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### Course: AP®︎ Calculus AB (2017 edition) > Unit 3

Lesson 13: Optional videos- Justifying the power rule
- Proof of power rule for positive integer powers
- Proof of power rule for square root function
- Limit of sin(x)/x as x approaches 0
- Limit of (1-cos(x))/x as x approaches 0
- Proof of the derivative of sin(x)
- Proof of the derivative of cos(x)
- Product rule proof
- Proof: Differentiability implies continuity
- If function u is continuous at x, then Δu→0 as Δx→0
- Chain rule proof
- Quotient rule from product & chain rules

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# If function u is continuous at x, then Δu→0 as Δx→0

Sal shows that if a function is continuous, the difference in the function's values approaches 0 as the difference in the x-values approaches 0. This is simply another way to define continuity.

## Want to join the conversation?

- Why wouldn't the change in X and the change in U not be the same in a discontinuous function? Towards the end Sal said it wouldn't for some, and the title specifies that it's true for continuous, but why isn't it true for discontinuous?(5 votes)
- The discontinuous function has break or "jump" somewhere. In the discontinuous function, if you move closer and closer to c from either negative or positive function, the change in x and change in U is not the same .

On the other hand at3:50, you can draw the graph without picking up the pencil, that's the example of continuous function. So, as you move closer and closer to c, the change in U and X is the same.(2 votes)

- as delta x approaches 0, delta u approaches zero. Doesn't this mean that derivative of a function at every point on the graph is 0/0(3 votes)
- Remember that we are dealing with limits. I found it useful to refresh my understanding of limits before going over this material ...

https://www.khanacademy.org/math/ap-calculus-ab/ab-limits-continuity(3 votes)

- Hi, I got lost at this part. lim[x->c]u(x)=u(c), how do you get to lim[x->c]{u(x)-u(c)} = 0?

I would get to {lim[x->c]u(x)} - u(c) = 0. Or they are actually the same thing?(1 vote)- The limit of a constant is the same constant, and the limit of the difference is the difference of the limits. Note that u(c) is considered a constant, since x is the quantity that is varying in the limits.

So if lim[x->c]u(x)=u(c), then we have

lim[x->c]{u(x)-u(c)} = {lim[x->c]u(x)} - {lim[x->c]u(c)} = {lim[x->c]u(x)} - u(c) = u(c) - u(c) = 0.

Have a blessed, wonderful day!(6 votes)

- what if the function oscillates?

as delta x approaches zero, wouldn't delta u get bigger and smaller and bigger and smaller etc.??(1 vote)- here, we consider delta x to be approaching very very very close to zero, so at any point on the oscillating function, however big the delta x maybe, after delta u getting bigger-smaller-bigger-smaller, it will reach a point where it can no longer get bigger i.e. it will be approaching zero.(4 votes)

- I am confused. What about a function with a negative slope. Like for example, y = -x +10.

Clearly, as the change in x approaches zero, the function gets bigger.5:50(2 votes)- If the function has a negative slope, as dx approaches zero, u(x) does get bigger. However, in this case u(c) > u(x), and our expression lim x→c {u(c) - u(x)} still goes to zero.(1 vote)

- Towards the end of the video, it is shown that given u(x) continuous at x=c, the limit of Δu as Δx approaches 0 is 0.

Wouldn't this imply that Δu=0 when Δx→0 and not that Δu→0 as Δx→0 which the reader seems to infer?

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leafers ultimate style avatar for user Andy Mainord

Andy(1 vote)- 𝛥𝑢 = 𝑢(𝑥 + 𝛥𝑥) − 𝑢(𝑥)

𝛥𝑥 approaches zero, but is not equal to zero.

Thereby, 𝑥 + 𝛥𝑥 approaches but is not equal to 𝑥,

𝑢(𝑥 + 𝛥𝑥) approaches but is not equal to 𝑢(𝑥),

and 𝛥𝑢 approaches but is not equal to 𝑢(𝑥) − 𝑢(𝑥) = 0

However, the limit of 𝛥𝑢 as 𝛥𝑥 → 0 is equal to zero.(2 votes)

- cant we prove this by using the delta epsilon thing?(1 vote)
- yes we can, but it will take a long time when you can do it straight forward. + the point of this is using variable"u" not delta right now(2 votes)

- What about point/removable discontinuities? Say for the function: f(x)=x, where x cannot equal 3. Even though as delta x decreases around 3, delta y also decreases, the function is still discontinuous at 3.(1 vote)
- Note that
`y=u(x)`

, so`∆y = ∆u = u(x) - u(c)`

. Consequently,`∆u`

will approach`[lim x→c u(x)] - u(c)`

— i.e. how far away the removable discontinuity is from the line.(1 vote)

- I don't get about x = c(M) (X) I'm Just confused with this.(0 votes)
- is it present in the video? if not, could you be a bit more specific.(2 votes)

## Video transcript

- [Voiceover] The result
that I hope to show you, or you give you an
intuition for, in this video is something that we will use in the proof of the chain rule, or in a proof of the chain rule actually. We may do more than
proof of the chain rule. But the result we're gonna look at is if we have some function u which is a function of x, and we know that it is continuous at x equals c, so if we know this, then
that's going to imply that the change in u goes to zero as our change in x in this region around c goes to zero. This is what I want to
get an intuition for. That if u is continuous at c, then as our change in x around c gets smaller, and smaller, and smaller as it approaches zero,
then our change in u approaches zero as well. To think about this, or
to even kind of prove it to ourselves a little bit more rigorously, let's think about what it means to be continuous at x equals c. Well the definition of continuity is, so this literally is
the same thing as saying that the limit as x approaches c of u of x is equal to u of c. The limit that our function approaches as x approaches c is equal to the value of the function at c. We don't have a point discontinuity or a jump discontinuity. If we had a jump discontinuity, then the limit wouldn't exist, and we've seen that in previous videos. Now I'm just gonna
manipulate this algebraically so it essentially gives us this conclusion right over here. So this we can rewrite. It's important to realize that u of c, this is just going to be some value. It looks like maybe this is a function of x or something, but no, this is just going to be some value. I've inputted c here and I've evaluated the function of that, and so this is going to be some number. It could be five, or seven, or pi, or negative one, but it's just going to be some value, some constant. So I can treat it like a constant. So this is going to be the same thing as saying the limit as x approaches c of u of x minus u of c is equal to zero. And actually, in the video where we prove that differentiability implies continuity, we started with this and we proved that right over there. We show that these two
are equivalent things. But hopefully you could even think about the intuition. This is just if the limit of u of x is x approaches c is equal to this, then when you evaluate this limit, the limit is x approaches c, well this thing is going
to approach u of c, because we saw it right up here, u of c minus u of c is indeed going to be equal to zero. So hopefully you don't
feel like it's too much of a stretch, and you can just subtract u of c from both sides
and apply properties of limits, and you can
get this result as well. But this is interesting
because this essentially can take us to this, the idea that as our change in x gets smaller, and smaller, and smaller as it approaches zero, then our change in our function is also
going to approach zero. Now let's just graph
this or visualize this to get a sense of that. So this is our x axis. Woops. That's our x axis. Let's call that our u axis maybe. And I did u intentionally because that's the variable we'll use in our proof of the chain rule video. Let's say this right over here is our function. Let's say this right over here, that is c. This right over here is u of c. U of c. Then let's just take some
arbitrary x over here. So some arbitrary x, and then this right over here is u of x. So if we define our change, let me do this. If we define our change in u is equal to u of x minus u of c, which makes sense because this is our change in u. So let's say this is going to be u of x minus u of c. If we define our change in x is equal to x minus c, which it is in this case, it is x minus c, then we can rewrite this limit right over here. Instead of saying the
limit as x approaches c, we could write the limit as delta x approaches zero, because
if x approaches c, then delta x is going to approach zero. So we could write this the limit as delta x approaches zero of delta u is going to be equal to zero. We define this as our change in u, and it is our change in u. So this is equal to zero. So another way of thinking about this is as delta x approaches zero, our change in u, our
change in the function, is going to approach zero. So as delta x approaches zero, delta u approaches zero. That's what we wrote over here. Delta u approaches zero as delta x approaches zero. In a lot of ways, this is
hopefully common sense. We're dealing with a continuous function. As you get smaller and smaller, and you can just think of it this way, as you get smaller and
smaller changes in x's, as our change in x gets smaller, and smaller, and smaller, and smaller, well it's because it's continuous. You wouldn't be able to say this for a discontinuous function, but because it's continuous, or you wouldn't be able to say this for some discontinuous functions, as our change in x gets smaller, and smaller, and smaller, then our change in u is
going to get smaller, and smaller, and smaller. So it makes intuitive sense, but hopefully this makes you feel even better about it
because we're going to use this idea to prove the chain rule in the next video.