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?
 A: This limit can be evaluated by applying l'Hospital's rule twice. For the first time we differentiate 
$$(1+x)^{\frac{1}{x}}=e^{\frac{\ln (1+x)}{x}}$$ to get 
$$\frac{\frac{x}{1+x}-\ln (1+x)}{x^2}
e^{\frac{\ln (1+x)}{x}}.$$ 
Now $$e^{\frac{\ln (1+x)}{x}}\rightarrow e$$ so we need the limit of 
$$\frac{\frac{x}{1+x}-\ln (1+x)}{x^2}$$ another application of l'Hosptial gives
$$\frac{\frac{1}{(1+x)^2}-\frac{1}{1+x}}{2x}=-\frac{1}{2}\frac{1}{(1+x)^2}\rightarrow -\frac{1}{2}$$ 
So the limit is $-\frac{e}{2}$.
A: After playing around a bit, it becomes clear that $f(x) = \frac{\log(1+x)}{x}$ is an important function for this problem; we can see that $f(0) = 1$ easily enough, but we are also interested in $f'(0)$. This function is analytic, and its Taylor expansion at $x = 0$ (using the Taylor expansion for $\log(1+x)$) is
$$f(x) = \frac{- \sum_{k=1}^{\infty} \frac{(-1)^k x^k}{k}}{x} = -\sum_{k=1}^{\infty} \frac{(-1)^k x^{k-1}}{k} = \sum_{k=0}^{\infty} \frac{(-1)^k x^k}{k+1}\text{.}$$
Dropping the constant term and calculating the derivative, we get
\begin{align*}
  f'(x) &= \sum_{k=1}^{\infty}\frac{k (-1)^k x^{k-1}}{k+1} \\
  f'(0) &= -\frac{1}{2}
\end{align*}
Back to the original problem.
\begin{align*}
\lim_{x \to 0} \frac{e - (1+x)^{\frac{1}{x}}}{x}
&= \lim_{x \to 0} \frac{e - e^{f(x)}}{x} \\
&= \lim_{x \to 0} \frac{-f'(x)e^{f(x)}}{1} \\
&= - f'(0) e^{f(0)} \\
&= \frac{e}{2}
\text{.}
\end{align*}
A: We can proceed as follows
\begin{align}
L &= \lim_{x \to 0}\frac{e - (1 + x)^{1/x}}{x}\notag\\
&= \lim_{x \to 0}\dfrac{\exp(1) - \exp\left(\dfrac{\log(1 + x)}{x}\right)}{x}\notag\\
&= -e\lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + x)}{x} - 1\right) - 1}{x}\notag\\
&= -e\lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + x)}{x} - 1\right) - 1}{\dfrac{\log(1 + x)}{x} - 1}\cdot\dfrac{\dfrac{\log(1 + x)}{x} - 1}{x}\notag\\
&= -e\lim_{x \to 0}\frac{\log(1 + x) - x}{x^{2}}\notag
\end{align}
The last limit can be easily evaluated either by Taylor series, or via L'Hospital's Rule or using integration and the value of this limit is $-1/2$ so that the answer to our question is $e/2$.
A: $g(x) = \dfrac{\ln(1+x)}{x} = 1 - \dfrac{x}{2} + \dfrac{x^2}{3} + \dots $ is analytic on $(-1,1)$ which implies $h(x) = \exp g(x)$ has a power series expansion around $0$ say
$$h(x) = a_0 + a_1 x + a_2 x^2 + \dots $$ 
We have $a_0 = h(0) = \exp g(0) = e$,
$a_1 = h'(0) = g'(0) \exp g(0) = -e/2$ 
so
$\dfrac{h(x) - e}{x} \to a_1$ as $ x \to 0$ i.e., $\lim_{x\to0}\dfrac{(1+x)^{1/x}-e}{x} = -\dfrac{e}{2}$ or  $\lim_{x \to 0}\dfrac{e - (1+x)^{1/x}}{x} = \dfrac{e}{2}.$
A: 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 $$ e/2-xK<\frac {e-(1+x)^{\frac {1}{x}}}{x}<e/2.$$
A: You can think of the limit  $$\lim_{x\to0}\frac{e-(1+x)^\frac1x}{x}$$
as the derivative of the function $f(x)=(1+x)^\frac1x$ at point $x=0$.
Did that idea gave you any help?
Edit: You should define the value of $f(x)$ at $x=0$ namely : $f(0)=e$
