Question

Can anybody do this?

Answer 1

`0`

Explanation:

`ln(ln(x))/x=ln(ln(x))*1/x`

`lim_(x->a)(f(x)*g(x))=lim_(x->a)(f(x))*lim_(x->a)(g(x))`

`lim_(x->oo)(ln(ln(x))=oo`

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

`lim_(x->oo)(ln(lnx))*lim_(x->oo)(1/x)=oo*0=0`

Answer 2

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

Explanation:

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

This limit is in the indeterminate form `oo/oo`, so we can use l'Hôpital's rule, which states that, if `lim_(x->c)f(x)/g(x)` is of the form `oo/oo` or `0/0`, then it is equal to `lim_(x->c)(f'(x))/(g'(x))`.

So, since `d/dx(ln(ln(x)))=1/(xln(x))` and `d/dx(x)=1`, we have the original limit equal to
`lim_(x->oo)1/(xln(x))`

Evaluating this demonstrates that
`lim_(x->oo)ln(ln(x))/x=0`

This can be verified by a graph:
graph{e^(xy)-ln(x)=0}