Question

Put limit in big O little O notation?

So for the two functions, lnx and x-lnx. Compare which one grows faster and put it in the big o little o notation ( slower = o (faster) , top = O (bottom))

Answer 1

`lnx = o(x-lnx)`

Explanation:

Supposing we are here required to evaluate the behavior for `x->oo`, by definition, given two real functions `f(x)` and `g(x)`:

`lim_(x->oo) (f(x))/(g(x)) = 0 <=> f(x) = o (g(x))`

and:

`lim _(x->oo) "sup"abs((f(x))/(g(x))) < oo <=> f(x) = O(g(x))`

In our case let `f(x) =lnx` and `g(x) =x-lnx`. Then:

`lim_(x->oo) (f(x))/(g(x)) =lim_(x->oo) lnx/(x-lnx)`

`lim_(x->oo) (f(x))/(g(x)) =lim_(x->oo) 1/(x/lnx-1)`

Evaluate now the limit:

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

it is in the indeterminate form `oo/oo` so we can use l'Hospital's rule:

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

`lim_(x->oo) x/lnx = lim_(x->oo) 1/(1/x) = lim_(x->oo) x = +oo`

Then:

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

which means:

`lnx = o(x-lnx)`