(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Let f be a continuos real function onR, 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?^{1}

2. Relevant equations

Mean Value Theorem

definition of f '(x)

3. The attempt at a solution

Fix ε > 0. Since lim_{x → 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).

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

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