I am having trouble showing that $\displaystyle \lim_{x\to \infty}(1+\frac{1}{x})^x = e$.
All I have gotten is that I somehow have to use the sequence $\displaystyle e_{n}=(1+\frac{1}{n})^n$. Im not really sure where to start though.
I am having trouble showing that $\displaystyle \lim_{x\to \infty}(1+\frac{1}{x})^x = e$.
All I have gotten is that I somehow have to use the sequence $\displaystyle e_{n}=(1+\frac{1}{n})^n$. Im not really sure where to start though.
One possible way is to valuate...
$\displaystyle \ln (1+\frac{1}{x})^{x} = x\cdot \ln (1+\frac{1}{x})= x \cdot (\frac{1}{x} - \frac{1}{2x^{2}} + \frac{1}{3x^{3}} - ...)$ (1)
From (1) is evident that...
$\displaystyle \lim_{x \rightarrow \infty} x\cdot \ln (1+\frac{1}{x}) =1$ (2)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$