A) \[-\frac{1}{4}\]
B) \[-\frac{1}{2}\]
C) \[0\]
D) \[\frac{2}{9}\]
E) \[-\frac{6}{5}\]
Correct Answer: D
Solution :
\[\underset{x\to \infty }{\mathop{\lim }}\,\left( \frac{{{x}^{3}}}{3{{x}^{2}}-4}-\frac{{{x}^{2}}}{3x+2} \right)\] \[=\underset{x\to \infty }{\mathop{\lim }}\,\left( \frac{2({{x}^{3}}+2{{x}^{2}})}{9{{x}^{3}}+6{{x}^{2}}-12x-8} \right)\] \[=\underset{x\to \infty }{\mathop{\lim }}\,\left( \frac{2\left( 1+\frac{2}{x} \right)}{9+\frac{6}{x}-\frac{12}{{{x}^{2}}}-\frac{8}{{{x}^{3}}}} \right)\] \[=\frac{2.1}{9}=\frac{2}{9}\]
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