Solution
= Solution
For real $x$ sufficiently close to zero, $1+x>0$. Taking the logarithm and using its <Maclaurin series> gives
$$
\log[(1+x)^{1/x}]=\frac{\log(1+x)}x
=1-\frac x2+\frac{x^2}{3}+O(x^3)\longrightarrow1.
$$
Continuity of the exponential then gives
$$
\boxed{\lim_{x\to0}(1+x)^{1/x}=e.}
$$
This is a two-sided real limit on the punctured neighbourhood where the base is positive.