The value of \[\underset{x\to \infty }{\mathop{\lim }}\,\frac{(x+1)(3x+4)}{{{x}^{2}}(x-8)}\] is equal to [Pb. CET 2002]

The value of \[\underset{x\to \infty }{\mathop{\lim }}\,\frac{(x+1)(

1.	The value of \[\underset{x\to \infty }{\mathop{\lim }}\,\frac{(x+1)(3x+4)}{{{x}^{2}}(x-8)}\] is equal to [Pb. CET 2002]
A.	2
B.	3
C.	1
D.	0
Answer» E.

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The value of \[\underset{x\to \infty }{\mathop{\lim }}\,\frac{(x+1)(3x+4)}{{{x}^{2}}(x-8)}\] is equal to [Pb. CET 2002]

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