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Question Number 58067 by mustakim420 last updated on 17/Apr/19

Commented by maxmathsup by imad last updated on 17/Apr/19
$let A(x) =x^((1+(1/x))^x −e) ⇒A(x) = e^({(1+(1/x))^x −e}ln(x)) ⇒ln(A(x))={(1+(1/x))^x −e)}ln(x) but (1+(1/x))^x −e=e^(xln(1+(1/x))) −e =_(x=(1/t)) e^((1/t)ln(1+t)) −e x→∞ ⇒t→0^+ we have ln(1+t)}^((1)) =(1/(1+t)) =1−t +o(t^2 ) ⇒ ln(1+t) =t−(t^2 /2) +o(t^3 ) ⇒((ln(1+t))/t) =1−(t/2) +o(t^2 ) ⇒e^((ln(1+t))/t) =e^(1−(t/2) +o(t^2 )) =e (1−(t/2) +o(t^2 )) ⇒(1+(1/x))^x −e =−(e/(2x)) +o((1/x^2 )) ⇒ln(A(x)) =(−(e/(2x)) +o((1/x^2 ))}ln(x) ⇒lim_(x→+∞) ln(A(x)) =0 ⇒lim_(x→∞) A(x) =1 .$
$l e t A (x) = x^{{(1 + \frac{1}{x})}^{x} - e} \Rightarrow A (x) = e^{{{(1 + \frac{1}{x})}^{x} - e} l n (x)} \Rightarrow l n (A (x)) = {{(1 + \frac{1}{x})}^{x} - e)} l n (x)$ $b u t {(1 + \frac{1}{x})}^{x} - e = e^{x l n (1 + \frac{1}{x})} - e =_{x = \frac{1}{t}} e^{\frac{1}{t} l n (1 + t)} - e$ ${x \to \infty \Rightarrow t \to 0^{+} w e h a v e l n (1 + t)}}^{(1)} = \frac{1}{1 + t} = 1 - t + o (t^{2}) \Rightarrow$ $l n (1 + t) = t - \frac{t^{2}}{2} + o (t^{3}) \Rightarrow \frac{l n (1 + t)}{t} = 1 - \frac{t}{2} + o (t^{2}) \Rightarrow e^{\frac{l n (1 + t)}{t}} = e^{1 - \frac{t}{2} + o (t^{2})}$ $= e (1 - \frac{t}{2} + o (t^{2})) \Rightarrow {(1 + \frac{1}{x})}^{x} - e = - \frac{e}{2 x} + o (\frac{1}{x^{2}}) \Rightarrow l n (A (x))$ $= (- \frac{e}{2 x} + o (\frac{1}{x^{2}})} l n (x) \Rightarrow {l i m}_{x \to + \infty} l n (A (x)) = 0 \Rightarrow {l i m}_{x \to \infty} A (x) = 1 .$
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