Five. (20 points) Let $a_{1}=\frac{1}{2}$,
$$
a_{n+1}=\frac{a_{n}}{(1-\sqrt{2})^{n+1} a_{n}+\sqrt{2}+1}(n=1,2, \cdots) \text {. }
$$

Find $\lim _{n \rightarrow \infty} \sqrt[n]{a_{n}}$.
Consider a more general sequence
$$
a_{1}=\frac{1}{p+q}, a_{n+1}=\frac{a_{n}}{p^{n+1} a_{n}+q} \text {. }
$$

Then $\frac{1}{a_{n+1}}=\frac{p^{n+1} a_{n}+q}{a_{n}}=p^{n+1}+\frac{q}{a_{n}}$.
So, $\frac{1}{a_{n+1}}-\frac{q}{a_{n}}=p^{n+1}$.
Therefore, $\left\{\frac{1}{a_{n+1}}-\frac{q}{a_{n}}\right\}$ is a geometric sequence with common ratio $p$.
Thus, $\frac{1}{a_{n+1}}-\frac{q}{a_{n}}=p\left(\frac{1}{a_{n}}-\frac{q}{a_{n-1}}\right)$.
Then $\frac{1}{a_{n+1}}-\frac{p}{a_{n}}=\frac{q}{a_{n}}-\frac{p q}{a_{n-1}}=q\left(\frac{1}{a_{n}}-\frac{p}{a_{n-1}}\right)$.
Hence, $\left\{\frac{1}{a_{n+1}}-\frac{p}{a_{n}}\right\}$ is a geometric sequence with common ratio $q$.
So, $\frac{1}{a_{n+1}}-\frac{p}{a_{n}}=\left(\frac{1}{a_{2}}-\frac{p}{a_{1}}\right) q^{n-1}$
$$
\begin{array}{l}
=\left(p^{2}+\frac{q}{a_{1}}-\frac{p}{a_{1}}\right) q^{n-1} \\
=\left[p^{2}+(q-p)(p+q)\right] q^{n-1}=q^{n+1} .
\end{array}
$$

From equations (1) and (2), eliminating $\frac{1}{a_{n+1}}$ gives
$\frac{1}{a_{n}}=\frac{p^{n+1}-q^{n+1}}{p-q}$, i.e., $a_{n}=\frac{p-q}{p^{n+1}-q^{n+1}}$.
Let $p=1-\sqrt{2}, q=1+\sqrt{2}$, then
$$
a_{n}=\frac{2 \sqrt{2}}{(1+\sqrt{2})^{n+1}-(1-\sqrt{2})^{n+1}} \text {. }
$$

Since $\frac{1}{2}(1+\sqrt{2})^{n+1}=(1+\sqrt{2})^{n+1}-\frac{1}{2}(1+\sqrt{2})^{n+1}$
$$
\begin{array}{l}
<(1+\sqrt{2})^{n+1}-\frac{1}{2}<(1+\sqrt{2})^{n+1}-(1-\sqrt{2})^{n+1} \\
<(1+\sqrt{2})^{n+1}+\frac{1}{2}<(1+\sqrt{2})^{n+1}+\frac{1}{2}(1+\sqrt{2})^{n+1} \\
=\frac{3}{2}(1+\sqrt{2})^{n+1},
\end{array}
$$

Therefore, $\frac{1}{4 \sqrt{2}}(1+\sqrt{2})^{n+1}<\frac{1}{a_{n}}<\frac{3}{4 \sqrt{2}}(1+\sqrt{2})^{n+1}$,
$$
\left(\frac{1}{4 \sqrt{2}}\right)^{\frac{1}{n}}(1+\sqrt{2})^{1+\frac{1}{n}}<\frac{1}{\sqrt[n]{a_{n}}}<\left(\frac{3}{4 \sqrt{2}}\right)^{\frac{1}{n}}(1+\sqrt{2})^{1+\frac{1}{n}} \text {. }
$$

Thus, $\lim _{n \rightarrow \infty} \frac{1}{\sqrt[n]{a_{n}}}=1+\sqrt{2}$, i.e., $\lim _{n \rightarrow \infty} \sqrt[n]{a_{n}}=\sqrt{2}-1$.