Question

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

Answer 1

1

Explanation:

`lim_{x to oo} (ln (e^x - 3x)) / x`

if we scope this out we know that

`(ln (e^(oo) - 3(oo))) / (oo) approx (ln (e^(oo) )) / (oo)` as `lim_{x to oo} e^(x) > > 3x` and so `Lim = oo/oo = 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 oo} (ln (e^x - 3x)) / x = lim_{x to oo} (1/ (e^x - 3x)*(e^x-3 ) )/ 1`

`= lim_{x to oo} (e^x-3 )/ (e^x - 3x)`

We can go again with L'Hopital as as `lim_{x to oo} e^(x) > > 3x` so this is still `oo/oo`

`= lim_{x to oo} (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 oo} (e^x )/ (e^x) = 1`