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I am asked to prove this statement $^{*}$. I am trying now, but it is getting to small and tiny steps that I even loose my way. my steps are as follows: $$\lim_{n\rightarrow \infty}(\sqrt[3]{n+\sqrt{n}}-\sqrt[3]{n})=0^{*}$$

$\lim_{n\rightarrow \infty}(\sqrt[3]{n+\sqrt{n}}-\sqrt[3]{n})=\dfrac{(\sqrt[3]{n+\sqrt{n}}-\sqrt[3]{n}) \cdot (\sqrt[3]{n+\sqrt{n}}+\sqrt[3]{n})}{(\sqrt[3]{n+\sqrt{n}}+\sqrt[3]{n})}=\dfrac{(\sqrt[3]{n+\sqrt{n}})^2-(\sqrt[3]{n})^2}{(\sqrt[3]{n+\sqrt{n}}+\sqrt[3]{n})}=\dfrac{\sqrt[3]{n^2+2n\sqrt{n}+n}}{\sqrt[3]{n+\sqrt{n}}+\sqrt[3]{n}}=\dfrac{(n+\sqrt{n})^{\frac{2}{3}}-n^{\frac{2}{3}}}{(n+\sqrt{n})^{\frac{1}{3}}+n^{\frac{1}{3}}}= .. help = 0$ $$if \quad n\rightarrow \infty$$

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Hint: $$a^3-b^3=(a-b)(a^2+ab+b^2)$$ Apply for $a=\sqrt[3]{n+\sqrt n}$ and $b=\sqrt[3] n$

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  • $\begingroup$ how? in my case, it is not cube, but it is cube root. can i easily replace it? $\endgroup$ – doniyor Jan 6 '13 at 11:37
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    $\begingroup$ But you can use the hint as $$ (a-b) = \frac{a^3-b^3}{a^2+ab+b^2} $$ $\endgroup$ – k1next Jan 6 '13 at 11:44
  • $\begingroup$ oh okay, thanks @macydanim $\endgroup$ – doniyor Jan 6 '13 at 11:45
  • $\begingroup$ now i came to conclusion that $\sqrt{n}<\sqrt[3]{n^2}$, so it converges to $0$, since $\dfrac{\sqrt{n}}{positive+\sqrt[3]{n^2}}$ $\endgroup$ – doniyor Jan 6 '13 at 11:51
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What Dennis wrote. Or, $$ \sqrt[3]{n+\sqrt{n}}-\sqrt[3]{n}=C\int_n^{n+\sqrt{n}}\frac{\mathrm dt}{t^{2/3}}\leqslant C\int_n^{n+\sqrt{n}}\frac{\mathrm dt}{C'n^{2/3}}=C''n^{1/2-2/3}\to0, $$ because $1/2\lt2/3$. Likewise, for every positive $A$ and $B$, $$ \sqrt[A]{n+\sqrt[B]{n}}-\sqrt[A]{n}\to0\iff\frac1A+\frac1B\lt1. $$

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