Question

Lim x->oo {x-(x^2)*ln((1+x)/x)} please explain all of phase...? thanks.

Answer 1

`Lim_(x->oo)x-x^2ln((1+x)/x)=1/2`

Explanation:

.

Let's factor out `x^2`:

`Lim_(x->oo)x-x^2ln((1+x)/x)=Lim_(x->oo)x^2(1/x-ln((1+x)/x))`

This is still in an indeterminate form. Let's rewrite it:

`=Lim_(x->oo)(1/x-ln((1+x)/x))/(1/x^2)`

Now, we can apply L'Hopital's rule and take the limit of the derivatives of the top and the bottom:

`=Lim_(x->oo)((-1/x^2-1/((1+x)/x)(-1/x^2)))/(-2/x^3)=Lim_(x->oo)(-1/x^2+1/(x(1+x)))/(-2/x^3)=Lim_(x->oo)((-1-x+x)/(x^2(1+x)))/(-2/x^3)=Lim_(x->oo)(-1/(x^2(1+x)))/(-2/x^3)=Lim_(x->oo)(x/(2(1+x)))`

Now, we apply L'Hopital's rule again and take the derivatives of the top and the bottom:

`=Lim_(x->oo)(1/2)=1/2`