We have determine: `lim_(x->oo)(2x^2 - 9x)/(4x^3 + x - 7)`
If we substitute x = inf. in the expression, both the numerator and the denominator are equal to inf. As ` `inf./inf. is indeterminate we can use l'Hopital's rule and substitute the numerator and the denominator with their derivatives.
=> `lim_(x->oo)(4x - 9)/(12x^2 + 1)`
Again, we see that substituting x with inf. gives the indefinite form inf./inf. The numerator and denominator are substitute by their derivatives again
=> `lim_(x->oo) 4/(24x)`
When x tends to inf. 1/6x tends to 0.
The value of `lim_(x->oo)(2x^2 - 9x)/(4x^3 + x - 7) = 0`