Mir ist, als habest du schon nicht verstanden zu haben, weshalb \(\lim\limits_{n\to\infty}\left (1+\dfrac{1}{n}\right)^n=e\) ist.
\(\lim\limits_{n\to\infty}\left (1+\dfrac{1}{2n}\right)^n \\
=\lim\limits_{n\to\infty}\exp \left [ n \ln \left (1+\dfrac{1}{2n}\right) \right ] \\
=\exp \left [ \lim\limits_{n\to\infty} \left( n \ln \left (1+\dfrac{1}{2n}\right)\right) \right ]\\
=\exp \left [ \lim\limits_{n\to\infty} \left( \dfrac{ \ln \left (1+\dfrac{1}{2n}\right)}{1/n} \right) \right ]\\
=\exp \left [ \lim\limits_{n\to\infty} \left( \dfrac{ \left[\ln \left (1+\dfrac{1}{2n}\right)\right]'}{[1/n]'} \right) \right ]\\
=\exp \left [ \lim\limits_{n\to\infty} \left( \dfrac{n}{2n+1} \right) \right ]\\
=\exp \left ( \dfrac{1}{2} \right )=\sqrt{e}\)
I.Ü. gilt \(\lim\limits_{n\to\infty}\left (1+\dfrac{1}{\alpha n}\right)^n=\sqrt[\alpha]{e}\).