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  • - [Voiceover] Let's see if we can figure out the limit

  • of x over natural log of x as x approaches one.

  • And like always pause this video and see if you can

  • figure it out on your own.

  • Well we know from out limit properties this is going to

  • be the same thing as the limit

  • as x approaches one of x over

  • over

  • the limit,

  • the limit

  • as x approaches one

  • of the natural log

  • of x.

  • Now this top limit, the one I have in magenta,

  • this is pretty straight forward,

  • if we had the graph of y equals x

  • that would be continuous everywhere

  • it's defined for all real numbers and it's continuous

  • at all real numbers.

  • So it's continuous to limit as x approaches one of x.

  • It's just gonna be this evaluated x equals one.

  • So this is just going to be one.

  • We just put a one in for this x.

  • For the numerator here we just evaluate to a one.

  • And then the denominator,

  • natural log of x is not defined for all x's,

  • therefore it isn't continuous everywhere.

  • But it is continuous at x equals one.

  • And since it is continuous at x equals one,

  • then the limit here is just gonna be the natural log

  • evaluated at x equals one.

  • So this is just going to be the natural log

  • the natural log of one.

  • Which of course

  • is zero.

  • E to the zero power is one.

  • So this is all going to be equal to

  • this is going to be equal to

  • we just evaluate it

  • one

  • over

  • one

  • over

  • zero.

  • And now we face a bit of a conundrum.

  • One over zero is not defined.

  • It is was zero over zero, we wouldn't necessarily be

  • done yet but it's indeterminate form

  • as we will learn in the future there are tools we can apply

  • when we're trying to find limits and we evaluate it

  • like this and we get zero over zero.

  • But one over zero.

  • This is undefined

  • which tells us that this limit

  • does not exist.

  • So does

  • not

  • exist.

  • And

  • we are done.

- [Voiceover] Let's see if we can figure out the limit

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A2 初級 美國腔

1.52 直接帶入法的未定義(Undefined limits by direct substitution | Limits and continuity | AP Calculus AB | Khan Academy)

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    yukang920108 發佈於 2022 年 07 月 02 日
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