Why Can L'Hôpital's Rule Not be Applied to the Sum or Difference of Limits?
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1
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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 49 mins ago
snorepion
31
asked 5 hours ago
Danielle
1457
add a comment |
up vote
1
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 49 mins ago
snorepion
31
asked 5 hours ago
Danielle
1457
1
Limit operator is not a linear transformation, that is why.
– Bertrand Wittgenstein's Ghost
4 hours ago
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
4 hours ago
add a comment |
up vote
1
down vote
favorite
up vote
1
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 49 mins ago
snorepion
31
asked 5 hours ago
Danielle
1457
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 49 mins ago
snorepion
31
asked 5 hours ago
Danielle
1457
edited 49 mins ago
snorepion
31
asked 5 hours ago
Danielle
1457
edited 49 mins ago
snorepion
31
edited 49 mins ago
snorepion
31
edited 49 mins ago
snorepion
31
31
asked 5 hours ago
Danielle
1457
asked 5 hours ago
Danielle
1457
asked 5 hours ago
Danielle
1457
1457
1
Limit operator is not a linear transformation, that is why.
– Bertrand Wittgenstein's Ghost
4 hours ago
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
4 hours ago
add a comment |
1
Limit operator is not a linear transformation, that is why.
– Bertrand Wittgenstein's Ghost
4 hours ago
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
4 hours ago
1
1
Limit operator is not a linear transformation, that is why.
– Bertrand Wittgenstein's Ghost
4 hours ago
Limit operator is not a linear transformation, that is why.
– Bertrand Wittgenstein's Ghost
4 hours ago
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
4 hours ago
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
4 hours ago
add a comment |
2 Answers
2
active
oldest
votes
up vote
8
down vote
accepted
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 8 mins ago
Tanner Swett
3,9461638
answered 4 hours ago
JDMan4444
2143
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.
add a comment |
up vote
1
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 4 hours ago
gimusi
88k74393
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
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 8 mins ago
Tanner Swett
3,9461638
answered 4 hours ago
JDMan4444
2143
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.
add a comment |
up vote
8
down vote
accepted
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 8 mins ago
Tanner Swett
3,9461638
answered 4 hours ago
JDMan4444
2143
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.
add a comment |
up vote
8
down vote
accepted
up vote
8
down vote
accepted
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 8 mins ago
Tanner Swett
3,9461638
answered 4 hours ago
JDMan4444
2143
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 8 mins ago
Tanner Swett
3,9461638
answered 4 hours ago
JDMan4444
2143
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 8 mins ago
Tanner Swett
3,9461638
edited 8 mins ago
Tanner Swett
3,9461638
edited 8 mins ago
Tanner Swett
3,9461638
3,9461638
answered 4 hours ago
JDMan4444
2143
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 4 hours ago
JDMan4444
2143
answered 4 hours ago
JDMan4444
2143
2143
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.
add a comment |
add a comment |
up vote
1
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 4 hours ago
gimusi
88k74393
add a comment |
up vote
1
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 4 hours ago
gimusi
88k74393
add a comment |
up vote
1
down vote
up vote
1
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 4 hours ago
gimusi
88k74393
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 4 hours ago
gimusi
88k74393
answered 4 hours ago
gimusi
88k74393
answered 4 hours ago
gimusi
88k74393
answered 4 hours ago
gimusi
88k74393
88k74393
add a comment |
add a comment |
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Limit operator is not a linear transformation, that is why.
– Bertrand Wittgenstein's Ghost
4 hours ago
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
4 hours ago