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

Jul 4, 2016

1

#### Explanation:

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

if we scope this out we know that

$\frac{\ln \left({e}^{\infty} - 3 \left(\infty\right)\right)}{\infty} \approx \frac{\ln \left({e}^{\infty}\right)}{\infty}$ as ${\lim}_{x \to \infty} {e}^{x} > > 3 x$ and so $L i m = \frac{\infty}{\infty} = 1$. that's a first order guess but it's pretty straightforward.

but clearly we have the indeterminate form here so we can use L'Hopital's Rule

so

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

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

We can go again with L'Hopital as as ${\lim}_{x \to \infty} {e}^{x} > > 3 x$ so this is still $\frac{\infty}{\infty}$

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

it's starting to becomes obvious but we can go one final time with L'Hopital's Rule

$= {\lim}_{x \to \infty} \frac{{e}^{x}}{{e}^{x}} = 1$