What is
$$\lim_{x\to 0}\left(\frac{x}{e^{-x}+x-1}\right)^x$$
Using the expansion of $e^x$, I get that the function
$$y=\left(\frac{x}{e^{-x}+x-1}\right)^x$$
is not defined for negative numbers.
Hence the limit at $0^{-}$ must not exist.$\implies$The limit at $0$ does not exist.
However WA says that it should be $1$. :(
Am I wrong?