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Question Number 75230 by aliesam last updated on 08/Dec/19

Commented by mathmax by abdo last updated on 08/Dec/19
$e^(1/x) =1+(1/x) +o((1/x^2 )) ⇒x e^(1/x) =x+1+o((1/x)) (x→+∞) e^(1/(x+1)) =1+(1/(x+1)) +o((1/x^2 )) ⇒(x+1)e^(1/(x+1)) =x+1+1+o((1/x)) ⇒ (x+1)e^(1/(x+1)) −xe^(1/x) =x+2 −x−1 +o((1/x))=1+o((1/x)) ⇒ lim_(x→+∞) { (x+1)e^(1/(x+1)) −x e^(1/x) } =1$
$e^{\frac{1}{x}} = 1 + \frac{1}{x} + o (\frac{1}{x^{2}}) \Rightarrow x e^{\frac{1}{x}} = x + 1 + o (\frac{1}{x}) (x \to + \infty)$ $e^{\frac{1}{x + 1}} = 1 + \frac{1}{x + 1} + o (\frac{1}{x^{2}}) \Rightarrow (x + 1) e^{\frac{1}{x + 1}} = x + 1 + 1 + o (\frac{1}{x}) \Rightarrow$ $(x + 1) e^{\frac{1}{x + 1}} - {x e}^{\frac{1}{x}} = x + 2 - x - 1 + o (\frac{1}{x}) = 1 + o (\frac{1}{x}) \Rightarrow$ ${l i m}_{x \to + \infty} {(x + 1) e^{\frac{1}{x + 1}} - x e^{\frac{1}{x}}} = 1$
	Answered by mind is power last updated on 08/Dec/19

	$let f (z) = {ze}^{\frac{1}{z}}$ $b y m e a n v a l u e t h e o r e m$ $\exists c \in] x, x + 1 [$ $f (x + 1) - f (x) = f^{'} (c)$ $f^{'} (c) = e^{\frac{1}{c}} - \frac{e^{\frac{1}{c}}}{c}$ $since c > x \Rightarrow x \to + \infty, c \to + \infty$ $f^{'} (c) \to 1$ $\Rightarrow \lim ((x + 1) e^{\frac{1}{x + 1}} - {xe}^{\frac{1}{x}}) \to 1$
	Commented by aliesam last updated on 08/Dec/19

	$t h a n k y o u s i r$
	Commented by mind is power last updated on 08/Dec/19

	$y^{'} re welcom$
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