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  • - [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.

- [Voiceover] So let's see if we can find the limit

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A2 初級

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

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    yukang920108 に公開 2022 年 07 月 02 日
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