So I didn't do a rigorous proof here, but just, it's conceptually not a big jump to say okay, well this is just a standard quadratic right over here, it's defined for all real numbers, and in fact it's continuous for all real numbers, and so we know that this expression, it could define a continuous function, so that means that the limit as x approaches a for this expression is just the same thing as evaluating this expression at a. And in this case our a is negative one. So all I have to do is evaluate this at negative one. This is going to be six times negative one squared plus five times negative one minus one. So that's just one, this is negative five. So it's six minus five minus one, which is equal to zero. And we are done.