Calculating $\lim_{x \to 0} [(x+1)^x-1]^x$ How the following limit can be calculated: $$\lim_{x \to 0} [(x+1)^x-1]^x$$ ? I've estimated the $0^0$ limit by writing the function under the form $e^ {\ln\\{...\\}}$. Then, by applying twice l'Hospital rule, I've found the limit is 1. I need help in order to obviate these large calculations.

Note that by Taylor's expansions

* $\log (1+x)=x+o(x)$
* $e^x=1+x+o(x)$



we have

$$(1+x)^x=e^{x\log(1+x)}=e^{x^2+o(x^2)}=1+x^2+o(x^2)$$

thus

$$[(x+1)^x-1] ^x=[1+x^2+o(x^2)-1] ^x=(x^2+o(x^2))^x=e^{x(\log x^2+o(x^2))}\to 1$$

indeed

$$x[\log x^2+o(x^2)]=x[\log x^2+\log (1+o(1))]=2x\log x+x\log(1+o(1))\to0+0=0$$