5. (10th Hope Cup - Senior High School Mathematics Test) Given $n \in \mathbf{N}^{*}$, constants $p, q$ are both greater than 1, and neither equals 2, then $\lim _{n \rightarrow \infty} \frac{p^{n+1}-q^{n}}{p^{n+2}-2 q^{n+1}}$ equals
A. $\frac{1}{p}$ or $\frac{1}{2 q}$
B. $-\frac{1}{p}$ or $-\frac{1}{2 q}$
C. $\frac{1}{p}$ or $\frac{1}{2 q}$ or $\frac{p-1}{p^{2}-2 p}$
D. $-\frac{1}{p}$ or $-\frac{1}{2 q}$ or $\frac{1-p}{p^{2}-2 p}$
5. Choose C.

Brief solution: When $p>q$, $\lim _{n \rightarrow \infty}\left(\frac{q}{p}\right)^{n}=0$, so $\lim _{n \rightarrow \infty} \frac{p^{n+1}-q^{n}}{p^{n+2}-2 q^{n+1}}=\lim _{n \rightarrow \infty} \frac{p-\left(\frac{q}{p}\right)^{n}}{p^{2}-2\left(\frac{q}{p}\right)^{n} \cdot q}=$ $\frac{p-0}{p^{2}-0}=\frac{1}{p}$; When $p<q$, $\lim _{n \rightarrow \infty}\left(\frac{p}{q}\right)^{n}=0$, so $\lim _{n \rightarrow \infty} \frac{p^{n+1}-q^{n}}{p^{n+2}-2 q^{n+1}}=\lim _{n \rightarrow \infty} \frac{p\left(\frac{p}{q}\right)^{n}-1}{p^{2}\left(\frac{p}{q}\right)^{n}-2 q}=$ $\frac{0-1}{0-2 q}=\frac{1}{2 q}$; When $p=q$, $\lim _{n \rightarrow \infty} \frac{p^{n+1}-q^{n}}{p^{n+2}-2 q^{n+1}}=\frac{p-1}{p^{2}-2 q}$.