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Weird result in complex limit

Limit of integral gives incorrect outputIntegrate returns unexpected resultLimit problem calculating directional derivativeWhy won't Limit evaluate, and what can be done about itLimit of an inverse functionDoes Mathematica implement Risch algorithm? If it does, in which cases?Limit problem no longer works in Mathematica 11.1.0Evaluating integral seems incorrectReal integral giving complex resultHow to apply NIntegrate three times

2

$begingroup$

I am trying to evaluate a limit:

gamma[w_] = Sqrt[-(u*e)w^2 + I*(u*s)w];
Limit[Re[gamma[x]], x -> DirectedInfinity[1]]

I calculated the limit by hand, and the correct answer is (I also checked numerically for some examples of $u,e,s$ using the software):

$qquad frac s2 sqrtfrac ue$

But for some reason, when using Limit, I get

DirectedInfinity[(Sign[e]^2 Sign[u]^2)^(1/4)]

So my questions are:

What is going here?

What issues should I be aware of when using Limit?

calculus-and-analysis complex

share|improve this question

edited Apr 29 at 1:37

m_goldberg

89.4k873201

asked Apr 28 at 22:14

VillaVilla

1133

New contributor

Villa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

2

$begingroup$

I am trying to evaluate a limit:

gamma[w_] = Sqrt[-(u*e)w^2 + I*(u*s)w];
Limit[Re[gamma[x]], x -> DirectedInfinity[1]]

I calculated the limit by hand, and the correct answer is (I also checked numerically for some examples of $u,e,s$ using the software):

$qquad frac s2 sqrtfrac ue$

But for some reason, when using Limit, I get

DirectedInfinity[(Sign[e]^2 Sign[u]^2)^(1/4)]

So my questions are:

What is going here?

What issues should I be aware of when using Limit?

calculus-and-analysis complex

share|improve this question

edited Apr 29 at 1:37

m_goldberg

89.4k873201

asked Apr 28 at 22:14

VillaVilla

1133

New contributor

Villa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

I am trying to evaluate a limit:

gamma[w_] = Sqrt[-(u*e)w^2 + I*(u*s)w];
Limit[Re[gamma[x]], x -> DirectedInfinity[1]]

I calculated the limit by hand, and the correct answer is (I also checked numerically for some examples of $u,e,s$ using the software):

$qquad frac s2 sqrtfrac ue$

But for some reason, when using Limit, I get

DirectedInfinity[(Sign[e]^2 Sign[u]^2)^(1/4)]

So my questions are:

What is going here?

What issues should I be aware of when using Limit?

calculus-and-analysis complex

share|improve this question

edited Apr 29 at 1:37

m_goldberg

89.4k873201

asked Apr 28 at 22:14

VillaVilla

1133

New contributor

Villa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

gamma[w_] = Sqrt[-(u*e)w^2 + I*(u*s)w];
Limit[Re[gamma[x]], x -> DirectedInfinity[1]]

DirectedInfinity[(Sign[e]^2 Sign[u]^2)^(1/4)]

share|improve this question

edited Apr 29 at 1:37

m_goldberg

89.4k873201

asked Apr 28 at 22:14

VillaVilla

1133

New contributor

Villa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited Apr 29 at 1:37

m_goldberg

89.4k873201

asked Apr 28 at 22:14

VillaVilla

1133

New contributor

Villa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited Apr 29 at 1:37

m_goldberg

89.4k873201

edited Apr 29 at 1:37

m_goldberg

89.4k873201

asked Apr 28 at 22:14

VillaVilla

1133

New contributor

Villa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Apr 28 at 22:14

VillaVilla

1133

2 Answers
2

active

oldest

votes

6

$begingroup$

I think it's worth reporting the issue to support. If you give appropriate assumptions, then you get your expected result:

Limit[Re[gamma[x]], x -> Infinity, Assumptions -> u>0 && e>0]

(s u)/(2 Sqrt[e u])

share|improve this answer

edited Apr 29 at 2:13

m_goldberg

89.4k873201

answered Apr 28 at 22:31

Carl WollCarl Woll

76.8k3101201

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, it worked.
  $endgroup$
  – Villa
  Apr 28 at 22:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Using Assumptions -> u >= 0 && e > 0 gives the same form as the OP's hand calculation.
  $endgroup$
  – Bob Hanlon
  Apr 28 at 23:32
  

  
  
  
  

add a comment | 

2

$begingroup$

The biggest difference between your hand calculation and the computation performed by Mathematica is that your hand calculation assumes $u$ and $e$ are nonnegative reals. Examples of how this produces different results:

• 
  $u = 1$ and $e = -1$: The limit of Re[gamma[x]] is $infty$, but your formula gives an imaginary number. A similar thing happens with $u = -1$ and $e = 1$.


• 
  $u = 1$ and $e = 0$: The limit of Re[gamma[x]] is a directed infinity, directed along $mathrmRe sqrtmathrmis$, which could be $-infty$, $0$, or $infty$, depending on the complex argument of $s$. (Mathematica misses this case in the answer you are seeing. Some insight comes from looking at the leading order term in ComplexExpand[Re[Sqrt[-(u*e)w^2 + I*(u*s)s]]], which is $(e^2 u^2 w^4)^1/4$. Of course, when $e = 0$, this term is suppressed and then the leading term is $(s^2 u^2 w^2)^1/4$. Note that ComplexExpand assumes variables are real unless it is explicitly told otherwise, so it assumes more than we have explicitly established.)


• 
  $u = e = -1$: The limit of Re[gamma[x]] is $-s/2$, but your formula gives $s/2$.


• 
  $u = e = mathrmi$: The limit of Re[gamma[x]] is $infty$, but your formula gives $s/2$.


• 
  $u = mathrmi, e = 0, s = 1$: The limit of Re[gamma[x]] is $0$, but your formula involves division by zero.

share|improve this answer

answered 2 days ago

Eric TowersEric Towers

2,396713

$endgroup$

add a comment | 

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6

$begingroup$

I think it's worth reporting the issue to support. If you give appropriate assumptions, then you get your expected result:

Limit[Re[gamma[x]], x -> Infinity, Assumptions -> u>0 && e>0]

(s u)/(2 Sqrt[e u])

share|improve this answer

edited Apr 29 at 2:13

m_goldberg

89.4k873201

answered Apr 28 at 22:31

Carl WollCarl Woll

76.8k3101201

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, it worked.
  $endgroup$
  – Villa
  Apr 28 at 22:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Using Assumptions -> u >= 0 && e > 0 gives the same form as the OP's hand calculation.
  $endgroup$
  – Bob Hanlon
  Apr 28 at 23:32
  

  
  
  
  

add a comment | 

6

$begingroup$

I think it's worth reporting the issue to support. If you give appropriate assumptions, then you get your expected result:

Limit[Re[gamma[x]], x -> Infinity, Assumptions -> u>0 && e>0]

(s u)/(2 Sqrt[e u])

share|improve this answer

edited Apr 29 at 2:13

m_goldberg

89.4k873201

answered Apr 28 at 22:31

Carl WollCarl Woll

76.8k3101201

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, it worked.
  $endgroup$
  – Villa
  Apr 28 at 22:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Using Assumptions -> u >= 0 && e > 0 gives the same form as the OP's hand calculation.
  $endgroup$
  – Bob Hanlon
  Apr 28 at 23:32
  

  
  
  
  

add a comment | 

$begingroup$

I think it's worth reporting the issue to support. If you give appropriate assumptions, then you get your expected result:

Limit[Re[gamma[x]], x -> Infinity, Assumptions -> u>0 && e>0]

(s u)/(2 Sqrt[e u])

share|improve this answer

edited Apr 29 at 2:13

m_goldberg

89.4k873201

answered Apr 28 at 22:31

Carl WollCarl Woll

76.8k3101201

$endgroup$

Limit[Re[gamma[x]], x -> Infinity, Assumptions -> u>0 && e>0]

share|improve this answer

edited Apr 29 at 2:13

m_goldberg

89.4k873201

answered Apr 28 at 22:31

Carl WollCarl Woll

76.8k3101201

edited Apr 29 at 2:13

m_goldberg

89.4k873201

edited Apr 29 at 2:13

m_goldberg

89.4k873201

answered Apr 28 at 22:31

Carl WollCarl Woll

76.8k3101201

answered Apr 28 at 22:31

Carl WollCarl Woll

76.8k3101201

2

$begingroup$

The biggest difference between your hand calculation and the computation performed by Mathematica is that your hand calculation assumes $u$ and $e$ are nonnegative reals. Examples of how this produces different results:

• 
  $u = 1$ and $e = -1$: The limit of Re[gamma[x]] is $infty$, but your formula gives an imaginary number. A similar thing happens with $u = -1$ and $e = 1$.


• 
  $u = 1$ and $e = 0$: The limit of Re[gamma[x]] is a directed infinity, directed along $mathrmRe sqrtmathrmis$, which could be $-infty$, $0$, or $infty$, depending on the complex argument of $s$. (Mathematica misses this case in the answer you are seeing. Some insight comes from looking at the leading order term in ComplexExpand[Re[Sqrt[-(u*e)w^2 + I*(u*s)s]]], which is $(e^2 u^2 w^4)^1/4$. Of course, when $e = 0$, this term is suppressed and then the leading term is $(s^2 u^2 w^2)^1/4$. Note that ComplexExpand assumes variables are real unless it is explicitly told otherwise, so it assumes more than we have explicitly established.)


• 
  $u = e = -1$: The limit of Re[gamma[x]] is $-s/2$, but your formula gives $s/2$.


• 
  $u = e = mathrmi$: The limit of Re[gamma[x]] is $infty$, but your formula gives $s/2$.


• 
  $u = mathrmi, e = 0, s = 1$: The limit of Re[gamma[x]] is $0$, but your formula involves division by zero.

share|improve this answer

answered 2 days ago

Eric TowersEric Towers

2,396713

$endgroup$

add a comment | 

2

$begingroup$

The biggest difference between your hand calculation and the computation performed by Mathematica is that your hand calculation assumes $u$ and $e$ are nonnegative reals. Examples of how this produces different results:

• 
  $u = 1$ and $e = -1$: The limit of Re[gamma[x]] is $infty$, but your formula gives an imaginary number. A similar thing happens with $u = -1$ and $e = 1$.


• 
  $u = 1$ and $e = 0$: The limit of Re[gamma[x]] is a directed infinity, directed along $mathrmRe sqrtmathrmis$, which could be $-infty$, $0$, or $infty$, depending on the complex argument of $s$. (Mathematica misses this case in the answer you are seeing. Some insight comes from looking at the leading order term in ComplexExpand[Re[Sqrt[-(u*e)w^2 + I*(u*s)s]]], which is $(e^2 u^2 w^4)^1/4$. Of course, when $e = 0$, this term is suppressed and then the leading term is $(s^2 u^2 w^2)^1/4$. Note that ComplexExpand assumes variables are real unless it is explicitly told otherwise, so it assumes more than we have explicitly established.)


• 
  $u = e = -1$: The limit of Re[gamma[x]] is $-s/2$, but your formula gives $s/2$.


• 
  $u = e = mathrmi$: The limit of Re[gamma[x]] is $infty$, but your formula gives $s/2$.


• 
  $u = mathrmi, e = 0, s = 1$: The limit of Re[gamma[x]] is $0$, but your formula involves division by zero.

share|improve this answer

answered 2 days ago

Eric TowersEric Towers

2,396713

$endgroup$

add a comment | 

$begingroup$

The biggest difference between your hand calculation and the computation performed by Mathematica is that your hand calculation assumes $u$ and $e$ are nonnegative reals. Examples of how this produces different results:

• 
  $u = 1$ and $e = -1$: The limit of Re[gamma[x]] is $infty$, but your formula gives an imaginary number. A similar thing happens with $u = -1$ and $e = 1$.


• 
  $u = 1$ and $e = 0$: The limit of Re[gamma[x]] is a directed infinity, directed along $mathrmRe sqrtmathrmis$, which could be $-infty$, $0$, or $infty$, depending on the complex argument of $s$. (Mathematica misses this case in the answer you are seeing. Some insight comes from looking at the leading order term in ComplexExpand[Re[Sqrt[-(u*e)w^2 + I*(u*s)s]]], which is $(e^2 u^2 w^4)^1/4$. Of course, when $e = 0$, this term is suppressed and then the leading term is $(s^2 u^2 w^2)^1/4$. Note that ComplexExpand assumes variables are real unless it is explicitly told otherwise, so it assumes more than we have explicitly established.)


• 
  $u = e = -1$: The limit of Re[gamma[x]] is $-s/2$, but your formula gives $s/2$.


• 
  $u = e = mathrmi$: The limit of Re[gamma[x]] is $infty$, but your formula gives $s/2$.


• 
  $u = mathrmi, e = 0, s = 1$: The limit of Re[gamma[x]] is $0$, but your formula involves division by zero.

share|improve this answer

answered 2 days ago

Eric TowersEric Towers

2,396713

$endgroup$

share|improve this answer

answered 2 days ago

Eric TowersEric Towers

2,396713

answered 2 days ago

Eric TowersEric Towers

2,396713

answered 2 days ago

Eric TowersEric Towers

2,396713