\displaystyle\mathop {\lim }\limits_{n \to \infty } \left(\sup_{k\in\{0,\ldots,n\} \,\,\mathrm{t.d.} \,\, a \le x_k \le b}
{\left| \frac{{\sqrt {2\pi npq} P\left( {X_n = k} \right)}}{{e^{ - \frac{{x_k^2 }}{2}} }} - 1\right|}\right)=0