A) 2/9
B) 1/9
C) 7/9
D) None of these
Correct Answer: A
Solution :
\[\underset{x\to \infty }{\mathop{\lim }}\,(-x)\left[ {{({{x}^{3}}+{{x}^{2}}+1)}^{1/3}}\,\,+\,\,{{({{x}^{3}}-{{x}^{2}}+1)}^{1/3}}-2x \right]\] \[=\,\,\underset{x\to \infty }{\mathop{\lim }}\,(-x)\,\left[ x{{\left( 1+\frac{1}{x}+\frac{1}{{{x}^{3}}} \right)}^{1/3}}\,\,+\,\,x{{\left( 1-\frac{1}{x}+\frac{1}{{{x}^{3}}} \right)}^{1/3}}-2x \right]\] \[=\,\,\underset{x\to \infty }{\mathop{\lim }}\,(-x)\,\left[ {{(1+X)}^{1/3}}+{{(1-Y)}^{1/3}}-2 \right]x\] where \[X=\frac{1}{x}+\frac{1}{{{x}^{3}}}\,\,\,and\,\,\,Y=\frac{1}{x}-\frac{1}{{{x}^{3}}}\] \[M=\underset{x\to \infty }{\mathop{\lim }}\,(-{{x}^{2}})\left[ \begin{align} & \left( 1+\frac{X}{3}+\frac{1}{3}\left( \frac{1}{3}-1 \right)\frac{1}{2!}{{X}^{2}}+..... \right) \\ & +\left( 1-\frac{Y}{3}+\frac{1}{3}\left( \frac{1}{3}-1 \right)\frac{1}{2!}{{Y}^{2}}-... \right)-2 \\ \end{align} \right]\]\[=\,\,\underset{x\to \infty }{\mathop{\lim }}\,(-{{x}^{2}})\left[ \left( \frac{X-Y}{3} \right)+\frac{1}{3}\left( \frac{1}{3}-1 \right)\frac{1}{2!}({{X}^{2}}+{{Y}^{2}})+... \right]\]But \[X-Y=\frac{2}{{{x}^{3}}}\] \[{{X}^{2}}+{{Y}^{2}}={{\left( \frac{1}{x}+\frac{1}{{{x}^{3}}} \right)}^{2}}+{{\left( \frac{1}{x}-\frac{1}{{{x}^{3}}} \right)}^{2}}=\left( \frac{2}{{{x}^{2}}}+\frac{2}{{{x}^{6}}} \right)\] \[M=\underset{x\to \infty }{\mathop{lim}}\,(-{{x}^{2}})\,\left[ \frac{2}{3{{x}^{3}}}-\frac{1}{9}\,\left( \frac{2}{{{x}^{2}}}+\frac{2}{{{x}^{6}}} \right)+... \right]=\frac{2}{9}\]
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