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Get the two missing terms when using Limit?

GROUPS:

I use the command Limit,

In[503]:= eq1 = 
 cinf/(s - \[Theta]c) + (
  Sqrt[Dc] E^((lh Sqrt[s])/Sqrt[Dc] - (Sqrt[s] x)/Sqrt[Dc]) F1)/(
  Sqrt[s] (s - \[Theta]c)) - (cinf \[Theta]c)/(s (s - \[Theta]c)) + (
  E^((Sqrt[s] x)/Sqrt[Dc]) s a[1][s])/(s - \[Theta]c) + (
  E^((2 lh Sqrt[s])/Sqrt[Dc] - (Sqrt[s] x)/Sqrt[Dc]) s a[1][s])/(
  s - \[Theta]c) - (E^((Sqrt[s] x)/Sqrt[Dc]) \[Theta]c a[1][s])/(
  s - \[Theta]c) - (
  E^((2 lh Sqrt[s])/Sqrt[Dc] - (Sqrt[s] x)/Sqrt[Dc]) \[Theta]c a[1][
    s])/(s - \[Theta]c)

Out[503]= cinf/(s - \[Theta]c) + (
 Sqrt[Dc] E^((lh Sqrt[s])/Sqrt[Dc] - (Sqrt[s] x)/Sqrt[Dc]) F1)/(
 Sqrt[s] (s - \[Theta]c)) - (cinf \[Theta]c)/(s (s - \[Theta]c)) + (
 E^((Sqrt[s] x)/Sqrt[Dc]) s a[1][s])/(s - \[Theta]c) + (
 E^((2 lh Sqrt[s])/Sqrt[Dc] - (Sqrt[s] x)/Sqrt[Dc]) s a[1][s])/(
 s - \[Theta]c) - (E^((Sqrt[s] x)/Sqrt[Dc]) \[Theta]c a[1][s])/(
 s - \[Theta]c) - (
 E^((2 lh Sqrt[s])/Sqrt[Dc] - (Sqrt[s] x)/Sqrt[Dc]) \[Theta]c a[1][
   s])/(s - \[Theta]c)

In[504]:= eq2 = (Limit[Expand@eq23, x -> + \[Infinity], 
     Assumptions -> {s > 0, Dc > 0}]) // Simplify // Normal

Out[504]= \[Infinity] a[1][s]

Please tell me where they are,

cinf/(s - \[Theta]c)-((cinf \[Theta]c)/(s (s - \[Theta]c)))
POSTED BY: Zhonghui Ou
Answer
4 months ago

This is very unclear. What specifically is the expected result?

POSTED BY: Daniel Lichtblau
Answer
4 months ago

I think that eq2 should include the constant number term eq3, but it seems that this term has been devoured by the infinity term.

I am sorry for the ambiguity. I don't know how to input mathematical expression explicitly.

enter image description here

POSTED BY: Zhonghui Ou
Answer
4 months ago

I'm not sure I follow, but maybe you mean something like this?

In[4]:= Infinity+xxx                                                            

Out[4]= Infinity

If so, then yes, that's how infinity arithmetic behaves in the Wolfram Language.

POSTED BY: Daniel Lichtblau
Answer
4 months ago

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