Question Number 136343 by snipers237 last updated on 21/Mar/21
$$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{{x}^{\mathrm{2021}} }{{x}−{ln}\left(\underset{{k}=\mathrm{0}} {\overset{\mathrm{2020}} {\sum}}\frac{{x}^{{k}} }{{k}!}\right)}\:\overset{?} {=}\:\mathrm{2021}!\: \\ $$
Answered by mindispower last updated on 21/Mar/21
$$\frac{{x}^{{n}+\mathrm{1}} }{{x}−{ln}\left(\underset{{k}\leqslant{n}} {\sum}\frac{{x}^{{k}} }{{k}!}\right)}=\left({n}+\mathrm{1}\right)!..{claim} \\ $$$$\partial^{{k}} {x}^{{n}+\mathrm{1}} =\left({n}+\mathrm{1}\right).....\left({n}+\mathrm{2}−{k}\right){x}^{{n}+\mathrm{1}−{k}} \\ $$$${x}−{ln}\left(\underset{{k}\leqslant{n}} {\sum}\frac{{x}^{{k}} }{{k}!}\right)={g}\left({x}\right) \\ $$$${g}'\left({x}\right)=\mathrm{1}−\frac{\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{{x}^{{k}} }{{k}}}{\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{x}^{{k}} }{{k}!}}=\frac{\frac{{x}^{{n}} }{{n}!}}{\underset{{k}\leqslant{n}} {\sum}\frac{{x}^{{k}} }{{k}!}} \\ $$$${hopital}\:{rull}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{g}\left({x}\right)}{{h}\left({x}\right)}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{g}'\left({x}\right)}{{h}'\left({x}\right)} \\ $$$${we}\:{get}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left({n}+\mathrm{1}\right){x}^{{n}} }{\frac{{x}^{{n}} }{\frac{{n}!}{\underset{{k}\leqslant{n}} {\sum}\frac{{x}^{{k}} }{{k}!}}}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left({n}+\mathrm{1}\right)!\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{x}^{{k}} }{{k}!}=\left({n}+\mathrm{1}\right)! \\ $$$$ \\ $$