Is function $(1+\frac{1}{x})^{x}$ defined on $(-1,0)$ interval? I have plotted function $(1+\frac{1}{x})^{x}$ using Maple and got following graph:

So it seems that function isn't defined on $(-1,0)$ interval , but if I take that $x=\frac{-1}{3}$ I can write:
$$y=(1+\frac{1}{\frac{-1}{3}})^{\frac{-1}{3}}=(1-3)^{\frac{-1}{3}}=(-2)^{\frac{-1}{3}}=\frac{1}{(-2)^{\frac{1}{3}}}=\frac{1}{\sqrt[3]{-2}}=\frac{-1}{\sqrt[3]{2}}$$ 
which is real number.So my question is :Is function $(1+\frac{1}{x})^{x}$ defined on $(-1,0)$ interval or not ?
 A: I guess here there is kind o a philosophical problem: for instance, if you take $x=-\frac{1}{2}$, you face the problem $\sqrt{-1}$. So one can say that there are some points in $(-1,0)$ where the function is defined and some other where is not. In general, it's quite difficult to say a priori which are the right points and so, in order to avoid contradiction, one prefers to exclude $(-1,0)$ from the domain.
This i by the way also what happens for the function $x^x$. One says that its domain is $(0,\infty)$, even if it is clear that it is well defined also for particular negative real numbers.
A: Another way to solve this is to allow complex values, and choose the principal value in the usual way.  Then your function IS defined and continuous on $(-1,0)$, but has non-real values everywhere there!  
Here it is, real part in green, imaginary part in red:

A: The expression $x^y$ is defined for $x>0$ if $y\not \in Z $ in the case of real numbers. Otherwise we could write, for example, $-\sqrt[3]{2}=\sqrt[3]{-2}=(-2)^{1/3}=(-2)^{2/6}=\sqrt[6]{(-2)^2}=\sqrt[6]{2^2}=\sqrt[3]{2}$. Contradiction.
Sincerely,
Tigran
