Calculus, derivative, limit, advanced mathematics, thank you. Look at this picture.

Your proof above is irrelevant to the topic. Your proof is correct in the whole process, but it has nothing to do with the topic. Your proof means that F(x) has a derivative at A, so there is g(a) at A, and this result is definitely correct. Because there are no restrictions on F (x) and G (x), it holds for any g(a).

However, f(x) is limited in the topic, and it can't be derived at A, so your proof doesn't match the meaning of the topic.

The correct proof can be proved by definition, so I won't write it.