How can I solve this limit using L'Hospital's rule somewhere? I have to solve this limit $\lim _{x\to 0+}\left(\left(\left(1+x\right)^x-1\right)^x\right)$ as $x$ approaches $0$ from the positive numbers. I have tried to change it to e to the limit of the exponent and apply L'Hospital's on it but to no avail. Can someone help me?
 A: We begin by writing for $x>0$
$$\begin{align}
\left((1+x)^x-1\right)^x&=e^{x\log\left((1+x)^x-1\right)}\\\\
&=e^{x\log\left(e^{x\log(1+x)}-1\right)}
\end{align}$$
Next, using $\log(1+x)=x+O(x^2)$ and $e^{x\log(1+x)}-1=x^2+O(x^3)$, we see that
$$\left((1+x)^x-1\right)^x=e^{2x\log(x)}\underbrace{e^{x\log(1+O(x))}}_{\to 1 \,\text{as} \,x\to 0}$$
Finally, since $\lim_{x\to 0^+}x\log(x)\overbrace{=}^{LHR}=-\lim_{x\to 0^+} x=0$
we conclude that 
$$\left((1+x)^x-1\right)^x=1$$
A: $$ \lim_{x \to 0^+} e^{{ln((1+x)^x-1)^x}} $$
$$  e^{\lim_{x \to 0^+} \frac{ln((1+x)^x-1)}{\frac {1}{x}}} $$
$$  e^{\lim_{x \to 0^+} \frac{ln((1+x)^x-1)}{\frac {1}{x}}} $$
$$  e^{\lim_{x \to 0^+} \frac{\frac {1}{(1+x)^x-1} [(1+x)^x(\frac{x}{x+1}+ln(1+x))]}{\frac{-1}{x^2}}} $$
$$  e^{\lim_{x \to 0^+} \frac {-x^2}{(1+x^2+...-1)} [(1+x)^x(\frac{x}{x+1}+ln(1+x))]} =e^0=1$$

A: By the binomial formula, $$(1+x)^x-1=\sum_{k=1}^\infty\binom{x}{k}x^k=x+x^2+\frac{x^3(x-1)}{2}+\frac{x^4(x-1)(x-2)}{6}+\cdots,$$ so$$
\left((1+x)^x-1\right)^x=x^x\left(1+x+\frac{x^2(x-1)}{2}+\cdots\right)^x
$$
As $x\to0+$, both factors go to $1$.
A: We use binomial approximation $(1+h)^\nu \approx 1+\nu h$, if $h\to 0$.
$$L=\lim_{x\to 0^+} ((1+x)^x-1)^x= \lim_{x\to 0^+} ((1+x.x)-1)^x=\lim_{x\to 0^+} x^{2x}=
\exp[\lim_{x\to 0^+} 2x \log x]$$ $$=\exp[\lim_{x\to 0^+} 2 \frac{\log x}{1/x}]=\exp[\lim_{x\to 0^+} 2 \frac{1/x}{-1/x^2}]~~~ \text{(L-Hospital's Rulr)}=e^0=1.$$
