`lim_(x-gtoo)(x^2-3x+7)/(x^3+10x-4)`

If you try to evaluate the limit straight away, you will get oo/oo which is indeterminate. We can remove this by dividing both numerator and denominator by x^3.

`lim_(x-gtoo)((x^2-3x+7)/x^3)/((x^3+10x-4)/x^3)`

`lim_(x-gtoo)(1/x-3/x^2+7/x^3)/(1+10/x^2-4/x^3)`

Now we can evaluate the limit.

`lim_(x-gtoo)(1/x-3/x^2+7/x^3)/(1+10/x^2-4/x^3) = 0/1 = 0`

Therefore,

`lim_(x-gtoo)(x^2-3x+7)/(x^3+10x-4) = 0`

The limit `lim_(x->oo)(x^2-3x+7) / (x^3+10x-4)` has to be determined.

The numerator x^2-3x+7 and denominator x^3+10x-4 do not have any common factors that can be eliminated.

Substituting x = `oo` , gives the result `oo/oo` which is indeterminate and in a form that allows the use of l'Hospital's rule. We can substitute the numerator and denominator by their derivatives.

This gives:

`lim_(x->oo) (2x - 2)/(3x^2 + 10)`

Again, substituting x = `oo` gives the indeterminate form `oo/oo` . Substitute the numerator and denominator by their derivatives once again.

This gives:

`lim_(x->oo) 2/(6x)`

Now substituting x = `oo` gives the result `2/oo` . This is equal to 0.

The required limit `lim_(x->oo)(x^2-3x+7) / (x^3+10x-4) = 0`