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Question Number 125728 by Mammadli last updated on 13/Dec/20

$$\underset{\mathit{n}\to \mathrm{\infty }}{\mathit{l}\mathit{i}\mathit{m}}{(1+\frac{3}{\mathit{n}(\mathit{n}+3)})}^{{\mathit{n}}^2}=?$$

Answered by bramlexs22 last updated on 13/Dec/20

$$\underset{x\to \mathrm{\infty }}{\lim }{\left[{(1+\frac{3}{n^2+3n})}^{\frac{n^2+3n}{3}}\right]}^{\frac{3n^2}{n^2+3n}}$$$$=e^{\underset{x\to \mathrm{\infty }}{\lim }\left(\frac{3n^2}{n^2+3n}\right)}=e^3$$

Answered by Dwaipayan Shikari last updated on 13/Dec/20

$$\underset{n\to \mathrm{\infty }}{\lim }{(1+\frac{3}{n^2+3n})}^{n^2}={(1+\frac{3}{n^2+3n})}^{n^2+3n}{(1+\frac{3}{n^2+3n})}^{-3n}$$$$={(1+\frac{3}{n^2+3n})}^{\frac{n^2+3n}{3}.3}(1-\frac{9}{n+3})=e^3$$

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