Weird result in complex limit
$begingroup$
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
edited 57 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.
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$endgroup$
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
edited 57 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
edited 57 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.
$endgroup$
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
calculus-and-analysis complex
edited 57 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 57 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 57 mins ago
m_goldberg
89.3k873200
edited 57 mins ago
m_goldberg
89.3k873200
edited 57 mins ago
m_goldberg
89.3k873200
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
asked 4 hours ago
VillaVilla
1083
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.
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.
Villa 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 |
1 Answer
1
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votes
$begingroup$
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])
edited 21 mins ago
m_goldberg
89.3k873200
answered 4 hours ago
Carl WollCarl Woll
76.3k3100200
$endgroup$
$begingroup$
Thank you, it worked.
$endgroup$
– Villa
3 hours ago
$begingroup$
UsingAssumptions -> u >= 0 && e > 0gives the same form as the OP's hand calculation.
$endgroup$
– Bob Hanlon
3 hours ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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])
edited 21 mins ago
m_goldberg
89.3k873200
answered 4 hours ago
Carl WollCarl Woll
76.3k3100200
$endgroup$
$begingroup$
Thank you, it worked.
$endgroup$
– Villa
3 hours ago
$begingroup$
UsingAssumptions -> u >= 0 && e > 0gives the same form as the OP's hand calculation.
$endgroup$
– Bob Hanlon
3 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])
edited 21 mins ago
m_goldberg
89.3k873200
answered 4 hours ago
Carl WollCarl Woll
76.3k3100200
$endgroup$
$begingroup$
Thank you, it worked.
$endgroup$
– Villa
3 hours ago
$begingroup$
UsingAssumptions -> u >= 0 && e > 0gives the same form as the OP's hand calculation.
$endgroup$
– Bob Hanlon
3 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])
edited 21 mins ago
m_goldberg
89.3k873200
answered 4 hours ago
Carl WollCarl Woll
76.3k3100200
$endgroup$
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])
edited 21 mins ago
m_goldberg
89.3k873200
answered 4 hours ago
Carl WollCarl Woll
76.3k3100200
edited 21 mins ago
m_goldberg
89.3k873200
edited 21 mins ago
m_goldberg
89.3k873200
edited 21 mins ago
m_goldberg
89.3k873200
89.3k873200
answered 4 hours ago
Carl WollCarl Woll
76.3k3100200
answered 4 hours ago
Carl WollCarl Woll
76.3k3100200
answered 4 hours ago
Carl WollCarl Woll
76.3k3100200
76.3k3100200
$begingroup$
Thank you, it worked.
$endgroup$
– Villa
3 hours ago
$begingroup$
UsingAssumptions -> u >= 0 && e > 0gives the same form as the OP's hand calculation.
$endgroup$
– Bob Hanlon
3 hours ago
add a comment |
$begingroup$
Thank you, it worked.
$endgroup$
– Villa
3 hours ago
$begingroup$
UsingAssumptions -> u >= 0 && e > 0gives the same form as the OP's hand calculation.
$endgroup$
– Bob Hanlon
3 hours ago
$begingroup$
Thank you, it worked.
$endgroup$
– Villa
3 hours ago
$begingroup$
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
3 hours ago
$begingroup$
Using
Assumptions -> u >= 0 && e > 0 gives the same form as the OP's hand calculation.$endgroup$
– Bob Hanlon
3 hours ago
add a comment |
Villa is a new contributor. Be nice, and check out our Code of Conduct.
Villa is a new contributor. Be nice, and check out our Code of Conduct.
Villa is a new contributor. Be nice, and check out our Code of Conduct.
Villa is a new contributor. Be nice, and check out our Code of Conduct.
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