Solution
= Solution
For fixed real $a$, the base is positive for all sufficiently large positive $x$. Take its <logarithm> and use the <Taylor expansion> $\ln(1-u)=-u+O(u^2)$:
$$
\ln\left[\left(1-\frac ax\right)^x\right]=x\ln\left(1-\frac ax\right)=-a+O(x^{-1}).
$$
<Continuity> of the <exponential function> then gives $\boxed{\lim_{x\to\infty}(1-a/x)^x=e^{-a}}$. This includes $a=0$.