Evaluate-the-following-limit-lim-n-1-n-d-n-dx-n-e-x-1-x-1-x-0-

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

Evaluate the following limit :

Φ = \lim_{n \to \infty} - \frac{1}{n!} \cdot \frac{d^{n}}{{dx}^{n}} (\frac{e^{x - 1}}{x - 1}) ∣_{x = 0}

Answered by Mathspace last updated on 09/Jan/22

f (x) = \frac{e^{x - 1}}{x - 1} \Rightarrow Φ = {l i m}_{n \to \infty} - \frac{1}{n!} f^{(n)} (0)

f^{(n)} (x) = \frac{1}{e} {(\frac{e^{x}}{x - 1})}^{(n)}

{e f}^{(n)} (x) = \sum_{k = 0}^{n} C_{n}^{k} {(\frac{1}{x - 1})}^{(k)} {(e^{x})}^{(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^{(n)} (0) = - 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^{(n)} (0) = \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 {l i m}_{n \to + \infty} - \frac{1}{n!} f^{(n)} (0) =

\sum_{p = 0}^{\infty} \frac{1}{p!} = e (e^{x} = Σ \frac{x^{n}}{n!})

Commented by Mathspace last updated on 09/Jan/22

f^{(n)} (0) = \frac{1}{e} (- 1 - \dots) \Rightarrow

{l i m}_{n \to + \infty} - \frac{1}{n!} f^{(n)} (0) = \frac{1}{e} \sum_{p = 0}^{\infty} \frac{1}{p!} = 1

Commented by HongKing last updated on 09/Jan/22

perfect my dear Sir thank you so much

Commented by Mathspace last updated on 09/Jan/22

y o u a r e w e l c o m e s i r