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Question Number 32666 by naka3546 last updated on 30/Mar/18

Commented by abdo imad last updated on 30/Mar/18

$$letput{A}_n=\frac{{(n+6)}^{\frac{n+6}{n}}-n^{\frac{n+6}{n}}}{{(n+3)}^{\frac{n+3}{n}}-n^{\frac{n+3}{n}}}$$$$A_n=\frac{n^{\frac{n+6}{n}}({(1+\frac{6}{n})}^{\frac{n+6}{n}}-1)}{n^{\frac{n+3}{n}}({(1+\frac{3}{n})}^{\frac{n+3}{n}}-1)}=n^{\frac{3}{n}}\frac{{(1+\frac{6}{n})}^{\frac{n+6}{n}}-1}{{(1+\frac{3}{n})}^{\frac{n+3}{n}}-1}but$$$${(1+\frac{6}{n})}^{\frac{n+6}{n}}\sim 1+\frac{6(n+6)}{n^2}\Rightarrow {(1+\frac{6}{n})}^{\frac{n+6}{n}}-1\sim \frac{6(n+6)}{n^2}$$$${(1+\frac{3}{n})}^{\frac{n+3}{n}}\sim 1+\frac{3(n+3)}{n^2}\Rightarrow {(1+\frac{3}{n})}^{\frac{n+3}{n}}-1\sim \frac{3(n+3)}{n^2}$$$$\Rightarrow A_n\sim n^{\frac{3}{n}}\frac{6n+36}{3n+9}\Rightarrow A_n\sim 2e_{n\to \mathrm{\infty }}^{\frac{3}{n}ln\left(n\right)}\to 2so$$$${lim}_{n\to \mathrm{\infty }}{A}_n=2.$$$$$$$$$$

Commented by MJS last updated on 31/Mar/18

$${(n+a)}^{\frac{n+a}{n}}-n^{\frac{n+a}{n}}=$$$$={(n+a)}^{1+\frac{a}{n}}-n^{1+\frac{a}{n}}=$$$$=(n+a){(n+a)}^{\frac{a}{n}}-n\times n^{\frac{a}{n}}$$$${\mathrm{I}}^{\prime }\mathrm{m}\mathrm{very}\mathrm{bad}\mathrm{in}\mathrm{these}\mathrm{things},\mathrm{but}$$$$\mathrm{it}\mathrm{seems}\mathrm{obvious}\mathrm{that}{(n+a)}^{\frac{a}{n}}\mathrm{and}$$$$n^{\frac{a}{n}}\mathrm{both}\mathrm{have}\mathrm{the}\mathrm{limit}1({\mathrm{I}}^{\prime }\mathrm{m}$$$$\mathrm{sorry}\mathrm{I}\mathrm{cannot}\mathrm{show}\mathrm{this})\mathrm{and}\mathrm{we}$$$$\mathrm{would}\mathrm{get}(n+a)-n=a\Rightarrow \frac{6}{3}=2$$$$\mathrm{Please}\mathrm{be}\mathrm{so}\mathrm{kind}\mathrm{and}\mathrm{critizise}$$$$\mathrm{this}‘‘{\mathrm{method}}^{″}$$

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