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### Course: Differential Calculus>Unit 3

Lesson 12: Proof videos

# 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.

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• 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?
• 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 at , 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.
• 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!
• 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
• 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.
• 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.

• 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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Andy
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• 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.
(1 vote)
• 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)
• 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
(1 vote)
• I don't get about x = c(M) (X) I'm Just confused with this.