The functions that we see in Calculus are usually quite well behaved and may fool us into thinking that this is the case for most functions. An example of a function for which the condition you mention fails, but which is not constant, is f(x) = sin(1/x). Of course this function is not defined at x=0 (which is where it will take the same values an infinite amount of times), but this can be fixed by putting f(x) = x sin(1/x) for x not equal to 0, and f(0) = 0. In this way, it is even differentiable at 0. Many other examples exist - try asking ChatGPT :-)

​​@@janfreol1 Is there a way to ultimately prove the chain rule in calculus 1?

Ok. Let me give you a hint. The choice for epsilon_1 is straight-forward since M is a constant (that epsilon_1 is allowed to depend on). The choice for epsilon_2, however, requires some more work, as epsilon_2 is not allowed to depend on a_n (as it changes with n - which is the variable here). So, you need to figure out a way to replace the absolute value of a_n by some (possibly larger) constant in the inequality.

Yes, that‘s the kind of dedication I expect from all university teaching staff. Olsen for president!

@@ArthurZeuner3dwe need more Hawaiian shirt!

Is there any other way? ;-)

But how do you know (a priori) that N does not have an upper bound?

@@jan-fredrikolsen5369 You assumed going towards a contradiction that the set of natural numbers has an upper bound but how do you show that they also have a suppremum

Whether or not N contains 0 is a matter of definition. Some countries, like Sweden and France, tend to include 0 in N. In these countries, the notation N* is therefore used to denote the non-zero positive integers.

You multiply all the terms by n to get n(1-D) smaller/equal to one, and then divide both sides by (1-D)

Btw subscribed this an underated channel.

@@wowfmomf6126 Sorry, late answer, but no need to suppose any continuity of f'' anywhere. In terms of the second derivative, it is actually enough that f''(c) exists and is positive or negative. Here, I assume that f'' exists on (a,b) but outside of f''(c) existing, I actually just need to know that f' exists on (a,b). So, here, I am actually assuming more than I need.

Detta gäller om en funktion är kontinuerlig på ett slutet intervall. Dvs., för slutna intervaller där sin(1/x) är punktvis kontinuerlig, så är även sin(1/x) likformig kontinuerlig.

Oh crap, yes, decidedly a brainfart on my part :-)

@Jovannie Castillo Don't think that is needed? The assumption of differentiability implies f,g continuous in punctured neighborhood.

Yes, but in the course this definition comes before the definite integral is defined, so we need to express this definition without that machinery.

f.lux Plot Twist: He was actually on vacation when he filmed those videos.

Maybe I will have to come up with something for the next batch :-)

Aargh! :-O

I beräkningen innan använde vi att x var positiv (vid 3:32) vilket inte gäller när x går mot minus oändligheten. Alltså måste vi börja om igen. Man kan göra en motsvarande beräkning som innan (men man måste då komma ihåg att ta med ett extra minus-tecken när man bryter ut x^2 från kvadratroten), men det behövs inte - vilket är vad vi i princip observerar vid 5:02. Detta därför att oändligheten minus oändligheten måste man räkna vidare på medan oändligheten plus oändligheten vet man att direkt ger oändligheten.