A) \[\frac{1}{3}\]
B) \[\frac{2}{3}\]
C) \[\frac{-2}{3}\]
D) \[\frac{2}{9}\]
Correct Answer: D
Solution :
[d] Consider \[\underset{x\to \infty }{\mathop{\lim }}\,\left[ \frac{{{x}^{2}}}{3x-2}-\frac{x}{3} \right]\] \[=\underset{x\to \infty }{\mathop{\lim }}\,\left[ \frac{3{{x}^{2}}-x(3x-2)}{3(3x-2)} \right]=\underset{x\to \infty }{\mathop{\lim }}\,\frac{2x}{3(3x-2)}\] \[=\underset{x\to \infty }{\mathop{\lim }}\,\frac{2}{3}\frac{1}{\left( 3-\frac{2}{x} \right)}=\frac{2}{9}\]
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