Q67 / 170 · Manipal (MET) 2016 · Mathematics

The value of $\lim _{n \rightarrow \infty}\left[\sqrt[3]{n^2-n^3}+n\right]$ is

  • Correct answer
    A
    $\frac{1}{3}$
  • B
    $-\frac{1}{3}$
  • C
    $\frac{2}{3}$
  • D
    $-\frac{2}{3}$

Solution

$\lim _{n \rightarrow \infty}\left[\sqrt[3]{n^2-n^3}+n\right]=\lim _{n \rightarrow \infty} n\left[\left(-1+\frac{1}{n}\right)^{1 / 3}+1^{1 / 3}\right]$ $=\lim _{n \rightarrow \infty} n \cdot \frac{\left(\frac{1}{n}-1\right)+1}{\left(\frac{1}{n}-1\right)^{2 / 3}+1-\left(\frac{1}{n}-1\right)^{1 / 3}}$ $\left[\because a+b=\frac{a^3+b^3}{a^2-a b+b^2}\right]$ $=\lim _{n \rightarrow \infty} \frac{1}{\left(\frac{1}{n}-1\right)^{2 / 3}+1-\left(\frac{1}{n}-1\right)^{1 / 3}}$ $=\frac{1}{1+1+1}=\frac{1}{3}$

Hence, the correct answer is option: 'A'.

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