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

1. Jan 6, 2012

### Jamin2112

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. Jan 6, 2012

### micromass

Staff Emeritus
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

$$\lim_{t\rightarrow 0}{\frac{f(t)-f(0)}{t-0}}=3$$

Overall, the proof seems to be nice.

3. Jan 6, 2012

### Jamin2112

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.