1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

If f is continuous at c and f(c)>1

  1. Mar 20, 2009 #1
    1. The problem statement, all variables and given/known data

    Hi everyone, this is probably an easy question but I'm having trouble on the wording of the proof.

    Let f be continuous at x=c and f(c) > 1

    Show that there exists an r > 0 such that [tex]\forall x \in B(c,r) \bigcap D : f(x) > 1[/tex]

    2. Relevant equations

    [tex]\forall \epsilon > 0, \exists r > 0, \forall x \in B(c,r) \bigcap D \Rightarrow |f(x) - f(c)| < \epsilon [/tex]

    f(c) > 1

    3. The attempt at a solution

    I'd put something here but I don't really have any attempts. I originally thought of using the limit rule that lim f(x) = f(c) but that doesn't work since c could be an isolated point.

    If I used a graph it just seems so obvious, but I have to use words to prove it. I tried also to use a contradiction, but using just what I put down in (2) didn't lead to any direct contradiction that I could see.

    A nudge in the right direction would be helpful.
     
  2. jcsd
  3. Mar 20, 2009 #2

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    If in doubt try an example.

    Let's say for the sake of argument that f(c)=2. What choice of epsilon, and r, from 2. above will mean that f(x) > 1 in the ball around c?
     
  4. Mar 20, 2009 #3

    MathematicalPhysicist

    User Avatar
    Gold Member

    -e<f(x)-f(c)<e
    f(x)>f(c)-e
    f(c)=1+h^2 for some h real, do you understand how to pick e such that f(x)>1.
     
  5. Mar 20, 2009 #4
    by this reasoning couldn't I just claim that f(c) = 1 + e?
     
  6. Mar 20, 2009 #5

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    Yes, you can do that.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook