Question 1

In the proof of differentiability implies continuity, you separate the limits saying that the limit of the products is the same as the product of the limits. But the limit of x*1/x at zero cannot be divided as the limit of x times the limit of 1/x as the latter one does not exist.
I do understand that this works in this case because both the limits exist.

Accepted Answer

Good! You have found where the theorem fails: that is the limit of the product is the product of the limit remains true only when both limits actually exist! Clearly if one of the limits fails to exist then any equality would be meaningless as you have argued above!!

Question 2

In the proof of the chain rule by multiplying delta u by delta y over delta x it assumes that delta u is nonzero when it is possible for delta u to be 0 (if for example u(x) =2 then the derivative of u at x would be 0) and then delta y over delta u would be undefined? This would be multiplying by 0 as we never proved that delta u was nonzero and only that it was approaching 0 in the proof above the chain ule proof. The limit of delta u as delta x approaches 0 is still 0 is delta u =0 but we then couldn’t multiply by delta u. How is it allowed to multiply delta y over delta x by delta u over delta u if we do not know that delta u is nonzero?

Accepted Answer

Good question — I looked around and it appears that for this proof Sal should have specified that u(x)≠u(c) for x close to c.

See for example:
https://math.stackexchange.com/questions/1180686/how-to-prove-the-chain-rule

And:
https://en.wikipedia.org/wiki/Chain_rule

Question 3

In the second video at 2:23, shouldn't it be:
[lim x->c (u(x))] - u(c) ?

Accepted Answer

Actually, with limit rules  if you had lim(x-3) you could turn it into lim(x)-3, so Sal is using that property in reverse.  He also says around 2:00 that you can treat u(c) as a constant.  Though I think it works even if you do it the way you explain.

Question 4

In the second video, we suppose that function is differentiable when rewriting the limit? Because continuity does not imply differentiability.

Accepted Answer

There is no need to worry about differentiability in the second video because Sal never differentiates anything during the proof.

Question 5

In video 1, why is it okay to divide by (x-c)? Isn't that 0 when x -> c?

Accepted Answer

Only if x = c. If it's approaching, then it's not quite equal yet.

Question 6

I didn't understand why if function u is continuous at x, then Δu→0 as Δx→0.
 Can't a continuous function u(x) have a vertical line somewhere in its graph? In the graph of a vertical line, as the change in x approaches 0, the change in u(x) approaches infinity.

Accepted Answer

A function cannot contain a vertical line in its graph because a vertical line would imply that one input value is being mapped to multiple output values (which would violate the definition of a function).

Question 7

It seems that using the same "reasoning" ("trick") used in the first video to demonstrate that differentiability implies continuity, we can prove statements which aren't (generally) true. Consider the following application to f(x) and an arbitrary number "a":

limit[x->c](f(x)-a)=
limit[x->c]((x-c)(f(x)-a)/(x-c))=
limit[x->c](x-c)limit[x->c](f(x)-a)=
0limit[x->c](f(x)-a)=
0

Is this a proof that limit of f(x) at an arbitrary "c" equals any arbitrary "a"?

Accepted Answer

The third step is wrong.
lim(𝑥 → 𝑐) [(𝑓(𝑥) − 𝑎)∕(𝑥 − 𝑐)] = ±∞ ∙ lim(𝑥 → 𝑐) [𝑓(𝑥) − 𝑎]

Question 8

Video 2 is supposed to prove that as the change in x APPROACHES 0, the change in u APPROACHES 0.

But at the end, Sal writes that the limit of the change in u is EQUAL to 0 as the change in x APPROACHES 0.

EQUAL and APPROACHES are not the same thing, are they? How can we conclude EQUALS from APPROACHING?

Accepted Answer

The sentences 'The limit as x goes c of Δu is 0' means the same thing as 'Δu approaches 0 as x approaches c'.

The quantity u is approaching 0, because the limit of u equals 0.

Course: AP®︎/College Calculus BC > Unit 3

Proving the chain rule

First, we would like to prove two smaller claims that we are going to use in our proof of the chain rule.

1. If a function is differentiable, then it is also continuous.

2. If function $u$ ‍ is continuous at $x$ ‍, then $Δ u \to 0$ ‍ as $Δ x \to 0$ ‍.

Now we are ready to prove the chain rule!

Bonus: We can use the chain rule and the product rule to prove the quotient rule.

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AP®︎/College Calculus BC