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L'Hospital for $ infty- infty$
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I want to find out the value of this expression.
$lim_xtoinfty(frac12ln^2(x)+xlnpi)-lim_xto0(frac12ln^2(x)+xlnpi)$
By plugging in the values, I get $infty-infty$.
According to wolframalpha, the value is undefined.
By just seeing that it's $infty-infty$, can we then conclude that it's undefined? Won't L'hospitale or other tricks work here? How to really know whether it's undefined?
EDIT:
Thanks for all the answers.
Actually I was trying to find whether the $int_0^inftyfracln(xpi^x)dxx$ converges or not. And after integrating I got the above-mentioned results. So I was lost there.
EDIT2: Sorry for messing this question up. I asked another question concerning the integral here:Improper Integration $int_0^inftyfracln(xpi^x),dxx$ converges or not.
Thanks alot.
calculus limits
edited Sep 9 at 10:06
quid♦
36.4k85091
asked Sep 9 at 8:03
JoisBack
1355
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add a comment |Â
up vote
2
down vote
favorite
I want to find out the value of this expression.
$lim_xtoinfty(frac12ln^2(x)+xlnpi)-lim_xto0(frac12ln^2(x)+xlnpi)$
By plugging in the values, I get $infty-infty$.
According to wolframalpha, the value is undefined.
By just seeing that it's $infty-infty$, can we then conclude that it's undefined? Won't L'hospitale or other tricks work here? How to really know whether it's undefined?
EDIT:
Thanks for all the answers.
Actually I was trying to find whether the $int_0^inftyfracln(xpi^x)dxx$ converges or not. And after integrating I got the above-mentioned results. So I was lost there.
EDIT2: Sorry for messing this question up. I asked another question concerning the integral here:Improper Integration $int_0^inftyfracln(xpi^x),dxx$ converges or not.
Thanks alot.
calculus limits
edited Sep 9 at 10:06
quid♦
36.4k85091
asked Sep 9 at 8:03
JoisBack
1355
New contributor
JoisBack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Note that by getting $∞−∞$ from a particular method, the only thing you can conclude is that you used a wrong method. It doesn't tell anything about the function.
– IllidanS4
Sep 9 at 9:54
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I want to find out the value of this expression.
$lim_xtoinfty(frac12ln^2(x)+xlnpi)-lim_xto0(frac12ln^2(x)+xlnpi)$
By plugging in the values, I get $infty-infty$.
According to wolframalpha, the value is undefined.
By just seeing that it's $infty-infty$, can we then conclude that it's undefined? Won't L'hospitale or other tricks work here? How to really know whether it's undefined?
EDIT:
Thanks for all the answers.
Actually I was trying to find whether the $int_0^inftyfracln(xpi^x)dxx$ converges or not. And after integrating I got the above-mentioned results. So I was lost there.
EDIT2: Sorry for messing this question up. I asked another question concerning the integral here:Improper Integration $int_0^inftyfracln(xpi^x),dxx$ converges or not.
Thanks alot.
calculus limits
edited Sep 9 at 10:06
quid♦
36.4k85091
asked Sep 9 at 8:03
JoisBack
1355
New contributor
JoisBack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I want to find out the value of this expression.
$lim_xtoinfty(frac12ln^2(x)+xlnpi)-lim_xto0(frac12ln^2(x)+xlnpi)$
By plugging in the values, I get $infty-infty$.
According to wolframalpha, the value is undefined.
By just seeing that it's $infty-infty$, can we then conclude that it's undefined? Won't L'hospitale or other tricks work here? How to really know whether it's undefined?
EDIT:
Thanks for all the answers.
Actually I was trying to find whether the $int_0^inftyfracln(xpi^x)dxx$ converges or not. And after integrating I got the above-mentioned results. So I was lost there.
EDIT2: Sorry for messing this question up. I asked another question concerning the integral here:Improper Integration $int_0^inftyfracln(xpi^x),dxx$ converges or not.
Thanks alot.
calculus limits
calculus limits
edited Sep 9 at 10:06
quid♦
36.4k85091
asked Sep 9 at 8:03
JoisBack
1355
New contributor
JoisBack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Sep 9 at 10:06
quid♦
36.4k85091
asked Sep 9 at 8:03
JoisBack
1355
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JoisBack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Sep 9 at 10:06
quid♦
36.4k85091
edited Sep 9 at 10:06
quid♦
36.4k85091
edited Sep 9 at 10:06
quid♦
36.4k85091
36.4k85091
asked Sep 9 at 8:03
JoisBack
1355
New contributor
JoisBack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Sep 9 at 8:03
JoisBack
1355
asked Sep 9 at 8:03
JoisBack
1355
1355
New contributor
JoisBack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
JoisBack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
JoisBack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Note that by getting $∞−∞$ from a particular method, the only thing you can conclude is that you used a wrong method. It doesn't tell anything about the function.
– IllidanS4
Sep 9 at 9:54
add a comment |Â
Note that by getting $∞−∞$ from a particular method, the only thing you can conclude is that you used a wrong method. It doesn't tell anything about the function.
– IllidanS4
Sep 9 at 9:54
Note that by getting $∞−∞$ from a particular method, the only thing you can conclude is that you used a wrong method. It doesn't tell anything about the function.
– IllidanS4
Sep 9 at 9:54
Note that by getting $∞−∞$ from a particular method, the only thing you can conclude is that you used a wrong method. It doesn't tell anything about the function.
– IllidanS4
Sep 9 at 9:54
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
4
down vote
accepted
We have that
$$lim_xtoinfty frac12ln^2(x)+xlnpi =infty$$
$$lim_xto 0 frac12ln^2(x)+xlnpi =infty$$
then we can't evaluate
$$lim_xtoinfty frac12ln^2(x)+xlnpi-lim_xtoinfty frac12ln^2(x)+xlnpi$$
since $infty$ is not a finite number but it is just used as a symbol to express the fact that the two expression are larger than any fixed number.
Otherwise for finite limit we have
$$lim_xto a f(x) =L_1 land lim_xto b g(x) =L_2 implies lim_xto a f(x) pm lim_xto b g(x) =L_1 pm L_2$$
What we can evaluate is
$$lim_xtoinftyleft[ left(frac12ln^2(x)+xlnpiright) - left(frac12ln^2(1/x)+(1/x)lnpiright)right]$$
answered Sep 9 at 8:16
gimusi
71.4k73786
add a comment |Â
up vote
4
down vote
An indeterminate form $infty - infty$ might be something like $lim_x to +infty (a(x)-b(x))$. But here you have two different limits: $lim_x to infty a(x) - lim_t to 0 b(t)$, with no connection between the two variables $x$ and $t$ (you called them both $x$, but they are really different). So there's no reason to assign any particular value to this expression.
answered Sep 9 at 8:10
Robert Israel
307k22201443
Thank you for pointing out the difference between t and x. Actually I was trying to find whether the integral of ln(xpi^x) over x from x=0 to x=inf is converging or diverging. So is it still meaningless? Also, does that mean that the mentioned integral is diverging (to something apart from inf and -inf)
– JoisBack
Sep 9 at 8:17
add a comment |Â
up vote
1
down vote
In writing
$$lim_xtoinftyleft(frac12ln^2(x)+xlnpiright)-lim_xto0left(frac12ln^2(x)+xlnpiright),$$
you presuppose that both of the involved limits exist, which is not the case, so that the expression is meaningless (or undefined, if you wish).
answered Sep 9 at 8:09
Sobi
2,778415
Actually I was trying to find whether the integral of ln(xpi^x) over x from x=0 to x=inf is converging or diverging. So is it still meaningless? Also, does that mean that the mentioned integral is diverging (to something apart from inf and -inf)
– JoisBack
Sep 9 at 8:13
1
@JoisBack Your question regarding the integral depends on how you interpret your improper integral. Take a look at Wikipedia: Improper integral.
– Sobi
Sep 9 at 8:24
$frac12 ln^2(x)+xln pi$ isn't an antiderivative of $ln(x pi^x)$.
– Robert Israel
2 days ago
add a comment |Â
up vote
1
down vote
There are two separate limits here, not just one. Let us put it in abstract terms. You have two functions $f$ and $g$ and two values $a$ and $b$ such that $lim_xto a f(x)=lim_xto b g(x)=infty$.
When you ask what is $lim_xto af(x)-lim_xto bg(x)$ you are asking what is the value of $infty-infty$, the difference between two actual infinities. As Wolfram tells you correctly, this is not well-defined.
If you had $a=b$, then you could ask whether $lim_xto a(f(x)-g(x))$ exists (and if so, the value it has). In this case, we sometimes say that it is a limit of the type "$infty-infty$". But that is very different: in here, the $infty$ symbol is only a potential infinity, and the expression "$infty-infty$" is only a convenient description, a description of a process, if you want. For every $x$, $f(x)-g(x)$ is an actual number, so it makes sense to ask whether these numbers converge somewhere (using any manner of methods).
But this is not the case here. Not only are the two limits separate, you don't even have $a=b$.
answered Sep 9 at 8:17
tomasz
22.9k23077
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
We have that
$$lim_xtoinfty frac12ln^2(x)+xlnpi =infty$$
$$lim_xto 0 frac12ln^2(x)+xlnpi =infty$$
then we can't evaluate
$$lim_xtoinfty frac12ln^2(x)+xlnpi-lim_xtoinfty frac12ln^2(x)+xlnpi$$
since $infty$ is not a finite number but it is just used as a symbol to express the fact that the two expression are larger than any fixed number.
Otherwise for finite limit we have
$$lim_xto a f(x) =L_1 land lim_xto b g(x) =L_2 implies lim_xto a f(x) pm lim_xto b g(x) =L_1 pm L_2$$
What we can evaluate is
$$lim_xtoinftyleft[ left(frac12ln^2(x)+xlnpiright) - left(frac12ln^2(1/x)+(1/x)lnpiright)right]$$
answered Sep 9 at 8:16
gimusi
71.4k73786
add a comment |Â
up vote
4
down vote
accepted
We have that
$$lim_xtoinfty frac12ln^2(x)+xlnpi =infty$$
$$lim_xto 0 frac12ln^2(x)+xlnpi =infty$$
then we can't evaluate
$$lim_xtoinfty frac12ln^2(x)+xlnpi-lim_xtoinfty frac12ln^2(x)+xlnpi$$
since $infty$ is not a finite number but it is just used as a symbol to express the fact that the two expression are larger than any fixed number.
Otherwise for finite limit we have
$$lim_xto a f(x) =L_1 land lim_xto b g(x) =L_2 implies lim_xto a f(x) pm lim_xto b g(x) =L_1 pm L_2$$
What we can evaluate is
$$lim_xtoinftyleft[ left(frac12ln^2(x)+xlnpiright) - left(frac12ln^2(1/x)+(1/x)lnpiright)right]$$
answered Sep 9 at 8:16
gimusi
71.4k73786
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
We have that
$$lim_xtoinfty frac12ln^2(x)+xlnpi =infty$$
$$lim_xto 0 frac12ln^2(x)+xlnpi =infty$$
then we can't evaluate
$$lim_xtoinfty frac12ln^2(x)+xlnpi-lim_xtoinfty frac12ln^2(x)+xlnpi$$
since $infty$ is not a finite number but it is just used as a symbol to express the fact that the two expression are larger than any fixed number.
Otherwise for finite limit we have
$$lim_xto a f(x) =L_1 land lim_xto b g(x) =L_2 implies lim_xto a f(x) pm lim_xto b g(x) =L_1 pm L_2$$
What we can evaluate is
$$lim_xtoinftyleft[ left(frac12ln^2(x)+xlnpiright) - left(frac12ln^2(1/x)+(1/x)lnpiright)right]$$
answered Sep 9 at 8:16
gimusi
71.4k73786
We have that
$$lim_xtoinfty frac12ln^2(x)+xlnpi =infty$$
$$lim_xto 0 frac12ln^2(x)+xlnpi =infty$$
then we can't evaluate
$$lim_xtoinfty frac12ln^2(x)+xlnpi-lim_xtoinfty frac12ln^2(x)+xlnpi$$
since $infty$ is not a finite number but it is just used as a symbol to express the fact that the two expression are larger than any fixed number.
Otherwise for finite limit we have
$$lim_xto a f(x) =L_1 land lim_xto b g(x) =L_2 implies lim_xto a f(x) pm lim_xto b g(x) =L_1 pm L_2$$
What we can evaluate is
$$lim_xtoinftyleft[ left(frac12ln^2(x)+xlnpiright) - left(frac12ln^2(1/x)+(1/x)lnpiright)right]$$
answered Sep 9 at 8:16
gimusi
71.4k73786
edited Sep 9 at 8:23
answered Sep 9 at 8:16
gimusi
71.4k73786
answered Sep 9 at 8:16
gimusi
71.4k73786
answered Sep 9 at 8:16
gimusi
71.4k73786
71.4k73786
add a comment |Â
add a comment |Â
up vote
4
down vote
An indeterminate form $infty - infty$ might be something like $lim_x to +infty (a(x)-b(x))$. But here you have two different limits: $lim_x to infty a(x) - lim_t to 0 b(t)$, with no connection between the two variables $x$ and $t$ (you called them both $x$, but they are really different). So there's no reason to assign any particular value to this expression.
answered Sep 9 at 8:10
Robert Israel
307k22201443
Thank you for pointing out the difference between t and x. Actually I was trying to find whether the integral of ln(xpi^x) over x from x=0 to x=inf is converging or diverging. So is it still meaningless? Also, does that mean that the mentioned integral is diverging (to something apart from inf and -inf)
– JoisBack
Sep 9 at 8:17
add a comment |Â
up vote
4
down vote
An indeterminate form $infty - infty$ might be something like $lim_x to +infty (a(x)-b(x))$. But here you have two different limits: $lim_x to infty a(x) - lim_t to 0 b(t)$, with no connection between the two variables $x$ and $t$ (you called them both $x$, but they are really different). So there's no reason to assign any particular value to this expression.
answered Sep 9 at 8:10
Robert Israel
307k22201443
Thank you for pointing out the difference between t and x. Actually I was trying to find whether the integral of ln(xpi^x) over x from x=0 to x=inf is converging or diverging. So is it still meaningless? Also, does that mean that the mentioned integral is diverging (to something apart from inf and -inf)
– JoisBack
Sep 9 at 8:17
add a comment |Â
up vote
4
down vote
up vote
4
down vote
An indeterminate form $infty - infty$ might be something like $lim_x to +infty (a(x)-b(x))$. But here you have two different limits: $lim_x to infty a(x) - lim_t to 0 b(t)$, with no connection between the two variables $x$ and $t$ (you called them both $x$, but they are really different). So there's no reason to assign any particular value to this expression.
answered Sep 9 at 8:10
Robert Israel
307k22201443
An indeterminate form $infty - infty$ might be something like $lim_x to +infty (a(x)-b(x))$. But here you have two different limits: $lim_x to infty a(x) - lim_t to 0 b(t)$, with no connection between the two variables $x$ and $t$ (you called them both $x$, but they are really different). So there's no reason to assign any particular value to this expression.
answered Sep 9 at 8:10
Robert Israel
307k22201443
answered Sep 9 at 8:10
Robert Israel
307k22201443
answered Sep 9 at 8:10
Robert Israel
307k22201443
answered Sep 9 at 8:10
Robert Israel
307k22201443
307k22201443
Thank you for pointing out the difference between t and x. Actually I was trying to find whether the integral of ln(xpi^x) over x from x=0 to x=inf is converging or diverging. So is it still meaningless? Also, does that mean that the mentioned integral is diverging (to something apart from inf and -inf)
– JoisBack
Sep 9 at 8:17
add a comment |Â
Thank you for pointing out the difference between t and x. Actually I was trying to find whether the integral of ln(xpi^x) over x from x=0 to x=inf is converging or diverging. So is it still meaningless? Also, does that mean that the mentioned integral is diverging (to something apart from inf and -inf)
– JoisBack
Sep 9 at 8:17
Thank you for pointing out the difference between t and x. Actually I was trying to find whether the integral of ln(xpi^x) over x from x=0 to x=inf is converging or diverging. So is it still meaningless? Also, does that mean that the mentioned integral is diverging (to something apart from inf and -inf)
– JoisBack
Sep 9 at 8:17
Thank you for pointing out the difference between t and x. Actually I was trying to find whether the integral of ln(xpi^x) over x from x=0 to x=inf is converging or diverging. So is it still meaningless? Also, does that mean that the mentioned integral is diverging (to something apart from inf and -inf)
– JoisBack
Sep 9 at 8:17
add a comment |Â
up vote
1
down vote
In writing
$$lim_xtoinftyleft(frac12ln^2(x)+xlnpiright)-lim_xto0left(frac12ln^2(x)+xlnpiright),$$
you presuppose that both of the involved limits exist, which is not the case, so that the expression is meaningless (or undefined, if you wish).
answered Sep 9 at 8:09
Sobi
2,778415
Actually I was trying to find whether the integral of ln(xpi^x) over x from x=0 to x=inf is converging or diverging. So is it still meaningless? Also, does that mean that the mentioned integral is diverging (to something apart from inf and -inf)
– JoisBack
Sep 9 at 8:13
1
@JoisBack Your question regarding the integral depends on how you interpret your improper integral. Take a look at Wikipedia: Improper integral.
– Sobi
Sep 9 at 8:24
$frac12 ln^2(x)+xln pi$ isn't an antiderivative of $ln(x pi^x)$.
– Robert Israel
2 days ago
add a comment |Â
up vote
1
down vote
In writing
$$lim_xtoinftyleft(frac12ln^2(x)+xlnpiright)-lim_xto0left(frac12ln^2(x)+xlnpiright),$$
you presuppose that both of the involved limits exist, which is not the case, so that the expression is meaningless (or undefined, if you wish).
answered Sep 9 at 8:09
Sobi
2,778415
Actually I was trying to find whether the integral of ln(xpi^x) over x from x=0 to x=inf is converging or diverging. So is it still meaningless? Also, does that mean that the mentioned integral is diverging (to something apart from inf and -inf)
– JoisBack
Sep 9 at 8:13
1
@JoisBack Your question regarding the integral depends on how you interpret your improper integral. Take a look at Wikipedia: Improper integral.
– Sobi
Sep 9 at 8:24
$frac12 ln^2(x)+xln pi$ isn't an antiderivative of $ln(x pi^x)$.
– Robert Israel
2 days ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
In writing
$$lim_xtoinftyleft(frac12ln^2(x)+xlnpiright)-lim_xto0left(frac12ln^2(x)+xlnpiright),$$
you presuppose that both of the involved limits exist, which is not the case, so that the expression is meaningless (or undefined, if you wish).
answered Sep 9 at 8:09
Sobi
2,778415
In writing
$$lim_xtoinftyleft(frac12ln^2(x)+xlnpiright)-lim_xto0left(frac12ln^2(x)+xlnpiright),$$
you presuppose that both of the involved limits exist, which is not the case, so that the expression is meaningless (or undefined, if you wish).
answered Sep 9 at 8:09
Sobi
2,778415
answered Sep 9 at 8:09
Sobi
2,778415
answered Sep 9 at 8:09
Sobi
2,778415
answered Sep 9 at 8:09
Sobi
2,778415
2,778415
Actually I was trying to find whether the integral of ln(xpi^x) over x from x=0 to x=inf is converging or diverging. So is it still meaningless? Also, does that mean that the mentioned integral is diverging (to something apart from inf and -inf)
– JoisBack
Sep 9 at 8:13
1
@JoisBack Your question regarding the integral depends on how you interpret your improper integral. Take a look at Wikipedia: Improper integral.
– Sobi
Sep 9 at 8:24
$frac12 ln^2(x)+xln pi$ isn't an antiderivative of $ln(x pi^x)$.
– Robert Israel
2 days ago
add a comment |Â
Actually I was trying to find whether the integral of ln(xpi^x) over x from x=0 to x=inf is converging or diverging. So is it still meaningless? Also, does that mean that the mentioned integral is diverging (to something apart from inf and -inf)
– JoisBack
Sep 9 at 8:13
1
@JoisBack Your question regarding the integral depends on how you interpret your improper integral. Take a look at Wikipedia: Improper integral.
– Sobi
Sep 9 at 8:24
$frac12 ln^2(x)+xln pi$ isn't an antiderivative of $ln(x pi^x)$.
– Robert Israel
2 days ago
Actually I was trying to find whether the integral of ln(xpi^x) over x from x=0 to x=inf is converging or diverging. So is it still meaningless? Also, does that mean that the mentioned integral is diverging (to something apart from inf and -inf)
– JoisBack
Sep 9 at 8:13
Actually I was trying to find whether the integral of ln(xpi^x) over x from x=0 to x=inf is converging or diverging. So is it still meaningless? Also, does that mean that the mentioned integral is diverging (to something apart from inf and -inf)
– JoisBack
Sep 9 at 8:13
1
1
@JoisBack Your question regarding the integral depends on how you interpret your improper integral. Take a look at Wikipedia: Improper integral.
– Sobi
Sep 9 at 8:24
@JoisBack Your question regarding the integral depends on how you interpret your improper integral. Take a look at Wikipedia: Improper integral.
– Sobi
Sep 9 at 8:24
$frac12 ln^2(x)+xln pi$ isn't an antiderivative of $ln(x pi^x)$.
– Robert Israel
2 days ago
$frac12 ln^2(x)+xln pi$ isn't an antiderivative of $ln(x pi^x)$.
– Robert Israel
2 days ago
add a comment |Â
up vote
1
down vote
There are two separate limits here, not just one. Let us put it in abstract terms. You have two functions $f$ and $g$ and two values $a$ and $b$ such that $lim_xto a f(x)=lim_xto b g(x)=infty$.
When you ask what is $lim_xto af(x)-lim_xto bg(x)$ you are asking what is the value of $infty-infty$, the difference between two actual infinities. As Wolfram tells you correctly, this is not well-defined.
If you had $a=b$, then you could ask whether $lim_xto a(f(x)-g(x))$ exists (and if so, the value it has). In this case, we sometimes say that it is a limit of the type "$infty-infty$". But that is very different: in here, the $infty$ symbol is only a potential infinity, and the expression "$infty-infty$" is only a convenient description, a description of a process, if you want. For every $x$, $f(x)-g(x)$ is an actual number, so it makes sense to ask whether these numbers converge somewhere (using any manner of methods).
But this is not the case here. Not only are the two limits separate, you don't even have $a=b$.
answered Sep 9 at 8:17
tomasz
22.9k23077
add a comment |Â
up vote
1
down vote
There are two separate limits here, not just one. Let us put it in abstract terms. You have two functions $f$ and $g$ and two values $a$ and $b$ such that $lim_xto a f(x)=lim_xto b g(x)=infty$.
When you ask what is $lim_xto af(x)-lim_xto bg(x)$ you are asking what is the value of $infty-infty$, the difference between two actual infinities. As Wolfram tells you correctly, this is not well-defined.
If you had $a=b$, then you could ask whether $lim_xto a(f(x)-g(x))$ exists (and if so, the value it has). In this case, we sometimes say that it is a limit of the type "$infty-infty$". But that is very different: in here, the $infty$ symbol is only a potential infinity, and the expression "$infty-infty$" is only a convenient description, a description of a process, if you want. For every $x$, $f(x)-g(x)$ is an actual number, so it makes sense to ask whether these numbers converge somewhere (using any manner of methods).
But this is not the case here. Not only are the two limits separate, you don't even have $a=b$.
answered Sep 9 at 8:17
tomasz
22.9k23077
add a comment |Â
up vote
1
down vote
up vote
1
down vote
There are two separate limits here, not just one. Let us put it in abstract terms. You have two functions $f$ and $g$ and two values $a$ and $b$ such that $lim_xto a f(x)=lim_xto b g(x)=infty$.
When you ask what is $lim_xto af(x)-lim_xto bg(x)$ you are asking what is the value of $infty-infty$, the difference between two actual infinities. As Wolfram tells you correctly, this is not well-defined.
If you had $a=b$, then you could ask whether $lim_xto a(f(x)-g(x))$ exists (and if so, the value it has). In this case, we sometimes say that it is a limit of the type "$infty-infty$". But that is very different: in here, the $infty$ symbol is only a potential infinity, and the expression "$infty-infty$" is only a convenient description, a description of a process, if you want. For every $x$, $f(x)-g(x)$ is an actual number, so it makes sense to ask whether these numbers converge somewhere (using any manner of methods).
But this is not the case here. Not only are the two limits separate, you don't even have $a=b$.
answered Sep 9 at 8:17
tomasz
22.9k23077
There are two separate limits here, not just one. Let us put it in abstract terms. You have two functions $f$ and $g$ and two values $a$ and $b$ such that $lim_xto a f(x)=lim_xto b g(x)=infty$.
When you ask what is $lim_xto af(x)-lim_xto bg(x)$ you are asking what is the value of $infty-infty$, the difference between two actual infinities. As Wolfram tells you correctly, this is not well-defined.
If you had $a=b$, then you could ask whether $lim_xto a(f(x)-g(x))$ exists (and if so, the value it has). In this case, we sometimes say that it is a limit of the type "$infty-infty$". But that is very different: in here, the $infty$ symbol is only a potential infinity, and the expression "$infty-infty$" is only a convenient description, a description of a process, if you want. For every $x$, $f(x)-g(x)$ is an actual number, so it makes sense to ask whether these numbers converge somewhere (using any manner of methods).
But this is not the case here. Not only are the two limits separate, you don't even have $a=b$.
answered Sep 9 at 8:17
tomasz
22.9k23077
answered Sep 9 at 8:17
tomasz
22.9k23077
answered Sep 9 at 8:17
tomasz
22.9k23077
answered Sep 9 at 8:17
tomasz
22.9k23077
22.9k23077
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Note that by getting $∞−∞$ from a particular method, the only thing you can conclude is that you used a wrong method. It doesn't tell anything about the function.
– IllidanS4
Sep 9 at 9:54