Limits by direct substitution | Limits and continuity | AP Calculus AB | Khan Academy

So let's see if we can find the limit as x approaches negative one of six x squared plus five x minus one.

Now, the first thing that might jump out at you is this right over here. This expression could be used to define the graph of a parabola. When you think about this, I'm not doing a rigorous proof here; a parabola would look something like this.

This would be an upward opening parabola. It looks something like this; this graph visually is continuous. You don't see any jumps or gaps in it. In general, a part of a quadratic like this is going to be defined for all values of x, for all real numbers, and it's going to be continuous for all real numbers.

So, something that is continuous for all real numbers—well then, the limit as x approaches some real number is going to be the same thing as just evaluating the expression at that real number. So what am I saying? I'm just going to say it another way: We know that some function is continuous at some x value, at x equals a, if and only if—that is, if or if if and only if—the limit as x approaches a of f of x is equal to f of a.

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.

So we know that this expression 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. In this case, our a is negative 1.

So all I have to do is evaluate this at negative 1. This is going to be 6 times negative 1 squared plus 5 times negative 1 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.