1
$\begingroup$

I hope this question ha not been asked since the limit is evaluated in 1 not at 0 or infinity.

I do not know how to ask Mathematica to find the limit as n -> 1 of the following function:

(A Kh n (n/( 1 - n))^ξ α (-A ((3 (1 - n))/2 + n) - δ + 
   A (n + 1/2 (1 - n) (1 + θ))) (1 - ξ/(1 + (n/(1 - n))^ξ α))) /
   ((1 + (n/(1 - n))^ξ α) (-A ((3 (1 - n))/2 + n) 
     + A (1 - ((1 - n) (n/(1 - n))^ξ α)/(1 + (n/(1 - n))^ξ α)) - δ))

It is not a problem of assumption since the following image shows that one can find the limit without any assumption:

explanation

$\endgroup$
  • 2
    $\begingroup$ Perhaps Assuming[ξ > 1, Limit[...yourexpression..., n->1]] which then returns A Kh. Otherwise Mathematica defaults to assume Xi is complex or possibly even zero or negative and cannot give a specific limit. $\endgroup$ – Bill Apr 1 '16 at 16:30
  • 1
    $\begingroup$ As @Bill said, it seems that you are making an assumption yourself when you substitute $(1-n)^\xi$ with $0$, i.e. you are assuming that $\xi>1$. Unless you state your assumption explicitly to Limit, the system is not allowed to use it and cannot get to your desired result. $\endgroup$ – MarcoB Apr 1 '16 at 17:24