6.61 If $x>\frac{2}{3}$, then $\lg \left(x^{2}+3\right)-2 \lg x$ is
(A) negative.
(B) 0.
(C) 1.
(D) less than any specified positive number.
(E) greater than any specified positive number.
(16th American High School Mathematics Examination, 1965)
[Solution] From $\lg \left(x^{2}+3\right)-2 \lg x=\lg \frac{x^{2}+3}{x^{2}}=\lg \left(1+\frac{3}{x^{2}}\right)$, since $x>\frac{2}{3}$, when $x$ is sufficiently large, $\frac{3}{x^{2}}$ can be less than any given positive number $\varepsilon$, then $\lg \left(1+\frac{3}{x^{2}}\right)$ can be less than $\lg (1+\varepsilon)$, i.e., less than any given positive number. Therefore, the answer is $(D)$.