# What is the limit of(1+(1/x))^x as x approaches infinity?

Aug 6, 2016

Make the limit of (1+(1/x))^x as x approaches infinity equal to any variable e.g. y, k. and take the natural logarithm of both sides.

#### Explanation:

$y = {\lim}_{x - \infty} {\left(1 + \left(\frac{1}{x}\right)\right)}^{x}$
$\ln y = {\lim}_{x - \infty} \ln {\left(1 + \left(\frac{1}{x}\right)\right)}^{x}$
$\ln y = {\lim}_{x - \infty} x \ln \left(1 + \left(\frac{1}{x}\right)\right)$
$\ln y = {\lim}_{x - \infty} \ln \frac{1 + \left(\frac{1}{x}\right)}{x} ^ - 1$
if x is substituted directly, the value will be undefined, so
l'hopital's rule is applied.
l'hopital's rule says that if ${\lim}_{x - a} f \left(x\right) = 0 = {\lim}_{x - a} g \left(x\right)$,
then ${\lim}_{x - a} \left(f \frac{x}{g} \left(x\right)\right) = {\lim}_{x - a} \left(\frac{f ' \left(x\right)}{g ' \left(x\right)}\right)$
$\ln y = {\lim}_{x - \infty} \frac{\left(\frac{1}{1 + \left(\frac{1}{x}\right)}\right) \left(0 - 1 {x}^{-} 2\right)}{- 1 {x}^{-} 2}$
$\ln y = {\lim}_{x - \infty} \left(\frac{1}{1 + \left(\frac{1}{x}\right)}\right)$
substitute for x
$\ln y = \left(\frac{1}{1 + 0}\right)$
$\ln y = 1$
introduce exponential $e$
${e}^{\ln} y = {e}^{1}$
$y = e$
$y = e = {\lim}_{x - \infty} {\left(1 + \left(\frac{1}{x}\right)\right)}^{x}$
${\lim}_{x - \infty} {\left(1 + \left(\frac{1}{x}\right)\right)}^{x} = e$