I have an array of c (10 elements) and I want to calculate the limit of the equation below when c(i)->0 (i = 1...10). How Can I express that c(i)->0 (i = 1...10) in the equation?

$$ \lim_{c(i)\to 0} \, \frac{1}{\frac{1000 \rho -\sum _{i=1}^{10} c(i) M(i)}{\text{Ms}}+\sum _{i=1}^{10} c(i)} $$

The code I have is listed below.

Array[c, 10]
Limit[1/(\!\(\*UnderoverscriptBox[\(\[Sum]\), \(i = 1\), \(10\)]\(c[i]\)\) + (1000 \[Rho] - \!\(\*UnderoverscriptBox[\(\[Sum]\), \(i = 1\), \(10\)]\(c[i]\ M[i]\)\))/Ms), c[i] -> 0] 

The output I got is below.

-(Ms/(-1000 \[Rho] - Ms (c[1] + c[2] + c[3] + c[4] + c[5] + c[6] + c[7] + c[8] + c[9] + c[10]) + c[1] M[1] + c[2] M[2] + c[3] M[3] + c[4] M[4] + c[5] M[5] + c[6] M[6] + c[7] M[7] + c[8] M[8] + c[9] M[9] + c[10] M[10]))

Let's first define an expression for what you are interested in:

expr = 1/(Sum[c[i], {i, 1, 10}] + (1000*\[Rho] - Sum[c[i] M[i], {i, 1, 10}])/Ms);

In your case the limit is trivial, since you can just set all c[i] to zero and get a finite result:


Ms/(1000 \[Rho])

(Here the underscore _ means that all c[x] expressions will get replaced by the same 0 regardless of what x is.) If you want to take all c[i] to zero carefully, you could do:

Series[expr/.c[x_]->eps c[x],{eps,0,0}]//Normal

Ms/(1000 \[Rho])

Here we first replaced each c[i] by eps c[i] and then made a series expansion around eps=0 to leading order. (The Limit function is less flexible than Series. Try to get used to using Series for your limit calculation needs.)

  • $\begingroup$ Many thanks! This is what I need! $\endgroup$ Mar 25 '17 at 17:57

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