Answer:
Option D
Explanation:
Consider $\lim_{x \rightarrow \infty}\left(\frac{x^{2}}{3x-2}-\frac{x}{3}\right)$
=$\lim_{x \rightarrow \infty}\left[\frac{3x^{2}- x(3x-2)}{3(3x-2)}\right]$
= $\lim_{x \rightarrow \infty}\frac{2x}{3(3x-2)}$
= $\lim_{x \rightarrow \infty}\frac{2x}{3x(3-\frac{2}{x})}$
= $\lim_{x \rightarrow \infty}\frac{2}{3}\frac{1}{(3-\frac{2}{x})}$
= $\frac{2}{3}\times\frac{1}{(3-0)}$ = 2/9
