You need to open the brackets to denominator such that:

`lim_(x->oo)(3x^2+ln x)/(2-3x+x^2)`

Substituting `oo` for x yields:

`lim_(x->oo)(3x^2+ln x)/(2-3x+x^2) = (oo + ln oo)/(2 - oo + oo) = oo/oo`

Since the result is indeterminate, you may use l'Hospital's theorem such that:

`lim_(x->oo) (3x^2+ln x)/(2-3x+x^2) = lim_(x->oo) ((3x^2+ln x)')/((2-3x+x^2)') `

`lim_(x->oo) (3x^2+ln x)/(2-3x+x^2) = lim_(x->oo) (6x + 1/x)/(-3+2x)`

You need to force factor 6x to numerator and 2x to denominator such that:

`lim_(x->oo) (6x + 1/x)/(-3+2x) = lim_(x->oo) (6x(1 + 1/(6x^2)))/(2x(-3/(2x)+1))`

Reducing by `2x` yields:

`lim_(x->oo) (6x(1 + 1/(6x^2)))/(2x(-3/(2x)+1)) =lim_(x->oo) (3(1 + 1/(6x^2)))/((-3/(2x)+1))`

`lim_(x->oo) (3(1 + 1/(6x^2)))/((-3/(2x)+1)) = (3+ lim_(x->oo) 3/(6x^2))/(lim_(x->oo) -3/(2x) + 1)`

`lim_(x->oo) (3(1 + 1/(6x^2)))/((-3/(2x)+1)) = (3+0)/(0+1) = 3`

**Hence, evaluating the limit of the given function yields `lim_(x->oo) (3x^2+ln x)/(2-3x+x^2) = 3.` **