# Question #8e58a

Nov 26, 2016

${\lim}_{n \to \infty} {\left(1 + \frac{1}{n}\right)}^{2 n} = {e}^{2}$

#### Explanation:

Using the limit definition of $e$:

One way to define $e$ is as $e : = {\lim}_{n \to \infty} {\left(1 + \frac{1}{n}\right)}^{n}$

Using this definition, we have

${\lim}_{n \to \infty} {\left(1 + \frac{1}{n}\right)}^{2 n} = {\lim}_{n \to \infty} {\left({\left(1 + \frac{1}{n}\right)}^{n}\right)}^{2}$

$= {\left({\lim}_{n \to \infty} {\left(1 + \frac{1}{n}\right)}^{n}\right)}^{2}$

(The above step follows from $f \left(x\right) = {x}^{2}$ being continuous)

$= {e}^{2}$

Using logarithms and L'Hopital's rule

${\lim}_{n \to \infty} {\left(1 + \frac{1}{n}\right)}^{2 n} = {\lim}_{n \to \infty} {e}^{\ln \left({\left(1 + \frac{1}{n}\right)}^{2 n}\right)}$

$= {\lim}_{n \to \infty} {e}^{2 n \ln \left(1 + \frac{1}{n}\right)}$

$= {e}^{{\lim}_{n \to \infty} 2 n \ln \left(1 + \frac{1}{n}\right)}$

(The above step follows from $f \left(x\right) = {e}^{x}$ being continuous)

${\lim}_{n \to \infty} 2 n \ln \left(1 + \frac{1}{n}\right) = {\lim}_{n \to \infty} \frac{2 \ln \left(1 + \frac{1}{n}\right)}{\frac{1}{n}}$

$= {\lim}_{n \to \infty} \frac{\frac{d}{\mathrm{dn}} 2 \ln \left(1 + \frac{1}{n}\right)}{\frac{d}{\mathrm{dn}} \frac{1}{n}}$

*(The above step follows from L'Hopital's rule. While it does not technically apply to discrete sequences, we could consider the continuous analog and apply L'Hopital's rule to get the same result.)

$= {\lim}_{n \to \infty} \frac{\frac{2}{1 + \frac{1}{n}} \left(- \frac{1}{n} ^ 2\right)}{- \frac{1}{n} ^ 2}$

$= {\lim}_{n \to \infty} \frac{2}{1 + \frac{1}{n}}$

$= \frac{2}{1 + 0}$

$= 2$

$\therefore {\lim}_{n \to \infty} {\left(1 + \frac{1}{n}\right)}^{2 n} = {e}^{{\lim}_{n \to \infty} 2 n \ln \left(1 + \frac{1}{n}\right)}$

$= {e}^{2}$