# How do you find the limit of (1-[6/x])^x as x approaches infinity using l'hospital's rule?

Apr 21, 2016

${\lim}_{x \to \infty} {\left(1 - \frac{6}{x}\right)}^{x} = \frac{1}{e} ^ 6$

#### Explanation:

${\lim}_{x \to \infty} {\left(1 - \frac{6}{x}\right)}^{x} = {\lim}_{x \to \infty} {e}^{\ln} \left({\left(1 - \frac{6}{x}\right)}^{x}\right)$

$= {\lim}_{x \to \infty} {e}^{x \ln \left(1 - \frac{6}{x}\right)}$

$= {e}^{{\lim}_{x \to \infty} x \ln \left(1 - \frac{6}{x}\right)} \text{ (*)}$

(Note that the above step holds as ${e}^{x}$ is a continuous function)

${\lim}_{x \to \infty} x \ln \left(1 - \frac{6}{x}\right) = {\lim}_{x \to 0} \ln \frac{1 - 6 x}{x}$

As the above is a $\frac{0}{0}$ indeterminate form , we can apply L'hopital's rule.

${\lim}_{x \to 0} \ln \frac{1 - 6 x}{x} = {\lim}_{x \to 0} \frac{\frac{d}{\mathrm{dx}} \ln \left(1 - 6 x\right)}{\frac{d}{\mathrm{dx}} x}$

$= {\lim}_{x \to 0} \frac{- \frac{6}{1 - 6 x}}{1}$

$= - 6$

Substituting this back in for ${\lim}_{x \to \infty} x \ln \left(1 - \frac{6}{x}\right)$ in $\text{(*)}$ gives us

${\lim}_{x \to \infty} {\left(1 - \frac{6}{x}\right)}^{x} = {e}^{- 6} = \frac{1}{e} ^ 6$