Question-60716

Question Number 60716 by peter frank last updated on 24/May/19

Commented by maxmathsup by imad last updated on 25/May/19

l e t A (x) = {(\frac{x}{x + 1})}^{x} \Rightarrow A (x) = {(1 - \frac{x + 1}{x})}^{x} = e^{x l n (1 - \frac{1}{x + 1})}

b u t w e h a v e f o r u \in V (0) l n (1 - u) \sim - u \Rightarrow l n (1 - \frac{1}{x + 1}) \sim - \frac{1}{x + 1} f o r x \in V (+ \infty)

\Rightarrow x l n (1 - \frac{1}{x + 1}) \sim - \frac{x}{x + 1} \to - 1 (x \to + \infty) \Rightarrow {l i m}_{x \to + \infty} A (x) = e^{- 1} = \frac{1}{e}

★ {l i m}_{x \to + \infty} {(\frac{x}{x + 1})}^{x} = \frac{1}{e} ★ .

Answered by Smail last updated on 24/May/19

{(\frac{x}{1 + x})}^{x} = {(\frac{1}{1 + 1 / x})}^{x} = {(1 + \frac{1}{x})}^{- x} = e^{- x l n (1 + 1 / x)}

L e t t = 1 / x

e^{- x l n (1 + 1 / x)} = e^{- \frac{l n (1 + t)}{t}}

l n (1 + t) \underset{0}{\sim} t

S o e^{- \frac{l n (1 + t)}{t}} \underset{0}{\sim} e^{- \frac{t}{t}} = \frac{1}{e}

T h u s, {\underset{x \to \infty}{l i m e}}^{- x l n (1 + 1 / x)} = \underset{x \to \infty}{l i m} {(\frac{x}{1 + x})}^{x} = \frac{1}{e}

Commented by peter frank last updated on 24/May/19

t h a n k y o u