A) \[\frac{3}{2}\]
B) \[\frac{2}{3}\]
C) 0
D) ¥
Correct Answer: B
Solution :
[b]\[=\underset{x\to \infty }{\mathop{\lim }}\,\frac{2{{x}^{3}}-4x+7}{3{{x}^{3}}+5x-4}\] Dividing numerator & denominator by \[{{x}^{3}},\]we have \[=\underset{x\to \infty }{\mathop{\lim }}\,\frac{2-\frac{4}{{{x}^{2}}}+\frac{7}{{{x}^{3}}}}{3+\frac{5}{x}-\frac{4}{{{x}^{3}}}}=\frac{2}{3}\] Hence, option [b]be current.You need to login to perform this action.
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