Why Can L'Hôpital's Rule Not be Applied to the Sum or Difference of Limits?
up vote
2
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 1 hour ago
snorepion
31
asked 5 hours ago
Danielle
1507
add a comment |
up vote
2
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 1 hour ago
snorepion
31
asked 5 hours ago
Danielle
1507
1
Limit operator is not a linear transformation, that is why.
– Bertrand Wittgenstein's Ghost
5 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
2
down vote
favorite
up vote
2
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 1 hour ago
snorepion
31
asked 5 hours ago
Danielle
1507
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 1 hour ago
snorepion
31
asked 5 hours ago
Danielle
1507
edited 1 hour ago
snorepion
31
asked 5 hours ago
Danielle
1507
edited 1 hour ago
snorepion
31
edited 1 hour ago
snorepion
31
edited 1 hour ago
snorepion
31
31
asked 5 hours ago
Danielle
1507
asked 5 hours ago
Danielle
1507
asked 5 hours ago
Danielle
1507
1507
1
Limit operator is not a linear transformation, that is why.
– Bertrand Wittgenstein's Ghost
5 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
5 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
5 hours ago
Limit operator is not a linear transformation, that is why.
– Bertrand Wittgenstein's Ghost
5 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 28 mins ago
Tanner Swett
3,9461638
answered 5 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
2
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 5 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 28 mins ago
Tanner Swett
3,9461638
answered 5 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 28 mins ago
Tanner Swett
3,9461638
answered 5 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 28 mins ago
Tanner Swett
3,9461638
answered 5 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 28 mins ago
Tanner Swett
3,9461638
answered 5 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 28 mins ago
Tanner Swett
3,9461638
edited 28 mins ago
Tanner Swett
3,9461638
edited 28 mins ago
Tanner Swett
3,9461638
3,9461638
answered 5 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 5 hours ago
JDMan4444
2143
answered 5 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
2
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 5 hours ago
gimusi
88k74393
add a comment |
up vote
2
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 5 hours ago
gimusi
88k74393
add a comment |
up vote
2
down vote
up vote
2
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 5 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 5 hours ago
gimusi
88k74393
answered 5 hours ago
gimusi
88k74393
answered 5 hours ago
gimusi
88k74393
answered 5 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
5 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