lim-x-0-1-x-1-x-e-ex-2-x-2-

Question Number 113742 by bemath last updated on 15/Sep/20

\lim_{x \to 0} \frac{{(1 + x)}^{\frac{1}{x}} - e - \frac{e x}{2}}{x^{2}} = ?

Answered by bobhans last updated on 15/Sep/20

L = \lim_{x \to 0} \frac{{(1 + x)}^{\frac{1}{x}} - e - \frac{e x}{2}}{x^{2}}

l e t y = {(1 + x)}^{\frac{1}{x}} \to \ln y = \frac{1}{x} \ln (1 + x)

\ln y = \frac{1}{x} (x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + \dots)

\ln y = 1 - \frac{x}{2} + \frac{x^{2}}{3} - \frac{x^{3}}{4} + \dots

y = e^{(1 - \frac{x}{2} + \frac{x^{2}}{3} - \frac{x^{3}}{4} + \dots)}

L = \lim_{x \to 0} \frac{e^{(1 - \frac{x}{2} + \frac{x^{2}}{3} - \frac{x^{3}}{4} + \dots)} - e - \frac{e x}{2}}{x^{2}}

L = \lim_{x \to 0} \frac{e (1 + (- \frac{x}{2} + \frac{x^{2}}{3} - \dots) + \frac{1}{2!} {(- \frac{x}{2} + \frac{x^{2}}{3} - \dots)}^{2} + \dots)}{x^{2}}

L = \lim_{x \to 0} \frac{(e + \frac{e x}{2} + \frac{11 {e x}^{2}}{24} + \dots) - e - \frac{e x}{2}}{x^{2}}

L = \lim_{x \to 0} \frac{\frac{11 {e x}^{2}}{24}}{x^{2}} = \frac{11 e}{24}

Answered by mathmax by abdo last updated on 16/Sep/20

let f (x) = \frac{{(1 + x)}^{\frac{1}{x}} - e - \frac{ex}{2}}{x^{2}} \Rightarrow f (x) = \frac{e^{\frac{1}{x} \ln (1 + x)} - e - \frac{ex}{2}}{x^{2}}

we have \frac{d}{dx} \ln (1 + x) = \frac{1}{1 + x} = 1 - x + x^{2} + o (x^{3}) \Rightarrow

\ln (1 + x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} + o (x^{4}) \Rightarrow \frac{\ln (1 + x)}{x} = 1 - \frac{x}{2} + \frac{x^{2}}{3} + o (x^{3}) \Rightarrow

e^{\frac{1}{x} \ln (1 + x)} = e e^{- \frac{x}{2} + \frac{x^{2}}{3} + o (x^{3})} \sim e (1 - \frac{x}{2} + \frac{x^{2}}{3}) = e - \frac{ex}{2} + \frac{{ex}^{2}}{3} \Rightarrow

e^{\frac{1}{x} \ln (1 + x)} - e + \frac{ex}{2} \sim \frac{{ex}^{2}}{3} \Rightarrow f (x) \sim \frac{e}{3} \Rightarrow \lim_{x \to 0} f (x) = \frac{e}{3}

i think the Q is \lim_{x \to 0} \frac{{(1 + x)}^{\frac{1}{x}} - e + \frac{ex}{2}}{x^{2}} \dots!