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Question Number 136343 by snipers237 last updated on 21/Mar/21

$$\underset{x\to 0}{\lim }\frac{x^{2021}}{x-ln\left(\underset{k=0}{\sum ^{2020}}\frac{x^k}{k!}\right)}\stackrel{?}{=}2021!$$

Answered by mindispower last updated on 21/Mar/21

$$\frac{x^{n+1}}{x-ln\left(\sum _{k⩽n}\frac{x^k}{k!}\right)}=(n+1)!..claim$$$${\partial }^k{x}^{n+1}=(n+1).....(n+2-k)x^{n+1-k}$$$$x-ln\left(\sum _{k⩽n}\frac{x^k}{k!}\right)=g\left(x\right)$$$$g^{\prime }\left(x\right)=1-\frac{\underset{k=0}{\sum ^{n-1}}\frac{x^k}{k}}{\underset{k=0}{\sum ^n}\frac{x^k}{k!}}=\frac{\frac{x^n}{n!}}{\sum _{k⩽n}\frac{x^k}{k!}}$$$$hopitalrull\underset{x\to 0}{\lim }\frac{g\left(x\right)}{h\left(x\right)}=\underset{x\to 0}{\lim }\frac{g^{\prime }\left(x\right)}{h^{\prime }\left(x\right)}$$$$weget=\underset{x\to 0}{\lim }\frac{(n+1)x^n}{\frac{x^n}{\frac{n!}{\sum _{k⩽n}\frac{x^k}{k!}}}}=\underset{x\to 0}{\lim }(n+1)!\underset{k=0}{\sum ^n}\frac{x^k}{k!}=(n+1)!$$$$$$

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