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1

$begingroup$

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):

$qquad frac s2 sqrt{frac ue}$

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 50 mins ago

m_goldberg

89.3k873200

asked 4 hours ago

VillaVilla

1083

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 | 

1

$begingroup$

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):

$qquad frac s2 sqrt{frac ue}$

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 50 mins ago

m_goldberg

89.3k873200

asked 4 hours ago

VillaVilla

1083

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):

$qquad frac s2 sqrt{frac ue}$

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 50 mins ago

m_goldberg

89.3k873200

asked 4 hours ago

VillaVilla

1083

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 50 mins ago

m_goldberg

89.3k873200

asked 4 hours ago

VillaVilla

1083

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 50 mins ago

m_goldberg

89.3k873200

asked 4 hours ago

VillaVilla

1083

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 50 mins ago

m_goldberg

89.3k873200

edited 50 mins ago

m_goldberg

89.3k873200

asked 4 hours ago

VillaVilla

1083

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 4 hours ago

VillaVilla

1083

1 Answer
1

active

oldest

votes

4

$begingroup$

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 14 mins ago

m_goldberg

89.3k873200

answered 3 hours ago

Carl WollCarl Woll

76.3k3100200

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, it worked.
  $endgroup$
  – Villa
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Using Assumptions -> u >= 0 && e > 0 gives the same form as the OP's hand calculation.
  $endgroup$
  – Bob Hanlon
  2 hours ago
  

  
  
  
  

add a comment | 

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4

$begingroup$

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 14 mins ago

m_goldberg

89.3k873200

answered 3 hours ago

Carl WollCarl Woll

76.3k3100200

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, it worked.
  $endgroup$
  – Villa
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Using Assumptions -> u >= 0 && e > 0 gives the same form as the OP's hand calculation.
  $endgroup$
  – Bob Hanlon
  2 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

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 14 mins ago

m_goldberg

89.3k873200

answered 3 hours ago

Carl WollCarl Woll

76.3k3100200

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, it worked.
  $endgroup$
  – Villa
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Using Assumptions -> u >= 0 && e > 0 gives the same form as the OP's hand calculation.
  $endgroup$
  – Bob Hanlon
  2 hours ago
  

  
  
  
  

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 14 mins ago

m_goldberg

89.3k873200

answered 3 hours ago

Carl WollCarl Woll

76.3k3100200

$endgroup$

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

share|improve this answer

edited 14 mins ago

m_goldberg

89.3k873200

answered 3 hours ago

Carl WollCarl Woll

76.3k3100200

edited 14 mins ago

m_goldberg

89.3k873200

edited 14 mins ago

m_goldberg

89.3k873200

answered 3 hours ago

Carl WollCarl Woll

76.3k3100200

answered 3 hours ago

Carl WollCarl Woll

76.3k3100200