Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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. =)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook