What is the natural domain of this function? I was wondering what the natural domain of $$(1+x)^{\frac{1}{x}}$$
was. I know it's defined on interval $x\in(-1,0)\cup (0,\infty)$. However, I'm not exactly sure what the rule is for the rest of the negative reals. Since $(1+x)$ would be negative, I know that I would need to avoid taking a square root, fourth root, etc., which would depend on the decimal that I end up choosing. Is there a rigorous way to define the domain?
 A: I guess the question is talking about $\mathbb R$ to $\mathbb R$ function. Your main concern should be why $x > -1$, the rest is obvious.
For real number, one natural definition of exponential function could be from the exponent of a sequence of rational numbers which converges to the real number, i.e.
$$b^x := \lim_{r \in \mathbb Q \rightarrow x} b^r \, (b \in \mathbb R^+, x \in \mathbb R)$$
But the limit above only exists for positive $b$ (and is unique when $b >0$), that's why we have $x > -1$.
Another definition could be via logarithm:
$$b^x:=e^{x \cdot \ln(b)}; \, e^x:=\sum_{k=0}^\infty\frac{x^k}{k!}$$
This is more straightforward why we need $(1+x)>0$.
If $b<0$, things are more complicated: we might get complex number as results, and also we get multiple values (you could define a principal value, but identity function $(b^r)^s=b^{r\cdot s}$ will fail).
A: Since $(1+x)^\frac{1}{x}$ undefined for $x=0$ and also the given function  $(1+x)^\frac{1}{x}$ gives only positive value so the domain of this function is
$x\in(-1,0)\cup (0,\infty)$.
You can also verify you domain using graph

