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Question Number 33168 by abdo imad last updated on 11/Apr/18

$$find{lim}_{n\to \mathrm{\infty }}{\int }_n^{n+1}\frac{{(t+n)}^{\frac{1}{3}}}{\sqrt{t}}dt.$$

Commented by abdo imad last updated on 13/Apr/18

$$\mathrm{\exists }\xi \in ]n,n+1[/I_n={\int }_n^{n+1}\frac{{(t+n)}^{\frac{1}{3}}}{\sqrt{t}}dt$$$${=}^3\sqrt{\xi +n}{\int }_n^{n+1}\frac{dt}{\sqrt{t}}={}^3\sqrt{\xi +n}[2\sqrt{t]}{}_n^{n+1}$$$$=2^3\sqrt{n+\xi }(\sqrt{n+1}-\sqrt{n})=\frac{2^3\sqrt{n+\xi }}{\sqrt{n+1}+\sqrt{n}}$$$$\sim \frac{2{\left(2n\right)}^{\frac{1}{3}}}{2\sqrt{n}}{=}^3\sqrt{3}{n}^{\frac{1}{3}-\frac{1}{2}}{=}^3\sqrt{2}{n}^{-\frac{1}{6}}\to 0(n\to \mathrm{\infty })so$$$${lim}_{n\to \mathrm{\infty }}{I}_n=0.$$

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