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I have the following equation, which equals 1 for positive values of $n$, but at $n = \inf$ the equation becomes indeterminate.

4 n (1 - 1/2) 1/2 (1/Sqrt[n])^2

If I require the input from the equation when $n$ is very large (and in principle infinitely large), is the best practice simply to set $n$ to an arbitrarily very large value, or does Mathematica have a standard, best practice, solution to this kind of problem?

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  • $\begingroup$ Your example evaluates to one for every n... $\endgroup$ – Ulrich Neumann Nov 6 '18 at 12:16
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This command gives one

4 n (1 - 1/2) 1/2 (1/Sqrt[n])^2 /. n :> \[Infinity]

which is a simple replacement rule.

So do the following two

4 n (1 - 1/2) 1/2 (1/Sqrt[n])^2 // Series[#, {n, \[Infinity], 10}] &

4 n (1 - 1/2) 1/2 (1/Sqrt[n])^2 // Limit[#, {n -> \[Infinity]}] &

which is what I would do if the replacement rule gave indeterminate.

Hope this helps.

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