Question

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

Answer 1

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

Explanation:

As we have:

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

The limit:

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

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/sqrt(x) = lim_(x->oo) (d/(dx)lnx)/(d/(dx)sqrtx) = lim_(x->oo) (1/x)/(1/(2sqrt(x))) = lim_(x->oo) (2sqrt(x))/x=lim_(x->oo) 2/sqrt(x) = 0`