Why is $\lim\limits_{n\to\infty} (1 + \frac{1}{2n})^n = e^{\frac{1}{2}}$ In my textbook it is stated that this is obvious: $\lim\limits_{n\to\infty} (1 + \frac{1}{2n})^n = e^{\frac{1}{2}}$.
However I feel stupid for not understanding why? What am I missing?
 A: If one knows the following Taylor series expansion, as $u \to 0$,
$$
\log(1+u)=u+O(u^2)
$$ then one may write, as $n \to \infty$,
$$
\left( 1+\frac{1}{2n} \right)^{n}=e^{\large n\log\left(1+\frac{1}{2n}\right)}=e^{\large n\left(\frac{1}{2n}+O\left(\frac{1}{n^2}\right)\right)}=e^{\frac{1}{2}+O\left(\frac{1}{n}\right)}
$$ which gives the announced result.
A: $$
(1+1/2n)^n = ((1+1/2n)^{2n})^{1/2} \to e^{1/2} \quad \text{when} \, n \to \infty 
$$
A: \begin{align}
  & t={{\left( 1+\frac{1}{2n} \right)}^{n}}\Rightarrow \,\,\,\ln \,t=\frac{\ln \left( 1+\frac{1}{2n} \right)}{\frac{1}{n}} \\ 
 & \ln \,t=\underset{n\to \infty }{\mathop{\lim }}\,\frac{\ln \left( 1+\frac{1}{2n} \right)}{\frac{1}{n}}=\underset{n\to \infty }{\mathop{\lim }}\frac{\frac{-\frac{1}{2{{n}^{2}}}}{1+\frac{1}{2n}}}{-\frac{1}{{{n}^{2}}}}=\frac{1}{2}\,\,\,\,\Rightarrow \,\,t=\sqrt{e} \\ 
\end{align}
A: Equivalently to Erik Joensson's answer, let m= 2n.  Then $(1+ 1/(2n))^n= (1+ 1/m)^{m/2}= ((1+ 1/m)^m)^{1/2}$.  Since x to the 1/2 is continuous for all positive x, the limit of this, as n goes to infinity, is the limit if $(1+ 1/m)^{1/2}$ to the 1/2 power- i.e. $e^{1/2}$.
A: $e^x=\lim_{n\to\infty}\left(1+\frac{x}n\right)^n$
$\lim_{n\to\infty}\left(1+\frac{1}{2n}\right)^n=\lim_{n\to\infty}\left(1+\frac{\tfrac{1}{2}}{n}\right)^n=e^{1/2}$
A: Puting $x=\frac{1}{2n}\iff xn=\frac1 2$, we have $$(1+\frac {1}{2n})^n=(1+x)^{\frac{x}{x}*n}=((1+x)^{\frac{1}{x}})^{xn}=((1+x)^{\frac{1}{x}})^{1/2}.$$ The well known limit of this expression when $x\to 0$ is $e^{1/2}$.
