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