3
$\begingroup$
Limit[(k/(k - 1)), k -> Infinity]

Gives correctly:

1

But,

L[k_Integer] := (k/(k - 1))
Limit[L[k], k -> Infinity]

leaves the limit unevaluated.

How can I tell Mathematica to evaluate this limit?

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2
  • 3
    $\begingroup$ Remove "Integer" $\endgroup$
    – ZaMoC
    Nov 5, 2018 at 20:02
  • 1
    $\begingroup$ Moreover, DiscreteLimit[L[k], k -> Infinity] is unevaluated too. $\endgroup$
    – user64494
    Nov 5, 2018 at 20:11

1 Answer 1

3
$\begingroup$

The input restriction means it is basically not amenable to analysis by Limit or even DiscreteLimit. If instead it is defined piecewise then the sequence limit can be found.

lL[k_] := 
 Piecewise[{{(k/(k - 1)), Element[k, Integers]}, {Indeterminate, 
    True}}]
DiscreteLimit[lL[k], k -> Infinity]

(* Out[113]= 1 *)
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