Question

My question is reffering to the proof of the theorem saying that a differentiable function at a point it is also continuous at that point(session 3). I do not believe that the proposed proof is correct. The proof starts from the definition of the continuous function and continues by dividing with x-x_0. The mistake is at the point that he extends the limit of f(x) to a limit of the whole fraction. Of course you can multiply and divide whith the same number but including the created fraction to the limit it seems quite arbitrary to me. I do not know if I was clear. What do you think?

Answer 1

links would help

Answer 2

Sorry, it is at session 5. See the attachement.

Answer 3

Ah yes, it might seem arbitrary to you, but mathematician resorts to tricks like that all the time, the most important thing is whether the 'trick' is legal or not, in the proof he divided by (x- x0) , it will not be legal if x=x0 which means he divided by zero, but when you take the limit, x approaches x0 (x→x0) but x never equals x0 (x ≠ x0) so (x-x0)≠ 0 so it is legal

Answer 4

also it is important to notice that after the limit is evaluated (x-x0) is practically zero, this might be confusing to you and may seem contradictory to what I said earlier, but that is the whole idea of calculus, and this kind of thing will often show up later

Answer 5

An didn't mention anything about division by zero. I am trying to be more clear in the following pdf. Thank you.

Answer 6

Ah so you didn't understand why the limit can be 'moved'? is that it?

Answer 7

Yes

Answer 8

Ah that confused me too the first time, another example of that would be \[\lim_{x \to 0^+} x^x = \lim_{x \to 0^+} e^{\ln x^x} = \lim_{x \to 0^+} e^{x \ln x} = e^{\lim_{x \to 0^+} (x \ln x)}\] , it is valid to move the limit as long as the function that we take the limit is continuous. in the example above \[f(x)=e^x\] is continuous.

Answer 9

\[\frac{\lim_{x\to x_0}f(x)-f(x_0)}{x-x_0} = \lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}\] is valid because f(x) is continuous, you might ask, if we're trying to prove that it is continuous then it's not good to assume that f(x) is continuous i the first place?, that is actually one of the many methods for proofs, you assume property A to f(x) and if the result agrees then f(x) is said to be proved to have property A, but if the result is contradictory then you have proved f(x) does not have property A, does this makes it clear to you?

Answer 10

Thank you.