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Does anyone know why the limit:

Limit[(α*(E^((t*ϵ)/(λ*τ))*t*ϵ*(-1 + 2*λ) - E^((t*ϵ)/(λ*τ))*ϵ*(-1 + 
    2*λ)*τ + E^((t*ϵ*(-1 + 2*λ))/((-1 + λ)*λ*τ))*ϵ*(-1 + λ)*(1 + (-2 + ϵ)*λ)*τ + 
    ϵ*λ*(-1 + ϵ + 2*λ - ϵ*λ)*τ))/(E^((t*ϵ)/(λ*τ))*ϵ*(-1 + 2*λ)), 
  λ -> 0
]

won't compute?

Sorry about the weird formatting.

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Thanks Halirutan for cleaning up that code. –  ThermalModeler Dec 18 '13 at 23:55
3  
I haven't looked in detail, but typically the result would depend on the parameters and it's difficult or impossible to get a result without knowing something about them. Example quiz: What's the limit $\lim_{x \rightarrow \infty} e^{-a x}$? If you said zero, that's not correct. The correct answer is $0$ if $a>0$, $\infty$ if $a < 0$ and $1$ if $a=0$. And that still assumed that $a$ is real. Correspondingly Limit[Exp[-a x], x -> Infinity] gives you nothing but Limit[Exp[-a x], x -> Infinity, Assumptions -> a > 0] does respond 0. –  Szabolcs Dec 19 '13 at 0:07
    
Keep in mind that by default Mathematica assumes that any parameter can take any complex value, which makes the problem very difficult. Generally one needs to specify assumptions about the parameters to get a solution. –  Szabolcs Dec 19 '13 at 0:09
1  
Just an additional comment: symbolic algebra systems make mistakes. It's wise not to trust them blindly. But they're also very useful, especially if the problem is straightforward but laborious. So what I'd do here to have some more confidence (other than numerical verification) is this: 1. get rid of α because it's just a constant factor. 2. Expand the expression and examine the terms. Notice that carrying out the limit is quite trivial, so Mma is unlikely to make mistakes. ... –  Szabolcs Dec 19 '13 at 0:28
1  
... 3. Map the limit onto the sum hoping that we won't get a cancelling sum of infinities: Assuming[\[Tau] > 0 && t > 0 && \[Epsilon] > 0, Limit[Expand[expr], \[Lambda] -> 0]]. This gives a good result (and the same as the direct question), which will improve your confidence in the answer. –  Szabolcs Dec 19 '13 at 0:29
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2 Answers

For the sake of having an answer (more or less given in the comments):

Limit[
  (α*(E^((t*ϵ)/(λ*τ))*t*ϵ*(-1 + 2*λ) -
       E^((t*ϵ)/(λ*τ))*ϵ*(-1 + 2*λ)*τ +
       E^((t*ϵ*(-1 + 2*λ))/((-1 + λ)*λ*τ))*ϵ*(-1 + λ)*(1 + (-2 + ϵ)*λ)*τ +
       ϵ*λ*(-1 + ϵ + 2*λ - ϵ*λ)*τ)) /
   (E^((t*ϵ)/(λ*τ))*ϵ*(-1 + 2*λ)),
  λ -> 0, Direction -> -1,
  Assumptions -> t ϵ τ > 0]
(*
   α (t + (-1 + E^(-((t ϵ)/τ))) τ)
*)

The OP is welcome to post their own answer, too!

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f=(α*(E^((t*ϵ)/(λ*τ))*t*ϵ*(-1 + 2*λ) -
       E^((t*ϵ)/(λ*τ))*ϵ*(-1 + 2*λ)*τ +
       E^((t*ϵ*(-1 + 2*λ))/((-1 + λ)*λ*τ))*ϵ*(-1 + λ)*(1 + (-2 + ϵ)*λ)*τ +
       ϵ*λ*(-1 + ϵ + 2*λ - ϵ*λ)*τ)) /
   (E^((t*ϵ)/(λ*τ))*ϵ*(-1 + 2*λ))//Expand;

One use

Limit[#, λ -> 0] & /@ f

and get $\lim_{\lambda \to 0} \, \frac{2 \alpha \lambda ^2 \tau e^{-\frac{t \epsilon }{\lambda \tau }}}{2 \lambda -1}+\lim_{\lambda \to 0} \, -\frac{\alpha \lambda ^2 \tau \epsilon e^{-\frac{t \epsilon }{\lambda \tau }}}{2 \lambda -1}+\lim_{\lambda \to 0} \, \frac{\alpha \lambda \tau \epsilon e^{-\frac{t \epsilon }{\lambda \tau }}}{2 \lambda -1}+\lim_{\lambda \to 0} \, -\frac{\alpha \lambda \tau e^{-\frac{t \epsilon }{\lambda \tau }}}{2 \lambda -1}-\alpha \tau +\alpha \tau e^{-\frac{t \epsilon }{\tau }}+\alpha t.$

One can see clearly the limit of each subexpression. As other guys have said before, some are indeterminate unless we make certain assumptions:

Limit[#, λ -> 0, Assumptions -> t ϵ/τ > 0] & /@f

$-\alpha \tau +\alpha \tau e^{-\frac{t \epsilon }{\tau }}+\alpha t$.

By the way. When I directly copy the mma code here from my notebook, it just show \[Alpha] but not $\alpha$ ? So I copy the latex here. Do you know some better mathods to copy mma code and keep the style?

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