Question

How do you find the limit of `(8lnx)/x` as x approaches infinity?

Answer 1

`Lt_(x->oo)(8lnx)/x=0`

Explanation:

We can use L'Hopital's rule here which states that

`Lt_(x->k)(f(x))/(g(x))=Lt_(x->k)(f'(x))/(g'(x))`

Here we have to find `Lt_(x->oo)(8lnx)/x`

which is equal to `Lt_(x->oo)(d/dx(8lnx))/(d/dx(x))`

= `Lt_(x->oo)(8/x)/1=Lt_(x->k)8/x=0`