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}