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Question Number 142668 by mathdanisur last updated on 03/Jun/21

$$\underset{n\to \mathrm{\infty }}{lim}{\left(\frac{n+6}{n}\right)}^{\frac{6}{n}}=?$$

Commented by iloveisrael last updated on 04/Jun/21

$$\underset{x\to \mathrm{\infty }}{\lim }{\left(\frac{n+6}{n}\right)}^{\frac{6}{n}}=$$$$e^{\underset{x\to \mathrm{\infty }}{\lim }(\frac{n+6}{n}-1).\frac{6}{n}}=$$$$e^{\underset{x\to \mathrm{\infty }}{\lim }\left(\frac{36}{n^2}\right)}=e^0=1$$

Commented by mathdanisur last updated on 05/Jun/21

$$thankssir$$

Answered by ajfour last updated on 03/Jun/21

$$\ln L=\underset{n\to \mathrm{\infty }}{\lim }\frac{6}{n}\ln (1+\frac{6}{n})=0$$$$L=1.$$

Answered by Ankushkumarparcha last updated on 03/Jun/21

$$Answer:e$$

Answered by mr W last updated on 03/Jun/21

$$\underset{n\to \mathrm{\infty }}{lim}{\left(\frac{n+6}{n}\right)}^{\frac{6}{n}}$$$$=\underset{n\to \mathrm{\infty }}{lim}{\left[{(1+\frac{1}{\frac{n}{6}})}^{\frac{n}{6}}\right]}^{\frac{36}{n^2}}$$$$=e^0=1$$

Commented by mathdanisur last updated on 05/Jun/21

$$thankssir$$

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