Evaluate $\lim_{x\to0} \big((1+x)^x-1\big)^x$ How can one evaluate this limit?
$$\lim_{x\to0} \big((1+x)^x-1\big)^x$$
I've already tried writing that using $f(x) = e^{\ln f(x)}$, but I don't dare going any further with this approach.
 A: Binomial expansion:
$$(1+x)^x=1+x^2+\frac12(x-1)x^3+\mathcal O(x^4)$$
Thus, we have
$$(1+x)^x-1=x^2+\frac12(x-1)x^3+\mathcal O(x^4)$$
Finally:
$$(x^2+\frac12(x-1)x^3+\mathcal O(x^4))^x=(x^x)^2(1+\frac12(x-1)x+\mathcal O(x^2))^x\\\to1\times1=1$$
Since $\lim_{x\to0}x^x=1$ and direct substitution into the second product.
A: $\lim_{x\rightarrow 0}((1+x)^x-1)^x = \lim_{x\rightarrow 0}\bigg(e^{x\ln(1+x)}-1\bigg)^x$
$\displaystyle  = \lim_{x\rightarrow 0}\bigg(1+\frac{x\ln(1+x)}{1!}+\frac{x^2\ln^2(1+x)}{2!}+\cdots \infty - 1\bigg)^x $
$\displaystyle = \lim_{x\rightarrow 0}x^x(\ln(1+x))^x\bigg(1+\ln(1+x)+\frac{\ln(1+x)}{2!}+\cdots \cdots \bigg)^{x} =  \lim_{x\rightarrow 0}x^x(\ln(1+x))^x$
$\displaystyle \lim_{x\rightarrow 0}x^x \times \lim_{x\rightarrow 0}(\ln(1+x))^x = 1\times \lim_{x\rightarrow 0}\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots \infty\right)^x$
$\displaystyle  = \lim_{x\rightarrow 0}x^x\times \lim_{x\rightarrow 0}\left(1-\frac{x}{2}+\frac{x^2}{3}+\cdots \right)^x = 1\times 1 = 1$
A: Here is another solution to this old posting bases in L'Hospital rule:
$$x\log((1+x)^x-1)=\frac{\log(1+x)^x-1)}{\tfrac{1}{x}}\sim (1+x)^x\Big(\frac{x}{1+x} +\log(1+x)\Big)\frac{x^2}{(1+x)^x -1}$$
The factor $x^2/((1+x)^x -1)$ may be estimated by L'Hopital rule:
$$\frac{x^2}{(1+x)^x-1}\sim \frac{1}{(1+x)^x}\frac{2x}{1-\tfrac{1}{1+x}+\log(1+x)}\xrightarrow{x\rightarrow0}1$$
since
$$ 
\frac{2x}{1-\frac{1}{1+x}+\log(1+x)}\sim \frac{2}{\tfrac{1}{(1+x)^2} +\tfrac{1}{1+x}}\xrightarrow{x\rightarrow0}1
$$
Putting things together we get
$$x\log((1+x)^x-1)\xrightarrow{x\rightarrow0} 0$$
