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I'm posting my whole thought process, but I'd like to ask specifically about whether is my expansion of $\log(1+y)$ into Taylor series good & allowed in this situation? Is there an easier (not saying this isn't an easy way) to compute this limit?

$$\\ \lim_{x\rightarrow \infty}x(e-(1+\frac{1}{x})^x) = \lim_{y \rightarrow 0}\frac{e-(1+y)^{\frac{1}{y}}}{y} = \lim_{y \rightarrow 0}\ \frac{e-e^\frac{\log(1+y)}{y}}{y} = \lim_{y \rightarrow 0} \ \frac{e-e^\frac{y-\frac{y^2}{2}+O(y^3)}{y}}{y} = \\ =\lim_{y \rightarrow 0}\ \frac{e-e^{1-\frac{y}{2}+O(y^2)}}{y}= e\lim_{y \rightarrow 0}\ \frac{1-e^{-\frac{y}{2}+O(y^2)}}{y} = (-e)\lim_{y \rightarrow 0}\ \frac{e^{-\frac{y}{2}+O(y^2)} - 1}{y} =\\ (-e)\lim_{y \rightarrow 0}\ \frac{e^{y (\frac{-1}{2})+O(y^2)} - 1}{y}$$

And now, from $$ \lim_{y \rightarrow 0}\ \frac{e^{ay} - 1}{y} = a$$

Follows

$$ (-e)*(\frac{-1}{2}) = \frac{e}{2}$$

Bag of Chips
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1 Answers1

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Considering $$A=\frac{e-(1+y)^{\frac{1}{y}}}{y}$$ let me focus first on $$B=(1+y)^{\frac{1}{y}}$$ $$\log(B)=\frac{1}{y}\log(1+y)=1-\frac{y}{2}+\frac{y^2}{3}+O\left(y^3\right)$$ $$B=e\times e^{\log(B)-1}=e\times\big(1-\frac{1 }{2}y+\frac{11 }{24}y^2+O\left(y^3\right)\big)$$ Now, replacing in $A$, $$A=\frac{e-B}y=\frac{e}{2}-\frac{11 e }{24}y+O\left(y^2\right)$$ which shows the limit and how it is approached.

This is very similar to what you did.

Claude Leibovici
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