needed around both fractions together, otherwise it's +∞ because the limit applies only to the left fraction, the right remains -x^2/x. But that was obvious from the context. More importantly, this applies to the original question as much as here, so comment should be there.
– user3445853
Nov 29 at 15:17
@user21820 It could be convention but since the expression $frac{x^2}{x}+frac{-x^2}{x}$ is identically equal to zero for all $xneq 0$ I think that it is a noce assumption read that as $$lim_{xrightarrowinfty} left(frac{x^2}{x}+frac{-x^2}{x}right) =lim_{xrightarrowinfty} 0= 0$$
– gimusi
Nov 29 at 15:29
@user21820 It could be convention but since the expression $frac{x^2}{x}+frac{-x^2}{x}$ is identically equal to zero for all $xneq 0$ I think that it is a noce assumption read that as $$lim_{xrightarrowinfty} left(frac{x^2}{x}+frac{-x^2}{x}right) =lim_{xrightarrowinfty} 0= 0$$
– gimusi
Nov 29 at 15:29
@user3445853: No. Teachers making errors are one big reason why students don't learn. I've nothing more to say here.
– user21820
Nov 29 at 16:47
@user3445853: No. Teachers making errors are one big reason why students don't learn. I've nothing more to say here.
– user21820
Nov 29 at 16:47
|
show 7 more comments
up vote
3
down vote
The following identity
$$lim_{xto infty}left(frac{f(x)}{g(x)} +frac{h(x)}{i(x)}right)=lim_{xto infty}frac{f(x)}{g(x)} + lim_{xto infty}frac{h(x)}{i(x)}$$
doesn't hold in general and to solve the LHS limit by l'Hopital, if necessary, we need to put it in the form
$$lim_{xto infty}left(frac{f(x)}{g(x)} +frac{h(x)}{i(x)}right)=lim_{xto infty}frac{f(x)i(x)+h(x)g(x)}{g(x)i(x)} $$
the reason is that the case you are referring to is not among the cases considered by l'Hopital theorem.
answered Nov 29 at 0:23
gimusi
89k74495
add a comment |
up vote
3
down vote
The following identity
$$lim_{xto infty}left(frac{f(x)}{g(x)} +frac{h(x)}{i(x)}right)=lim_{xto infty}frac{f(x)}{g(x)} + lim_{xto infty}frac{h(x)}{i(x)}$$
doesn't hold in general and to solve the LHS limit by l'Hopital, if necessary, we need to put it in the form
$$lim_{xto infty}left(frac{f(x)}{g(x)} +frac{h(x)}{i(x)}right)=lim_{xto infty}frac{f(x)i(x)+h(x)g(x)}{g(x)i(x)} $$
the reason is that the case you are referring to is not among the cases considered by l'Hopital theorem.
answered Nov 29 at 0:23
gimusi
89k74495
add a comment |
up vote
3
down vote
up vote
3
down vote
The following identity
$$lim_{xto infty}left(frac{f(x)}{g(x)} +frac{h(x)}{i(x)}right)=lim_{xto infty}frac{f(x)}{g(x)} + lim_{xto infty}frac{h(x)}{i(x)}$$
doesn't hold in general and to solve the LHS limit by l'Hopital, if necessary, we need to put it in the form
$$lim_{xto infty}left(frac{f(x)}{g(x)} +frac{h(x)}{i(x)}right)=lim_{xto infty}frac{f(x)i(x)+h(x)g(x)}{g(x)i(x)} $$
the reason is that the case you are referring to is not among the cases considered by l'Hopital theorem.
answered Nov 29 at 0:23
gimusi
89k74495
The following identity
$$lim_{xto infty}left(frac{f(x)}{g(x)} +frac{h(x)}{i(x)}right)=lim_{xto infty}frac{f(x)}{g(x)} + lim_{xto infty}frac{h(x)}{i(x)}$$
doesn't hold in general and to solve the LHS limit by l'Hopital, if necessary, we need to put it in the form
$$lim_{xto infty}left(frac{f(x)}{g(x)} +frac{h(x)}{i(x)}right)=lim_{xto infty}frac{f(x)i(x)+h(x)g(x)}{g(x)i(x)} $$
the reason is that the case you are referring to is not among the cases considered by l'Hopital theorem.
answered Nov 29 at 0:23
gimusi
89k74495
answered Nov 29 at 0:23
gimusi
89k74495
answered Nov 29 at 0:23
gimusi
89k74495
answered Nov 29 at 0:23
gimusi
89k74495
89k74495
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add a comment |
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It seems to me you got some bad, or incomplete, advice. It's not incorrect if both of those limits exist and are finite. This doesn't have much to do with L'Hopital. The $infty - infty$ or $-infty + infty$ cases are problematic whether you're contemplating L'Hopital or not.
answered Nov 29 at 7:28
zhw.
70.6k43075
The advice was given HERE for that specific example and it was a completely different case to the one you are referring to.
– gimusi
Nov 29 at 7:41
@gimusi I responded to the OP's question here. She wrote "I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits" and then asked why this is incorrect. I answered that. Why are you sending me to some other question?
– zhw.
Nov 29 at 19:19
My aim was solely give you a better context for the question posed.
– gimusi
Nov 29 at 19:23
add a comment |
up vote
1
down vote
It seems to me you got some bad, or incomplete, advice. It's not incorrect if both of those limits exist and are finite. This doesn't have much to do with L'Hopital. The $infty - infty$ or $-infty + infty$ cases are problematic whether you're contemplating L'Hopital or not.
answered Nov 29 at 7:28
zhw.
70.6k43075
The advice was given HERE for that specific example and it was a completely different case to the one you are referring to.
– gimusi
Nov 29 at 7:41
@gimusi I responded to the OP's question here. She wrote "I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits" and then asked why this is incorrect. I answered that. Why are you sending me to some other question?
– zhw.
Nov 29 at 19:19
My aim was solely give you a better context for the question posed.
– gimusi
Nov 29 at 19:23
add a comment |
up vote
1
down vote
up vote
1
down vote
It seems to me you got some bad, or incomplete, advice. It's not incorrect if both of those limits exist and are finite. This doesn't have much to do with L'Hopital. The $infty - infty$ or $-infty + infty$ cases are problematic whether you're contemplating L'Hopital or not.
answered Nov 29 at 7:28
zhw.
70.6k43075
It seems to me you got some bad, or incomplete, advice. It's not incorrect if both of those limits exist and are finite. This doesn't have much to do with L'Hopital. The $infty - infty$ or $-infty + infty$ cases are problematic whether you're contemplating L'Hopital or not.
answered Nov 29 at 7:28
zhw.
70.6k43075
answered Nov 29 at 7:28
zhw.
70.6k43075
answered Nov 29 at 7:28
zhw.
70.6k43075
answered Nov 29 at 7:28
zhw.
70.6k43075
70.6k43075
The advice was given HERE for that specific example and it was a completely different case to the one you are referring to.
– gimusi
Nov 29 at 7:41
@gimusi I responded to the OP's question here. She wrote "I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits" and then asked why this is incorrect. I answered that. Why are you sending me to some other question?
– zhw.
Nov 29 at 19:19
My aim was solely give you a better context for the question posed.
– gimusi
Nov 29 at 19:23
add a comment |
The advice was given HERE for that specific example and it was a completely different case to the one you are referring to.
– gimusi
Nov 29 at 7:41
@gimusi I responded to the OP's question here. She wrote "I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits" and then asked why this is incorrect. I answered that. Why are you sending me to some other question?
– zhw.
Nov 29 at 19:19
My aim was solely give you a better context for the question posed.
– gimusi
Nov 29 at 19:23
The advice was given HERE for that specific example and it was a completely different case to the one you are referring to.
– gimusi
Nov 29 at 7:41
The advice was given HERE for that specific example and it was a completely different case to the one you are referring to.
– gimusi
Nov 29 at 7:41
@gimusi I responded to the OP's question here. She wrote "I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits" and then asked why this is incorrect. I answered that. Why are you sending me to some other question?
– zhw.
Nov 29 at 19:19
@gimusi I responded to the OP's question here. She wrote "I was told that I cannot apply L'Hôpital's rule to each individual limit and then join the limits" and then asked why this is incorrect. I answered that. Why are you sending me to some other question?
– zhw.
Nov 29 at 19:19
My aim was solely give you a better context for the question posed.
– gimusi
Nov 29 at 19:23
My aim was solely give you a better context for the question posed.
– gimusi
Nov 29 at 19:23
add a comment |
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1
Limit operator is not a linear transformation, that is why.
– Bertrand Wittgenstein's Ghost
Nov 29 at 0:20
1
I think it only goes wrong when you will end up with something of the form $+infty - infty$. So, if you avoid these cases you can apply it if I am correct.
– Stan Tendijck
Nov 29 at 0:54