Limit $(e^x+x)^{1/x}$, when $x\to 0$ Can I expand e^x in the limit $$\large{\lim _{x\to 0}(e^x+x)^{1/x}},$$ just as $1+x$ according to the Taylor expansion? I mean is it normal to think about limits of a form $(1+x+o(x))^{1/x}$ just as about $e$? Explain please.
 A: let $$y = (e^x + x)^{1/x}.  $$  then 
$$\begin{align}\ln y &= \frac 1x \ln(e^x + x)\\
&=\frac1x\ln(1+x+\cdots + x)\\
&= \frac1x\ln(1 + 2x + \cdots)\\
&=\frac1x\left(2x+\cdots\right) = 2 + \cdots  \text{ as } x \to 0.\end{align}$$
therefore $$\lim_{x \to 0}(e^x + x)^{1/x} =e^2. $$
A: HINT:
Use
$$\left[(1+e^x-1+x)^{\dfrac1{e^x-1+x}}\right]^{\dfrac{e^x-1+x}x}$$
Use $\lim_{h\to0}\dfrac{e^h-1}h=1$ and $\lim_{n\to\infty}\left(1+\dfrac1n\right)^n=e$

Alternatively, $e^x=1+x+O(x^2)$
$(e^x+x)^{1/x}=(1+2x+O(x^2))^{1/x}$
$=\left[\{1+2x+O(x^2)\}^{\dfrac1{2x+O(x^2)}}\right]^{\dfrac{2x+O(x^2)}x}$

Alternatively,
Let $A=\lim_{x\to0}(e^x+x)^{1/x},\ln(A)=\lim_{x\to0}\dfrac{\ln(e^x+x)}x$ which is of the form $\dfrac00$
So, applying L'Hospital's Law,   $\ln(A)=\lim_{x\to0}\dfrac{e^x+1}{e^x+x}=?$
A: The limit is $e^2$ because of this one line computation: $$(e^x+x)^{1/x}=(1+2x+o(x))^{1/x}=(1+2x)^{1/x}(1+o(1)).$$
Elaborating where the last equality comes from:
$$
\begin{align*}
(1+o(x))^{1/x}&=1+o(1),\quad\text{by the Binomial Theorem}\\
\implies (1+2x)^{1/x}(1+o(x))^{1/x}&=(1+2x)^{1/x}(1+o(1))\\
\implies (1+2x+o(x))^{1/x}&=(1+2x)^{1/x}(1+o(1)).
\end{align*}
$$
A: Whenever we need to find limit of an expression of $\{f(x)\}^{g(x)}$ it is better to take logs. Thus if $L$ is the desired limit then
\begin{align}
\log L &= \log\left\{\lim_{x \to 0}(e^{x} + x)^{1/x}\right\}\notag\\
&= \lim_{x \to 0}\log(e^{x} + x)^{1/x}\text{ (via continuity of log)}\notag\\
&= \lim_{x \to 0}\frac{1}{x}\log(e^{x} + x)\notag\\
&= \lim_{x \to 0}\frac{1}{x}\log(1 + e^{x} + x - 1)\notag\\
&= \lim_{x \to 0}\frac{e^{x} + x - 1}{x}\cdot\frac{\log(1 + e^{x} + x - 1)}{e^{x} + x - 1}\notag\\
&= \lim_{x \to 0}\left(\frac{e^{x} - 1}{x} + 1\right)\cdot\lim_{t \to 0}\frac{\log(1 + t)}{t}\text{ (putting }t = e^{x} + x - 1)\notag\\
&= (1 + 1)\cdot 1 = 2\notag
\end{align}
Hence $L = e^{2}$.
