<s>[INST] Evaluate $\lim_{x\to0}\frac{e-(1+x)^\frac1x}{x}$ Somebody asked this and I think it's quite interesting as I couldn't figure out how to evaluate this but the Wolfram Alpha says its limit is $\frac e2$.
$$\lim_{x\to0}\frac{e-(1+x)^\frac1x}{x}$$
Could someone help here?
 [/INST] For $0<x<1$: We have $$\frac {1}{x}\ln (1+x)=1-x/2+(x^2/3-x^3/4)+(x^4/5-x^5/6)+...=$$ $$=1-x/2+x^2/3-(x^3-x^4/5)-(x^5-x^6/7)-...$$ Therefore $$1- x/2<\frac {1}{x}\ln (1+x)<1-x/2+x^2/3.$$ Therefore $$e(1-e^{-x/2+x^2/3})<e-(1+x)^{\frac {1}{x}} <e(1-e^{-x/2}).$$ 
Both $(-x/2+x^2/3)$ and $(-x/2)$ belong to the interval $(-1,0).$ When $y\in (-1,0)$ we have $$1+y<1+y+(y^2/2!+y^3/3!)+(y^4/4!+y^5/5!)+...=e^y$$ $$=1+y+y^2/2!+(y^3/3!+y^4/4)+..<1+y+y^2/2!.$$ So $$-y_1-y_1^2/2=1-(1+y_1+y_1^2/2!))<1-e^{y_1}$$ where $y_1=-x/2+x^2/3.$..... And also $$1-e^{-x/2}<1-(1-x/2!)=x/2.$$ Now $y_1+y_1^2/2=-x/2 +x^2F(x)$ where $F(x)$ is a polynomial, so for some $K>0$ we have  $x\in (0,1)\implies |F(x)|<K.$ Therefore $$ e/2-xK<\frac {e-(1+x)^{\frac {1}{x}}}{x}<e/2.$$
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