Hi I have a question about continuity as it is proven by the existence of a derivative.(adsbygoogle = window.adsbygoogle || []).push({});

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:

**Physics Forums - The Fusion of Science and Community**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Definition of Continuity?

Loading...

Similar Threads - Definition Continuity | Date |
---|---|

Limit epsilon-delta definition vs. continuity | Oct 25, 2014 |

Continuous function and definition | Sep 29, 2014 |

Equivalent continuity definition | Oct 17, 2013 |

Explain why this is no good as a definition of continuity | Dec 6, 2011 |

Rigorous definition of continuity on an open vs closed interval | Dec 1, 2011 |

**Physics Forums - The Fusion of Science and Community**