Why won't Limit evaluate, and what can be done about it

Question

I am trying to evaluate $$\lim_{n\to \infty }\sqrt{n}{{n}\choose {[np + \sqrt{np(1-p)}]}}p^{[np + \sqrt{np(1-p)}]}(1-p)^{(n-[np + \sqrt{np(1-p)}])}$$ where $0<p<1$ and $[\cdot ]$ denotes the nearest integer (if it greatly simplifies things at no cost, I don't mind if $[\cdot]$ is removed):

Limit[Sqrt[n ] Binomial[n, Round[n p + Sqrt[n p (1 - p)]]] p^
Round[n p + Sqrt[n p (1 - p)]] (1 - p)^(
n -  Round[n p - Sqrt[n p (1 - p)]] ), n -> Infinity]

Mathematica (version 11.0, Student Edition) gives the input as the answer. This limit is of the type $0\times \infty$.

In hopes of simplifying things, I removed the nearest integer function in both of the exponents, but the same issue occurs:

Limit[Sqrt[n ] Binomial[n, Round[n p + Sqrt[n p (1 - p)]]] p^(
n p + Sqrt[n p (1 - p)]) (1 - p)^(n -  n p - Sqrt[n p (1 - p)] ), 
n -> Infinity, Assumptions -> 0 < p < 1]

Other Ideas: (1) I tried computing limits of products separately, but $\sqrt{n}p^{np + \sqrt{np(1-p)}}\to 0$ while the remaining product tends ComplextInfinity.

(2) I tried using bounds on the binomial coeffiecient: ${{a}\choose{b}}\le \frac{a^b}{b!}\le (\frac{a *e}{b})^b$, but the same issue occurs here when I use either of these upper bounds in place of the binomial coefficient:

Limit[Sqrt[n ] (n ^(n p + Sqrt[n p (1 - p)])/
Gamma[1 + n p + Sqrt[n p (1 - p)]]) p^(
n p + Sqrt[n p (1 - p)]) (1 - p)^(n -  n p - Sqrt[n p (1 - p)] ), 
n -> Infinity, Assumptions -> 0 < p < 1]

and

Limit[Sqrt[n ] ((n E)/(n p + Sqrt[n p (1 - p)]))^(
n p + Sqrt[n p (1 - p)]) p^(n p + Sqrt[n p (1 - p)]) (1 - p)^(
n -  n p - Sqrt[n p (1 - p)] ), n -> Infinity, 
Assumptions -> 0 < p < 1]

Follow up: Is there other software that is better at evaluating limits?

Answer 1

If we remove Round, Series (around $n\to\infty$) does the trick:

$Assumptions = n \[Element] Integers && n > 2 && 0 < p < 1;
Series[Sqrt[n] Binomial[n,(n p+Sqrt[n p(1-p)])] p^(n p+Sqrt[n p(1-p)])(1-p)^(n-(n p-Sqrt[n p(1-p)])), {n, Infinity, 0}]
Simplify[%] // Normal
Limit[%, n -> \[Infinity]]
(*0*)

This works for all $0<p<1$. For $p\equiv0$,

Limit[Sqrt[n] Binomial[n,(n p+Sqrt[n p(1-p)])] p^(n p+Sqrt[n p(1-p)]) (1-p)^(n-(n p-Sqrt[n p(1-p)])), p -> 0]
(*Sqrt[n]*)

and therefore the limit diverges (the exact same thing happens for $p\equiv 1$).

Answer 2

Use NLimit

expr = Sqrt[n] *
   Binomial[n, Round[n p + Sqrt[n p (1 - p)]]] *
   p^Round[n p + Sqrt[n p (1 - p)]] *
   (1 - p)^(n - Round[n p - Sqrt[n p (1 - p)]]);

Needs["NumericalCalculus`"];

Table[
  {p, NLimit[expr, n -> Infinity,
       WorkingPrecision -> #] & /@
     {15, 20, 25}} // Flatten,
  {p, 1/10, 9/10, 1/20}] // Grid

Increasing the WorkingPrecision produces progressively smaller values which supports the Limit being zero.

Answer 3

Actually Limit works if you just get rid of the Round.

Limit[Sqrt[n] Binomial[n, n p + Sqrt[n p (1 - p)]] 
   p^(n p + Sqrt[n p (1 - p)]) (1 - p)^(n - (n p - Sqrt[n p (1 - p)])),
   n -> Infinity, Assumptions -> 0 < p < 1]

0

Note for p near zero the limit is approached very slowly (just looking at it numerically)