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!

Differentiability implies continuity proof (delta epsilon)

  1. Jun 21, 2008 #1
    1. The problem statement.

    Give a complete and accurate [tex]\delta[/tex] - [tex]\epsilon[/tex] proof of the thereom: If f is differentiable at a, then f is continuous at a.

    2. The attempt at a solution

    [tex]\forall\epsilon>0, \exists\delta>0, \forall x, |x-a|<\delta \implies \left|\frac{f(x) - f(a)}{x-a} - f'(a)\right|<\epsilon [/tex]

    Want to show:

    [tex]\forall\epsilon>0, \exists\delta>0, \forall x, |x-a|<\delta \implies |f(x) - f(a)|<\epsilon[/tex]

    So I start with the known info and cross multiply [tex]\left|\frac{f(x) - f(a)}{x-a} - f'(a)\right|[/tex] to get [tex]\left|\frac{f(x) - f(a) - (x-a)f'(a)}{x-a}\right|[/tex] which doesn't really help me in completing the proof, especially since x-a is in the denominator. =[

    And is my known and want to show info correct?
  2. jcsd
  3. Jun 21, 2008 #2
    The known and show info look correct. As for the proof, hint: [tex]f(x)-f(a)= \frac{f(x)-f(a)}{x-a}*(x-a)[/tex] and [tex]\lim_{x \to a}(x-a)=0[/tex]

    In the end you can should be able to prove that [tex]\lim_{x \to a}f(x)-f(a)=0[/tex] i.e. [tex]|f(x) - f(a)|<\epsilon [/tex]
  4. Jun 21, 2008 #3
    Does that mean we can assume [tex]\frac{f(x)-f(a)}{x-a}*(x-a) = 0[/tex]?
  5. Jun 21, 2008 #4
    Let me rephrase that.

    From what you know:
    [tex]\lim_{x \to a}\frac{f(x)-f(a)}{x-a}=f'(a)[/tex] which when written in [tex]\epsilon - \delta[/tex] definition,[tex]\left|\frac{f(x) - f(a)}{x-a} - f'(a) \right|<\epsilon [/tex]

    Also, since f is differentiable at a, then [tex]\lim_{x \to a}(x-a)=0[/tex]

    You are trying to show that [tex]\lim_{x \to a}f(x)=f(a)[/tex] , by the definition of continuity at a. Which is also written in [tex]\epsilon - \delta[/tex] definition as
    [tex]|f(x) - f(a)|<\epsilon [/tex]

    From the hint I gave you, [tex]\lim_{x \to a}f(x)-f(a)=blank[/tex]
    Last edited: Jun 21, 2008
  6. Jun 21, 2008 #5
    Okay so tell me if this is right:

    [tex]\left|f(x)-f(a)\right| = \left|\frac{f(x)-f(a)}{x-a}(x-a)\right| = \left|\frac{f(x)-f(a)}{x-a}(0)\right| = 0 < \epsilon[/tex]

    Since [tex]\lim_{x \to a}\frac{f(x)-f(a)}{x-a}*(x-a) = \lim_{x \to a}\frac{f(x)-f(a)}{x-a}*\lim_{x \to a}(x-a) = \lim_{x \to a}\frac{f(x)-f(a)}{x-a}*0 = 0 [/tex]
  7. Jun 21, 2008 #6
    Yes, this part is correct. You can further simplify that into:

    [tex]\lim_{x \to a}\frac{f(x)-f(a)}{x-a}*0 = f'(a)*0 = 0 [/tex]

    Then, you can just say that: [tex] \forall\epsilon>0, \exists\delta, 0<|x-a|<\delta \implies |[f(x) - f(a)]-0|<\epsilon \implies |f(x) - f(a)|<\epsilon
    since you've shown that [tex]\lim_{x \to a}[ f(x)-f(a)]=0[/tex]

    You can't say that [tex](x-a)=0[/tex] since the limit [tex]\lim_{x \to a}(x-a)=0[/tex]
    Last edited: Jun 21, 2008
  8. Jun 21, 2008 #7
    Okay thanks a lot. =)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Differentiability implies continuity proof (delta epsilon)
  1. Delta epsilon proof (Replies: 5)