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Question Number 162827 by saboorhalimi last updated on 01/Jan/22

Answered by Ar Brandon last updated on 01/Jan/22

$$\mathrm{g}\left(x\right)=\underset{r\to 0}{\lim }{({(x+1)}^{r+1}-x^{r+1})}^{\frac{1}{r}}$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{\mathrm{g}\left(x\right)}{x}=\underset{r\to 0,x\to \mathrm{\infty }}{\lim }{x}^{1+\frac{1}{r}-1}{({(1+\frac{1}{x})}^{r+1}-1)}^{\frac{1}{r}}$$$$=\underset{r\to 0,x\to \mathrm{\infty }}{\lim }{x}^{\frac{1}{r}}{(1+\frac{r+1}{x}-1)}^{\frac{1}{r}}=\underset{r\to 0}{\lim }{(r+1)}^{\frac{1}{r}}$$$$=\underset{r\to 0}{\lim }{e}^{\frac{1}{r}\ln (r+1)}=\underset{r\to 0}{\lim }{e}^{\frac{1}{r}\left(r\right)}=e$$

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