1. Let $a, b$ be constants satisfying $a>b>0$, for the sequence $\left\{\mathrm{x}_{\mathrm{n}}\right\}$ defined by the recurrence relation
$$
\cdots x_{1}=b, x_{n}-a x_{n-1}=b^{n}(n=2,3,
$$

answer the following questions:
(1) If $y_{n}=\frac{x_{n}}{b^{n^{2}}}$, find the general term of $\left\{y_{n}\right\}$;
(2) Find $\lim _{n \rightarrow \infty} \sqrt[n]{x_{n}}$
Solve: (1) From $x_{n}-a x_{n-1}=b^{n}$, we have $\frac{x_{n}}{b^{n}}-\frac{a}{b} \cdot \frac{x_{n-1}}{b^{n-1}}=1$,
thus $\mathrm{y}_{\mathrm{n}}-\frac{\mathrm{a}}{\mathrm{b}} \mathrm{y}_{\mathrm{n}-1}=1$.
Let $y-\frac{a}{b} y=1$, we get $y=\frac{b}{b-a}$,
hence $\frac{\mathrm{b}}{\mathrm{b}-\mathrm{a}}-\frac{\mathrm{a}}{\mathrm{b}}\left(\frac{\mathrm{b}}{\mathrm{b}-\mathrm{a}}\right)=1$.
Therefore, from (1)-(2), we have
$$
\begin{array}{l}
y_{n}-\frac{b}{b-a}-\frac{a}{b}\left(y_{n-1}-\frac{b}{b-a}\right)=0 . \\
\therefore y_{n}-\frac{b}{b-a}=\left(y_{1}-\frac{b}{b-a}\right)
\end{array}
$$
$$
\text { - }\left(\frac{\mathrm{a}}{\mathrm{b}}\right)^{\mathrm{n}-1} \text {. }
$$

Thus, we get the general term
$$
y_{n}=\frac{a}{a-b}\left(\frac{a}{b}\right)^{n-1}-\frac{b}{a-b} .
$$
(2) From (1)
$$
\begin{array}{l}
x_{n}=b^{n} y_{n}=\frac{a^{n}}{a-b} \cdot b \cdots \frac{b}{a-b} b^{n} \\
=\operatorname{lob}_{a-b}^{b}\left(a^{n}-b^{n}\right) \cdot \\
\therefore \lim _{n \rightarrow \infty} \sqrt[n]{x_{n}}=\lim \sqrt[n]{\frac{b}{a-b}} \\
\cdot \sqrt[n]{a^{n}-b^{n}} \\
=\lim _{n}^{\frac{b}{a-b}} \cdot a \cdot \sqrt[n]{1-\left(\frac{b}{a}\right)^{n}}=a .
\end{array}
$$