# Question #8d080

##### 1 Answer
Jun 11, 2017

${\lim}_{n \to \infty} {\left(1 + \frac{2}{n}\right)}^{4 n} = {e}^{8}$

#### Explanation:

Consider the sequence:

${a}_{n} = \ln \left({\left(1 + \frac{2}{n}\right)}^{4 n}\right) = 4 n \ln \left(1 + \frac{2}{n}\right) = 8 \left(\ln \frac{1 + \frac{2}{n}}{\frac{2}{n}}\right)$

Now we have :

${\lim}_{x \to 0} \ln \frac{1 + x}{x} = 1$

which means that the limit must be the same for any succession $\left\{{x}_{n}\right\}$ such that ${\lim}_{n \to \infty} {x}_{n} = 0$, so if we pose ${x}_{n} = \frac{2}{n}$ we have:

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

and then:

${\lim}_{n \to \infty} {a}_{n} = 8$

and since ${e}^{x}$ is a continuous function in all of $\mathbb{R}$:

${\lim}_{n \to \infty} {e}^{{a}_{n}} = {e}^{8}$

Then:

${e}^{{a}_{n}} = {e}^{\ln} \left({\left(1 + \frac{2}{n}\right)}^{4 n}\right) = {\left(1 + \frac{2}{n}\right)}^{4 n}$

${\lim}_{n \to \infty} {\left(1 + \frac{2}{n}\right)}^{4 n} = {e}^{8}$