You need to evaluate the limit, hence you may force factor `x^(5/3` ) to numerator and `x^(8/5)` to denominator such that:

`lim_(x-gtoo) (x^(5/3)*(7 - 2*x^(1/3-5/3) + 6/x^(5/3)))/(x^(8/5)*(-2 - 5*x^(1-8/5) + x^(1/2-8/5)))`

`lim_(x-gtoo) (x^(5/3)*(7 - 2/x^(4/3) + 6/x^(5/3)))/(x^(8/5)*(-2 - 5/x^(3/5) + 1/x^(11/5)))`

`lim_(x-gtoo) (x^(5/3-8/5))*lim_(x-gtoo) ((7 - 2/x^(4/3) + 6/x^(5/3)))/((-2 - 5/x^(3/5) + 1/x^(11/5)))`

`lim_(x-gtoo) (x^((25-24)/15))*lim_(x-gtoo) ((7 - 2/x^(4/3) + 6/x^(5/3)))/((-2 - 5/x^(3/5) + 1/x^(11/5)))`

`lim_(x-gtoo) (x^(1/15))*((7 - lim_(x-gtoo) 2/x^(4/3) + lim_(x-gtoo) 6/x^(5/3)))/((-2 - lim_(x-gtoo) 5/x^(3/5) + lim_(x-gtoo) 1/x^(11/5)))`

You need to substitute `oo` for x in equation under limit such that:

`lim_(x-gtoo) (x^(1/15))*((7 - lim_(x-gtoo) 2/x^(4/3) + lim_(x-gtoo) 6/x^(5/3)))/((-2 - lim_(x-gtoo) 5/x^(3/5) + lim_(x-gtoo) 1/x^(11/5))) = oo*(-7/2) = -oo`

**Hence, evaluating the limit yields `-oo.` **