Why does $a_n = (1+\frac{2}{n})^{n}$ converge to $e^2$? 
Determine whether $a_n =
 (1+\frac{2}{n})^{n}$ converges or
  diverges. If it converges, find the
  limit.

So I tried to say that $a_n = (1+\frac{2}{n})^{n} \Rightarrow \ln(a_n) = n\ln(1+\frac{2}{n})$.
Unfortunately I don't know what the next step is since I think that $n \rightarrow \infty$ as $n \rightarrow \infty$
and that $\ln(1+\frac{2}{n}) \rightarrow 0$ as $n\rightarrow \infty$, but somehow the solution is $e^{2}$...
Can someone please help fill me in on the steps in-between? Thanks! 
 A: If you already know that $\displaystyle \lim_{x\to 0^+}(1+x)^{\frac{1}{x}}= e$ and that $a_n\to a$ implies $a_n^2\to a^2$, then:
$$\lim_n \left( 1+\frac{2}{n}\right)^n =\lim_n \left[ \left( 1+\frac{2}{n}\right)^{\frac{n}{2}} \right]^2=\left[\lim_n \left( 1+\frac{2}{n}\right)^{\frac{n}{2}} \right]^2 =e^2 \; .$$
Obviously, one can also write:
$$\lim_n \left( 1+\frac{2}{n}\right)^n = \lim_n \left( 1+\frac{1}{\frac{n}{2}}\right)^n =\lim_n \left[ \left( 1+\frac{1}{\frac{n}{2}}\right)^{\frac{n}{2}} \right]^2=\left[\lim_n \left( 1+\frac{1}{\frac{n}{2}}\right)^{\frac{n}{2}} \right]^2\; ,$$
and use the limit $\displaystyle \lim_{y\to +\infty} (1+\tfrac{1}{y})^y =e$.
A: We know that the derivative of $\log x$ is $\frac{1}{x}$, so apply first principle differentiation to $\log x$:
$$
\begin{align}
\lim_{h\to0}\frac{\log(x+h)-\log(x)}{h}&=\frac{1}{x}\\
\lim_{h\to0}\log\left(1+\frac{h}{x}\right)^{\frac{1}{h}}&=\frac{1}{x}
\end{align}$$
Now replace $\frac{1}{h}$ with $n$:
$$\lim_{n\to\infty}\log\left(1+\frac{1}{nx}\right)^n=\frac{1}{x}$$
Now replace $\frac{1}{x}$ by $2$:
$$\begin{align}
\lim_{n\to\infty}\log\left(1+\frac{2}{n}\right)^n&=2\\
\lim_{n\to\infty}e^{\log\left(1+\frac{2}{n}\right)^n}&=e^2\\
\lim_{n\to\infty}\left(1+\frac{2}{n}\right)^n&=e^2
\end{align}$$
A: You are right that $\ln (1+ \frac{2}{n}) \to 0$. However, it approaches zero only like $1/n$. Therefore, $n \ln(1+ \frac{2}{n}) = 2 + O(1/n)$ (using Taylor) and you obtain
$$ a_n \to e^2.$$
