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

Evaluate the following limit: Φ =lim_(n→∞) - (1/(n!)) ∙ (d^n /dx^n ) ((e^(x-1) /(x - 1)))∣_(x=0)

$$\mathrm{Evaluate}\:\mathrm{the}\:\mathrm{following}\:\mathrm{limit}: \\ $$$$\Phi\:=\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:-\:\frac{\mathrm{1}}{\mathrm{n}!}\:\centerdot\:\frac{\mathrm{d}^{\boldsymbol{\mathrm{n}}} }{\mathrm{dx}^{\boldsymbol{\mathrm{n}}} }\:\left(\frac{\mathrm{e}^{\boldsymbol{\mathrm{x}}-\mathrm{1}} }{\mathrm{x}\:-\:\mathrm{1}}\right)\mid_{\boldsymbol{\mathrm{x}}=\mathrm{0}} \\ $$

Answered by Mathspace last updated on 09/Jan/22

f(x)=(e^(x−1) /(x−1)) ⇒Φ=lim_(n→∞) −(1/(n!))f^((n)) (0) f^((n)) (x)=(1/e)((e^x /(x−1)))^((n)) ef^((n)) (x)=Σ_(k=0) ^n C_n ^k ((1/(x−1)))^((k)) (e^x )^((n−k)) =(e^x /(x−1))+e^(x ) Σ_(k=1) ^n C_n ^k (((−1)^k k!)/((x−1)^(k+1) )) ⇒ f^((n)) (0)=−1+Σ_(k=1) ^(n ) C_n ^k (−1)^(k+1) (−1)^k k! =−1−Σ_(k=1) ^n ((n!)/((n−k)!)) ⇒ −(1/(n!))f^((n)) (0)=(1/(n!))+Σ_(k=1) ^n (1/((n−k)!)) =Σ_(k=0) ^n (1/((n−k)!))=_(n−k=p) Σ_(p=0) ^(n ) (1/(p!)) ⇒lim_(n→+∞) −(1/(n!))f^((n)) (0)= Σ_(p=0) ^∞ (1/(p!)) =e (e^x =Σ(x^n /(n!)))

$${f}\left({x}\right)=\frac{{e}^{{x}−\mathrm{1}} }{{x}−\mathrm{1}}\:\Rightarrow\Phi={lim}_{{n}\rightarrow\infty} −\frac{\mathrm{1}}{{n}!}{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$${f}^{\left({n}\right)} \left({x}\right)=\frac{\mathrm{1}}{{e}}\left(\frac{{e}^{{x}} }{{x}−\mathrm{1}}\right)^{\left({n}\right)} \\ $$$${ef}^{\left({n}\right)} \left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \left(\frac{\mathrm{1}}{{x}−\mathrm{1}}\right)^{\left({k}\right)} \left({e}^{{x}} \right)^{\left({n}−{k}\right)} \\ $$$$=\frac{{e}^{{x}} }{{x}−\mathrm{1}}+{e}^{{x}\:} \sum_{{k}=\mathrm{1}} ^{{n}} \:{C}_{{n}} ^{{k}} \frac{\left(−\mathrm{1}\right)^{{k}} {k}!}{\left({x}−\mathrm{1}\right)^{{k}+\mathrm{1}} }\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right)=−\mathrm{1}+\sum_{{k}=\mathrm{1}} ^{{n}\:} {C}_{{n}} ^{{k}} \left(−\mathrm{1}\right)^{{k}+\mathrm{1}} \left(−\mathrm{1}\right)^{{k}} {k}! \\ $$$$=−\mathrm{1}−\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{{n}!}{\left({n}−{k}\right)!}\:\Rightarrow \\ $$$$−\frac{\mathrm{1}}{{n}!}{f}^{\left({n}\right)} \left(\mathrm{0}\right)=\frac{\mathrm{1}}{{n}!}+\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{\left({n}−{k}\right)!} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \:\frac{\mathrm{1}}{\left({n}−{k}\right)!}=_{{n}−{k}={p}} \:\sum_{{p}=\mathrm{0}} ^{{n}\:} \frac{\mathrm{1}}{{p}!} \\ $$$$\Rightarrow{lim}_{{n}\rightarrow+\infty} −\frac{\mathrm{1}}{{n}!}{f}^{\left({n}\right)} \left(\mathrm{0}\right)= \\ $$$$\sum_{{p}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{{p}!}\:={e}\:\:\:\left({e}^{{x}} \:=\Sigma\frac{{x}^{{n}} }{{n}!}\right) \\ $$

Commented by Mathspace last updated on 09/Jan/22

f^((n)) (0)=(1/e)(−1−...) ⇒ lim_(n→+∞) −(1/(n!))f^((n)) (0)=(1/e)Σ_(p=0) ^∞ (1/(p!)) =1

$${f}^{\left({n}\right)} \left(\mathrm{0}\right)=\frac{\mathrm{1}}{{e}}\left(−\mathrm{1}−...\right)\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} −\frac{\mathrm{1}}{{n}!}{f}^{\left({n}\right)} \left(\mathrm{0}\right)=\frac{\mathrm{1}}{{e}}\sum_{{p}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{{p}!}\:=\mathrm{1} \\ $$

Commented by HongKing last updated on 09/Jan/22

perfect my dear Sir thank you so much

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

Commented by Mathspace last updated on 09/Jan/22

you are welcome sir

$${you}\:{are}\:{welcome}\:{sir} \\ $$

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