$sin({e}^{-n} )$ =$({e}^{-n}-{\frac{{e}^{-3n}}{3!}}+{\frac{{e}^{-5n}}{5!}}-{\frac{{e}^{-7n}}{7!}}+...)$
$\frac{sin({e}^{-n})}{{e}^{-n}}$=$(1-{\frac{{e}^{-2n}}{3!}}+{\frac{{e}^{-4n}}{5!}}-{\frac{{e}^{-6n}}{7!}}+...)$
$lim_{n->{\infty}}(\frac{sin({e}^{-n})}{{e}^{-n}})$ =$1$ dir. O halde limit durumunda $n$ ->$\infty$ iken , $sin({e}^{-n})$ yerine onun dengi olan ${e}^{-n}$ yazmak işimizi kolaylaştırır.
$=>$ $lim_{n->{\infty}}{[1-{sin({e}^{-n})}]}^{n}$=$lim_{n->{\infty}}{[1-{{e}^{-n}}]}^{n}$
$=>$ $lim_{n->{\infty}}{[1-{{e}^{-n}}]}^{n}$=$lim_{n->{\infty}}{[1-{\frac{1}{{e}^{n}}}]}^{n}$
$=>$ $lim_{n->{\infty}}{[1-{\frac{1}{{e}^{n}}}]}^{n}$=${e}^{lim_{n->{\infty}}[\frac{-n}{{e}^{n}}]}$
$lim_{n->{\infty}}[\frac{-n}{{e}^{n}}]$=$0$
$=>$ ${e}^{lim_{n->{\infty}}[\frac{-n}{{e}^{n}}]}$=${e}^{0}$=${1}$
Dolayısıyla , $lim_{n->{\infty}}{[1-{sin({e}^{-n})}]}^{n}$=${1}$