Ask Me Help Desk - Question about Calculus (Limit)..

The definition of e is
$e:=\lim_{n\rightarrow +\infty} \left(1+\frac{1}{n} \right)^n;$
this forces that for any sequence of integers $\{m_n\}\rightarrow \infty$ you have
$\lim_{n\rightarrow +\infty} \left(1+\frac{1}{m_n} \right)^{m_n}=e$

To prove $\lim_{x\rightarrow +\infty} \left(1+\frac{1}{x} \right)^x=e,$
it is enough to prove that this limit exists. First by direct computation you see that it is a monotone increasing function. Then given any sequence $\{x_n\}\rightarrow \infty$ you have
$[x_n]\leq x_n \leq [x_n]+1$ thus passing to the limits

$e=\lim_{n\rightarrow +\infty} \left(1+\frac{1}{[x_n]} \right)^{[x_n]}\leq<br /> \lim_{n\rightarrow +\infty} \left(1+\frac{1}{x_n} \right)^{x_n}\leq<br /> \lim_{n\rightarrow +\infty} \left(1+\frac{1}{[x_n]+1} \right)^{[x_n]+1}=e.$

To prove $\lim_{x\rightarrow -\infty} \left(1+\frac{1}{x} \right)^x=e,$ ,
use $\lim_{x\rightarrow +\infty} \left(1-\frac{1}{x} \right)^x=\frac{1}{e}$
(it comes from the more general characterization $\lim_{x\rightarrow +\infty} \left(1+\frac{y}{x} \right)^x=e^y$ ), thus

$\lim_{x\rightarrow -\infty} \left(1+\frac{1}{x} \right)^x=<br /> \lim_{x\rightarrow +\infty} \left(1-\frac{1}{x} \right)^{-x}=\left(\frac{1}{e}\right)^{-1}=e.$