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  4. The value of displaystyle lim n arrow ∞ √[3]n2-n3+n is

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

Limits and Derivatives

Solution:

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