Question

How do you find the limit of `lnx/x` as `x->oo`?

Answer 1

`lim_(x->oo) lnx/x = 0`

Explanation:

As we have:

`lim_(x->oo) lnx = +oo`
`lim_(x->oo) x = +oo`

The limit:

`lim_(x->oo) lnx/x`

presents itself in the indeterminate form `oo/oo` and we can use l'Hospital's rule:

`lim f(x)/g(x) = lim (f'(x))/(g'(x))`

so:

`lim_(x->oo) lnx/x = lim_(x->oo) (d/(dx)lnx)/(d/(dx)x) = lim_(x->oo) 1/x = 0`

Answer 2

On request, giving a demonstration without using l'Hospital:

For `x > 1` we have:

`ln x < x`

if we express `x` as `x = (sqrt(x))^2` then:

`ln x = ln (sqrt(x))^2 = 2ln sqrtx < 2sqrtx`

so:

`lnx/x < 2sqrtx/x = 2/sqrtx`

On the other hand for `x > 1` the logarithm is positive so:

`0 < lnx/x < 2/sqrtx`

Then for `x->oo`

`lim_(x->oo) lnx/x = 0`

based on the squeeze theorem.