# How do you use L'hospital's rule to find the limit lim_(x->oo)(1+a/x)^(bx) ?

Sep 11, 2014

By using logarithm and l'Hopital's Rule,
${\lim}_{x \to \infty} {\left(1 + \frac{a}{x}\right)}^{b x} = {e}^{a b}$.

By using the substitution $t = \frac{a}{x}$ or equivalently $x = \frac{a}{t}$,
${\left(1 + \frac{a}{x}\right)}^{b x} = {\left(1 + t\right)}^{\frac{a b}{t}}$

By using logarithmic properties,
$= {e}^{\ln \left[{\left(1 + t\right)}^{\frac{a b}{t}}\right]} = {e}^{\frac{a b}{t} \ln \left(1 + t\right)} = {e}^{a b \frac{\ln \left(1 + t\right)}{t}}$

By l'Hopital's Rule,
${\lim}_{t \to 0} \frac{\ln \left(1 + t\right)}{t} = {\lim}_{t \to 0} \frac{\frac{1}{1 + t}}{1} = 1$

Hence,
${\lim}_{x \to \infty} {\left(1 + \frac{a}{x}\right)}^{b x} = {e}^{a b {\lim}_{t \to 0} \frac{\ln \left(1 + t\right)}{t}} = {e}^{a b}$

(Note: $t \to 0$ as $x \to \infty$)