$10 \cdot 87$ If $x$ is a positive real number that increases without bound, then $\log _{3}(6 x-5)-\log _{3}(2 x+1)$ approaches
(A) 0 .
(B) 1 .
(C) 3 .
(D) 4 .
(E) not a finite number.
(18th American High School Mathematics Examination, 1967)
[Solution] From the given and properties of logarithms, we have
$$
\begin{array}{l}
\log _{3}(6 x-5)-\log _{3}(2 x+1)=\log _{3} \frac{6 x-5}{2 x+1}=\log _{3}\left(3-\frac{8}{2 x+1}\right) . \\
\because \quad \lim _{x \rightarrow+\infty} \frac{8}{2 x+1}=0, \\
\therefore \quad \lim _{x \rightarrow+\infty} \log _{3}\left(3-\frac{8}{2 x+1}\right)=\log _{3} 3=1 .
\end{array}
$$

Therefore, the correct choice is $(B)$.