Why $lim_{n \to \infty}(1-\frac{1}{n})^{n} = e^{-1} $ ?
OR How can i find this rule, whats it name ?

Question

anonymous · Answer

It is very well known that
$$
\lim_{n->\infty} \left ( (1+\frac x n  \right)^n = e^x
$$
put x=-1 and you are done.

anonymous · Answer

Let
$$
a_n= \left ( 1+\frac x n  \right)^n\
\ln(a_n) = n \ln \left ( (1+\frac x n  \right)\
\ln(a_n) = \frac{ \ln \left ( (1+\frac x n  \right)}{\frac 1 n}\

$$
Apply L'Hospital's Rule (Derivative with respect to n)
you get$$
\lim_{n->\infty} \ln a_n= \lim_{n->\infty} -\frac{x}{\frac{(-1) \left(n^2
   \left(\frac{x}{n}+1\right)\right)}{n
   ^2}}=\
\lim_{n->\infty} \frac{x}{\frac{x}{n}+1}=x\

\lim_{n->\infty}   a_n =e^x

$$