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# Limits by direct substitution

AP.CALC:
LIM‑1 (EU)
,
LIM‑1.D (LO)
,
LIM‑1.D.1 (EK)

## Video transcript

- [Voiceover] 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. And when you think about this, I'm not doing a rigorous proof here, parabola would look something like this and this would be an upward opening parabola, look something like this. It is a, this graph, visually, is continuous, you don't see any jumps or gaps in it. And in general, 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, oh, for all real numbers. And so if something is continuous for all real numbers, well then the limit as X approaches some real number is going to be the same thing, it's just a value within the expression at that real number. So what am I saying, I'm just gonna say it another way, we know that some function is continuous, is continuous at some X value, at X equals A if and only if, I'll write that as iff, or iff, if and only if the limit as X approaches A of F of X is equal to F of, 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, 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 to 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.
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