1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Critique my proof that f '(0) = 3.

  1. Jan 6, 2012 #1
    1. The problem statement, all variables and given/known data

    Let f be a continuos real function on R1, of which is its known that f '(x) exists for all x ≠ 0 and that f '(x) → 3 as x → 0. Does it follow that f '(0) exist?

    2. Relevant equations

    Mean Value Theorem
    definition of f '(x)

    3. The attempt at a solution

    Fix ε > 0. Since limx → 0 f '(x) = 3, there exists a ∂ > 0 such that |f '(t) - 3| < ε whenever 0 < |t - 0| = |t| < ∂. So choose t with 0 < |t| < ∂. Suppose without loss of generality that t > 0. Then f is continuous on [0, t] and differentiable on (0, t); hence by the Mean Value Theorem there exists an x in (0, t) for which f '(x) = [f(t) - f(0)] / [t -0]. Since 0 < |x| < ∂, we have

    |f '(x) - 3| = |[f(t) - f(0)] / [t -0] - 3| < ε.​

    Thus 0 < |t| = |t - 0| < ∂ implies |[f(t) - f(0)] / [t -0] - 3| < ε, which means f '(0) = 3.






    Now rip that apart. Pretend that you're my nitpicky TA grading my paper and in the process trying to degrade me (pun intended).
     
    Last edited: Jan 6, 2012
  2. jcsd
  3. Jan 6, 2012 #2
    That should be "whenever 0<|t|< ∂

    It's the other way around, it has to be continuous on [0,t] and differentiable on (0,t)

    You should mention the mean-value theorem here.

    Before "which mean..." I'dd add that it actually means that

    [tex]\lim_{t\rightarrow 0}{\frac{f(t)-f(0)}{t-0}}=3[/tex]


    Overall, the proof seems to be nice.
     
  4. Jan 6, 2012 #3
    Thanks, brah. When I write my final draft, I'll fancy it up by adding archaic words such as "whence" and "thereupon." This particular homework assignment will be golden.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook




Loading...