# How do you evaluate the limit (x/(x+2))^x as x approaches oo?

Aug 5, 2016

$= \frac{1}{e} ^ 2$

#### Explanation:

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

there is a well known limit ${\lim}_{n \to \infty} {\left(1 + \frac{1}{n}\right)}^{n} = e$ ....Bernoulli's compounding formula so we can aim for that maybe

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

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

with sub $\frac{2}{x} = \frac{1}{y}$ , $x = 2 y$

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

using the power law of limits
$= {\left({\lim}_{x \to \infty} \frac{1}{1 + \frac{1}{y}} ^ \left(y\right)\right)}^{2}$

using the division law
$= {\left(\frac{{\lim}_{x \to \infty} 1}{{\lim}_{x \to \infty} {\left(1 + \frac{1}{y}\right)}^{y}}\right)}^{2}$

$= \frac{1}{e} ^ 2$

the law quoted are summarised here

IACOBUS BERNOULLI
MATHEMATICUS INCOMPARABILIS