# How do you find the limit of (e^x - 3*x)^ (1/x)  as x approaches infinity?

Jul 3, 2016

e

#### Explanation:

${\lim}_{x \to \infty} {\left({e}^{x} - 3 x\right)}^{\frac{1}{x}}$

$= {\lim}_{x \to \infty} \exp \left(\ln {\left({e}^{x} - 3 x\right)}^{\frac{1}{x}}\right)$

$= \exp \left({\lim}_{x \to \infty} \textcolor{red}{\ln \frac{{e}^{x} - 3 x}{x}}\right\}$

applying L'Hopital to term in red as it amounts to $\frac{\infty}{\infty}$ because ${e}^{x} > > 3 x$ as $x \to \infty$

$= \exp \left({\lim}_{x \to \infty} \frac{\frac{1}{{e}^{x} - 3 x} \cdot \left({e}^{x} - 3\right)}{1}\right)$

$= \exp \left({\lim}_{x \to \infty} \frac{{e}^{x} - 3}{{e}^{x} - 3 x}\right)$

$= \exp \left({\lim}_{x \to \infty} \frac{1 - 3 {e}^{- x}}{1 - \textcolor{g r e e n}{3 x {e}^{- x}}}\right)$

looking at the green term, we can see already that the exponential term is the dominant term but we can apply L'Hopital to see this through

${\lim}_{x \to \infty} 3 x {e}^{- x} = {\lim}_{x \to \infty} \frac{3 x}{e} ^ x = {\lim}_{x \to \infty} \frac{3}{e} ^ x = 0$

so the limit is

$\exp \left(\frac{1}{1}\right)$

$= e$