Evaluate limit of $a_n = \bigg( 1 + \frac{2}{n} \bigg)^n$ $a_n = \bigg( 1 + \frac{2}{n} \bigg)^n$
$$\lim_{n\to\infty} \bigg(1+\frac{2}{\infty}\bigg)^\infty$$
so we need to use L'Hospital rule, I want to take the derivative of $\bigg( 1 + \frac{2}{n} \bigg)^n$, which I thought should be $(1 + \frac{2}{n})$, but am I doing this part wrong? Then I take the limit of the derivative and get 
$$\lim_{n\to\infty} 1 + \frac{2}{n} = 1$$
But the book says the answer is $e^2$. I am really lost where $e$ comes into play
 A: Look up how use l'Hopital's rule.  To use it on an indeterminate form $1^\infty$ as we have here, we need to take the logarithm, get an indeterminate form either $0/0$ or $\infty/\infty$, apply l'Hopital's rule to get the limit of that, then exponentiate.
\begin{align}
f(n) &= \left(1+\frac{2}{n}\right)^n,\qquad\text{indeterminate form }{1^\infty}
\\
\log f(n) &= n\log\left(1+\frac{2}{n}\right),\qquad\text{indeterminate form }\infty \times 0
\\
\log f(n) &= \frac{\log\left(1+2/n\right)}{1/n},\qquad\text{indeterminate form }\frac{0}{0}
\\
\log f(n) &\sim \frac{\frac{d}{dn}[\log\left(1+2/n\right)]}{\frac{d}{dn}[1/n]}\\
&\text{where I wrote $\sim$ for: "has the same limit as,}
\\
&\qquad\text{provided the limit on the right exists"}
\\
\log f(n) &\sim 
\frac{-2/n^2}{(-1/n^2)(1+2/n)} = \frac{2}{1+2/n}
\\
\lim\log f(n) &= 2
\\
\log \lim f(n) &= 2
\\
\lim f(n) &= e^2 .
\end{align}

Question: Why is $1^\infty$ indeterminate?  
Answer. Because we can get different results for limits of that form.  Example
$$
\lim_{n\to\infty}\left(1+\frac{2}{n}\right)^n = e^2
\\
\lim_{n\to\infty} 1^n = 1
\\
\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^{n^2} = +\infty
$$
and so on.
A: Write $$\left (1+\frac{2}{n} \right )^n = \left (1+ \frac{1}{\left (\frac{n}{2} \right )} \right )^{2\cdot\frac{n}{2}} $$
The result follows from
$$\left (1+\frac{1}{m}\right ) ^m \xrightarrow[m\to\infty]{}e $$
L'Hopital's rule unnecessary and it's not clear to me how one would use it to start with.
A: $\lim_{x\to a} (f(x))^{g(x)}$ $\\$ Where $\lim_{x\to a}f(x) =1\, and \lim_{x\to a}g(x)\,=\infty $ $\\$ 
Is $ e^{\lim_{x\to a}( g(x)(f(x)-1)) } $ $\\$ 
Hence  $ e^{\lim_{n\to \infty}({n}(\frac{2}{n}))}\, = \,e^2 $
