A) \[\frac{2}{3}\]
B) \[1\]
C) 0
D) \[\infty \]
Correct Answer: A
Solution :
\[\underset{x\to \,\infty }{\mathop{\lim }}\,\,\,\frac{2{{x}^{2}}+3x+4}{3{{x}^{2}}+3x+4}=\underset{x\to \infty }{\mathop{\lim }}\,\,\,\frac{2+\frac{3}{x}+\frac{4}{{{x}^{2}}}}{3+\frac{3}{x}+\frac{4}{{{x}^{2}}}}=\frac{2}{3}\].
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