A) 1
B) -1
C) 0
D) \[-\infty \]
Correct Answer: C
Solution :
[c] \[\underset{n\to \infty }{\mathop{\lim }}\,\left[ \sqrt[3]{{{(n+1)}^{2}}}-\sqrt[3]{{{(n-1)}^{2}}} \right]\] \[=\underset{n\to \infty }{\mathop{\lim }}\,{{n}^{2/3}}\left[ {{\left( 1+\frac{1}{n} \right)}^{2/3}}-{{\left( 1-\frac{1}{n} \right)}^{2/3}} \right]\] \[=\underset{n\to \infty }{\mathop{\lim }}\,{{n}^{2/3}}\left[ \left( 1+\frac{2}{3}.\frac{1}{n}+\frac{\frac{2}{3}\left( \frac{2}{2}-1 \right)}{2!}\frac{1}{{{n}^{2}}}... \right)-\left( 1-\frac{2}{3}.\frac{1}{n}+\frac{\frac{2}{3}\left( \frac{2}{3}-1 \right)}{2!}\frac{1}{{{n}^{2}}}... \right) \right]\] \[=\underset{n\to \infty }{\mathop{\lim }}\,{{n}^{2/3}}\left[ \frac{4}{3}.\frac{1}{n}+\frac{8}{81}.\frac{1}{{{n}^{3}}}+... \right]\] \[=\left[ \frac{4}{3}.\frac{1}{{{n}^{1/3}}}+\frac{8}{81}.\frac{1}{{{n}^{7/3}}}+... \right]=0\]
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