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  • Let's say that f(x) is equal to the absolute value of x minus three over x minus three

  • and what I'm curious about is the limit of f(x) as x approaches three

  • and just from an inspection you can see that the function is not defined when x is equal to three - you get zero over zero: it's not defined

  • So to answer this question let's try to re-write the same exact function definition slightly differently

  • So let's say f(x) is going to be equal to - and I'm going to think of two cases:

  • I'm going to think of the case when x is greater than three

  • and when x is less than three

  • So when x is - I'll do this in two different colors actually

  • When x - I'll do it in green - that's not green

  • When x is greater than three...

  • When x is greater than three, what does this function simplify to?

  • Well, whatever I get up here, I'm just taking the..

  • I'm going to get a positive value up here and then I'm...

  • Well, if I take the absolute value it's going to be the exact same thing, so let me...

  • For x is greater than three, this is going to be the exact same thing as x minus three over x minus three

  • because if x is greater than three, the numerator's going to be positive, you take the absolute value of that, you're not going to change its value

  • so you get this right over here or, if we were to re-write it...

  • ...if we were to re-write it, this is equal to, for x is greater than three, you're going to have f(x) is equal to one

  • for x is greater than three

  • Similarly, let's think about what happens when x is less than three

  • When x is less than three, well, x minus three is going to be a negative number

  • When you take the absolute value of that, you're essentially negating it

  • so it's going to be the negative of x minus three over x minus three

  • or if you were to simplify these two things, for any value as long as x doesn't equal three

  • this part right over her simplifies to one, so you are left with a negative one

  • negative one for x is less than three

  • I encourage you, if you don't believe what I just said, try it out with some numbers

  • Try out some numbers:

  • 3.1, 3.001, 3.5, 4, 7

  • Any number greater than three, you're going to get one

  • You're going to get the same thing divided by the same thing

  • and try values for x less than three:

  • you're going to get negative one no matter what you try

  • So let's visualise this function now

  • So, now you draw some axes...

  • That's my x- axis

  • and then this is my...

  • This is my f(x) axis - y is equal to f(x)

  • and what we care about is x is equal to three

  • so x is equal to one, two, three, four, five

  • and we could keep going...

  • and let's say this is positive one, two, so that's y is equal to one

  • this is y is equal to negative one and negative two

  • and we can keep going...

  • So this way that we have re-written the function

  • is the exact same function as this

  • we've just written [it] in a different way

  • and so what we're saying is...

  • is we're...

  • Our function is undefined at three

  • but if our x is greater than three, our function is equal to one

  • so if our x is greater than three, our function is equal to one

  • so it looks like...

  • It looks like that, and it's undefined at three

  • and if x is less than three our function is equal to negative one

  • so it looks like - I'll be doing that same color

  • It looks like this...

  • It looks like...

  • Looks like this...

  • Once again, it's undefined at three

  • So it looks like that

  • So now let's try to answer our question:

  • What is the limit as x approaches three?

  • Well, let's think about the limit as x approaches three

  • from the negative direction, from values less than three

  • So let's think about first the limit...

  • ...the limit, as x approaches three...

  • ...as x approaches three, the limit of f(x)...

  • ...as x approaches three from the negative direction

  • and all this notation here - I wrote this negative as a superscript right after the three - says

  • Let's think about the limit as we're approaching...

  • ...let me make this clear...

  • Let's think about the limit as we're approaching from the left

  • So in this case, if we get closer...

  • If we get...

  • If we start with values lower than three

  • as we get closer and closer and closer...

  • So, say we start at zero, f(x) is equal to negative one

  • We go to one, f(x) is equal to negative one

  • We go to two, f(x) is equal to negative one

  • If you go to 2.999999, f(x) is equal to negative one

  • So it looks like it is approaching negative one if you approach..

  • ...if you approach from the left-hand side

  • Now let's think about the limit...

  • ...the limit of f(x)...

  • ...the limit of f(x) as x approaches three from the positive direction, from values greater than three

  • So here we see, when x is equal to five, f(x) is equal to one

  • When x is equal to four, f(x) is equal to one

  • When x is equal to 3.0000001, f(x) is equal to one

  • So it seems to be approaching...

  • It seems to be approaching positive one

  • So now we have something strange

  • We seem to be approaching a different value when we approach from the left

  • than when we approach from the right

  • and if we are approaching two different values then the limit does not exist

  • So this limit right over here does not exist

  • or another way of saying it:

  • The limit...

  • ...the limit of...

  • (Let me write this in a new color - I have a little idea here)

  • ...the limit of a function f(x) as x approaches some value c is equal to L if and only if...

  • ...if and only if the limit of f(x) as x approaches c from the negative direction is equal to the limit

  • of f(x) as x approaches c from the positive direction which is equal to L

  • This did not happen here -

  • the limit when we approached the left was negative one,

  • the limit when we approached from the right was positive one,

  • So we did not get the same limits when we approached from either side

  • So the limit does not exist in this case

Let's say that f(x) is equal to the absolute value of x minus three over x minus three

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A2 初級 美國腔

1.55 極限在不連續的點(Limit at a point of discontinuity)

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    yukang920108 發佈於 2022 年 07 月 02 日
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