Question

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

Answer 1

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

Explanation:

If we evaluate the limit of the numerator and denominator separately we'll find that:

`*` As `ln(x)` goes to `oo` as `x` goes to `oo`: `ln(oo)=oo`

`*` `x` goes to `oo`

Therefore we have a ratio of two infinities `oo/oo` meaning that we will have to apply L'Hospital's Rule.

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

The limit approaches `0` because `1` divided over something approaching `oo` becomes closer and closer to `0`

For example, consider:

`1/10=0.1`

`1/100=0.01`

`1/10000=0.0001`

We can see that as the denominator gets larger and larger, approaching `oo`, the value gets smaller and smaller and more closer to `0`.

Answer 2

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

Explanation:

The question is to find the value of `lnx/x` where `x to oo`

If we let `x=oo` then `lnx/x=(oo)/(oo)` `=` Undefined

So now we can apply L'Hospital's rule by differentiating the numerator and denominator individually.

So, `d/dxlnx=1/x` and `d/dxx=1`

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

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

Now we let `x=oo`

Therefore `1/(oo)=0` because `oo` is a very large number and `1` divided by a very large number always approaches `0` and it is very close to `0` but never quite gets there.