= Solution
Use the <root limit of a finite sum of powers>. Put $M=\max\{1,a,a^2/2\}\geq1$. For every $x>0$, at least one of the three terms is $M^x$ and none is larger, so
$$
M^x\leq1+a^x+(a^2/2)^x\leq3M^x.
$$
Taking the positive $x$th root gives
$$
M\leq\left(1+a^x+(a^2/2)^x\right)^{1/x}\leq3^{1/x}M.
$$
Since $3^{1/x}\to1$, the <squeeze theorem> gives
$$
\boxed{\lim_{x\to\infty}\left(1+a^x+(a^2/2)^x\right)^{1/x}=\begin{cases}1,&0\leq a\leq1,\\a,&1<a\leq2,\\a^2/2,&a>2.\end{cases}}
$$
Indeed, for $a\leq1$ both remaining bases are at most $1$; for $1\leq a\leq2$, $a\geq a^2/2$; and for $a\geq2$, $a^2/2\geq a$. The endpoint cases $a=0,1,2$ are included in the same bound.
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