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Question Number 163719 by HongKing last updated on 09/Jan/22

$$\mathrm{Evaluate}\mathrm{the}\mathrm{following}\mathrm{limit}:$$$$\mathrm{\Phi }=\underset{\mathbf{n}\to \mathrm{\infty }}{\lim }-\frac{1}{\mathrm{n}!}\cdot \frac{{\mathrm{d}}^{\mathbf{n}}}{{\mathrm{dx}}^{\mathbf{n}}}\left(\frac{{\mathrm{e}}^{\mathbf{x}-1}}{\mathrm{x}-1}\right){\mid }_{\mathbf{x}=0}$$

Answered by Mathspace last updated on 09/Jan/22

$$f\left(x\right)=\frac{e^{x-1}}{x-1}\Rightarrow \mathrm{\Phi }={lim}_{n\to \mathrm{\infty }}-\frac{1}{n!}{f}^{\left(n\right)}\left(0\right)$$$$f^{\left(n\right)}\left(x\right)=\frac{1}{e}{\left(\frac{e^x}{x-1}\right)}^{\left(n\right)}$$$${ef}^{\left(n\right)}\left(x\right)=\sum _{k=0}^n{C}_n^k{\left(\frac{1}{x-1}\right)}^{\left(k\right)}{\left(e^x\right)}^{(n-k)}$$$$=\frac{e^x}{x-1}+e^x\sum _{k=1}^n{C}_n^k\frac{{(-1)}^{k}k!}{{(x-1)}^{k+1}}\Rightarrow$$$$f^{\left(n\right)}\left(0\right)=-1+\sum _{k=1}^n{C}_n^k{(-1)}^{k+1}{(-1)}^{k}k!$$$$=-1-\sum _{k=1}^n\frac{n!}{(n-k)!}\Rightarrow$$$$-\frac{1}{n!}{f}^{\left(n\right)}\left(0\right)=\frac{1}{n!}+\sum _{k=1}^n\frac{1}{(n-k)!}$$$$=\sum _{k=0}^n\frac{1}{(n-k)!}{=}_{n-k=p}\sum _{p=0}^n\frac{1}{p!}$$$$\Rightarrow {lim}_{n\to +\mathrm{\infty }}-\frac{1}{n!}{f}^{\left(n\right)}\left(0\right)=$$$$\sum _{p=0}^{\mathrm{\infty }}\frac{1}{p!}=e(e^x=\mathrm{\Sigma }\frac{x^n}{n!})$$

Commented by Mathspace last updated on 09/Jan/22

$$f^{\left(n\right)}\left(0\right)=\frac{1}{e}(-1-...)\Rightarrow$$$${lim}_{n\to +\mathrm{\infty }}-\frac{1}{n!}{f}^{\left(n\right)}\left(0\right)=\frac{1}{e}\sum _{p=0}^{\mathrm{\infty }}\frac{1}{p!}=1$$

Commented by HongKing last updated on 09/Jan/22

$$\mathrm{perfect}\mathrm{my}\mathrm{dear}\mathrm{Sir}\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}$$

Commented by Mathspace last updated on 09/Jan/22

$$youarewelcomesir$$

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