lim-x-x-e-1-1-x-x-

Question Number 79236 by jagoll last updated on 23/Jan/20

\lim_{x \to + \infty} x {e - {(1 + \frac{1}{x})}^{x}} = ?

Commented by mathmax by abdo last updated on 23/Jan/20

l e t A (x) = x {e - {(1 + \frac{1}{x})}^{x}} w e h a v e {(1 + \frac{1}{x})}^{x} = e^{x l n (1 + \frac{1}{x})}

w e h a v e {l n}^{^{'}} (1 + u) = \frac{1}{1 + u} = 1 - u + o (u^{2}) \Rightarrow

l n (1 + u) = u - \frac{u^{2}}{2} + o (u^{3}) (u \sim o) \Rightarrow l n (1 + \frac{1}{x}) = \frac{1}{x} - \frac{1}{2 x^{2}} + o (\frac{1}{x^{3}}) (x \sim + \infty) \Rightarrow

x l n (1 + \frac{1}{x}) = 1 - \frac{1}{2 x} + o (\frac{1}{x^{2}}) \Rightarrow e^{x l n (1 + \frac{1}{x})} = e^{1 - \frac{1}{2 x} + o (\frac{1}{x^{2}})}

= e (1 - \frac{1}{2 x} + o (\frac{1}{x^{2}})) \Rightarrow e - {(1 + \frac{1}{x})}^{x} = \frac{e}{2 x} + o (\frac{1}{x^{2}}) \Rightarrow

x {e - {(1 + \frac{1}{x})}^{x}} = \frac{e}{2} + o (\frac{1}{x}) \Rightarrow {l i m}_{x \to + \infty} A (x) = \frac{e}{2}

Commented by jagoll last updated on 23/Jan/20

thank you sir

Commented by mathmax by abdo last updated on 24/Jan/20

y o u a r e w e l c o m e .

Answered by Smail last updated on 23/Jan/20

x = 1 / t

\underset{t \to 0}{l i m} \frac{e - {(1 + t)}^{1 / t}}{t} = \underset{t \to 0}{l i m} \frac{e - e^{\frac{l n (1 + t)}{t}}}{t}

\underset{t \to 0}{l i m} \frac{e - e^{(t - \frac{t^{2}}{2}) \frac{1}{t}}}{t} = \underset{t \to 0}{l i m} \frac{e - e (1 - t / 2)}{t} = \frac{e}{2}

Commented by jagoll last updated on 23/Jan/20

thank you sir

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