# How do you find the limit of [1+(a/x)]^(bx) as x approaches infinity using l'hospital's rule?

Sep 1, 2016

$= {e}^{a b}$

#### Explanation:

${\lim}_{x \to \infty} {\left[1 + \left(\frac{a}{x}\right)\right]}^{b x}$

first you need to force it into an indeterminate form, and there's little to lose in trying to but the stuff in parentheses apart so we do this next step

${\lim}_{x \to \infty} \exp \left(\ln {\left[1 + \left(\frac{a}{x}\right)\right]}^{b x}\right)$

we can lift the $e$ out as it is a continuous function
$\exp \left({\lim}_{x \to \infty} b x \ln \left[1 + \left(\frac{a}{x}\right)\right]\right)$

we also notice that ${\lim}_{x \to \infty} \ln \left[1 + \left(\frac{a}{x}\right)\right] = \ln 1 = 0$

so the indeterminate form is

$\exp \left({\lim}_{x \to \infty} \frac{\ln \left[1 + \left(\frac{a}{x}\right)\right]}{\frac{1}{b x}}\right)$ and we can apply L'Hopital

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

$= \exp \left({\lim}_{x \to \infty} \frac{a b}{1 + \left(\frac{a}{x}\right)}\right)$

$= {e}^{a b}$