We are required to evaluate the limit:
\[
\lim_{x \to 1} x^{(1-x)}
\]
This is an indeterminate form of the type \(1^0\), so we can apply logarithms to simplify:
\[
y = x^{(1-x)} \quad \Rightarrow \quad \ln(y) = (1-x) \ln(x)
\]
As \(x \to 1\), we observe that \(\ln(x) \to 0\) and \(1 - x \to 0\), so applying L'Hôpital's rule to the limit of \(\frac{\ln(x)}{1-x}\) gives:
\[
\lim_{x \to 1} \frac{\ln(x)}{1-x} = -1
\]
Thus, \(\ln(y) = -1\), so \(y = e^{-1} = \frac{1}{e}\).