= Solution
For $|x|<1$, the <geometric series> gives $(1+x)^{-1}=\sum_{j=0}^\infty(-x)^j$. Integrating from zero to $x$ yields the <Maclaurin series>
$$
\boxed{\ln(1+x)=\sum_{j=1}^\infty\frac{(-1)^{j+1}x^j}{j}
=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots.}
$$
On the real axis the series also converges at $x=1$ to $\ln2$, but diverges at $x=-1$.
For fixed real $a$ and sufficiently large positive $x$, the base in the requested limit is positive. Apply the series with argument $-a^2/x^2$:
$$
x^2\ln\left(1-\frac{a^2}{x^2}\right)
=-a^2-\frac{a^4}{2x^2}+O(x^{-4})\longrightarrow-a^2.
$$
Continuity of the <exponential function> therefore gives
$$
\boxed{\lim_{x\to\infty}\left(1-\frac{a^2}{x^2}\right)^{x^2}=e^{-a^2}.}
$$
This includes $a=0$, when the expression is identically one.
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