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Question Number 152447 by fotosy2k last updated on 28/Aug/21

Commented by fotosy2k last updated on 28/Aug/21

$$plshelpme$$

Answered by EDWIN88 last updated on 28/Aug/21

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

Commented by fotosy2k last updated on 30/Aug/21

$$thankyou$$

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