# Finding the flaw in this continuity proof?

1. Mar 10, 2009

### Mathmos6

1. The problem statement, all variables and given/known data

“Let f ′ exist on (a, b) and let c ∈ (a, b) . If c + h ∈ (a, b) then (f (c + h) − f (c))/h = f ′(c+θh). Let h→0 ; then f ′ (c + θh) → f ′ (c) . Thus f ′ is continuous at c .” Is this argument correct?

3. The attempt at a solution

I'm pretty sure the argument's wrong - is the flaw simply that in saying f ′ (c + θh) → f ′ (c) you have assumed the result that f' is continuous at c in proving the result? I.e proving lim f'(x) as x→c=f'(c) by assuming lim f'(x) as x→c=f'(c)? Or is there a different error in the statement?

2. Mar 10, 2009

### CompuChip

What does that statement mean?
What is θ?

It is true that
$$\lim_{h \to 0} \frac{f(c+h) - f(c)}{h} = f'(c)$$
if f is differentiable at c (which I suppose you were assuming anyway, if you want to prove something about f') but that has a limit that the quoted statement does not have, and lacks a θ.

3. Mar 10, 2009

### Office_Shredder

Staff Emeritus
$$\theta$$ is a number between 0 and 1. In this case, you're not taking the limit, you're simply starting by looking at a small value of h, and noting by the mean value theorem,

$$\frac{f(c+h) - f(c)}{h} = f'(c + \theta h)$$ for some theta(endpoints c, c+h, c+h-c = h and somewhere between c and c+h the derivative equals the left hand side, we describe that point by adding a portion of h onto c).

Now when you take the limit, you see $$f'(c) = \lim_{h \rightarrow 0} f(c+ \theta h)$$

and since theta is in between 0 and 1 (it varies as h varies, but doesn't matter), we expect that to approach f(c). But wait! We can only pass the limit into the argument when f' is continuous (that's a definition of continuity in fact). So you're right