- #1

sponsoredwalk

- 532

- 5

The proof I've read is the following and I really just want to talk about it to be 100% sure I've understood it and I know where it comes from;

**1:**We'll take the equation of a line; [tex] f(x) \ - \ f(x_0) \ = \ m (x \ - \ x_0) \ [/tex]

where;

[tex] m \ = \ \frac{f(x) \ - \ f(x_0)}{x \ - \ x_0} [/tex]

and set it up like so;

**2:**[tex] f(x) \ = \ \frac{f(x) \ - \ f(x_0)}{x \ - \ x_0} (x \ - \ x_0 ) \ + \ f(x) [/tex]

*So, as I understand the proof, if we want to prove that the function f is continuous we will take the limit of the above expression and as x tends back onto itself and by getting the equality f(x_0) = f(x_0) we've proved the function continues along at this point*

**3:**[tex] \lim_{x \to x_0} f(x) \ = \ \lim_{x \to x_0} \frac{f(x) \ - \ f(x_0)}{x \ - \ x_0} (x \ - \ x_0 ) \ + \ \lim_{x \to x_0} f(x) [/tex]

**4:**[tex] f(x_0) \ = \ f ' (x_0) (x_0 \ - \ x_0 ) \ + \ f(x_0) [/tex]

**5:**[tex] f(x_0) \ = \ f ' (x_0) (0) \ + \ f(x_0) [/tex]

**6:**[tex] f(x_0) \ = \ f(x_0) [/tex]

I wonder how deep a proof this is, it is from an advanced calculus text. :tongue2: