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## Homework Statement

Let f: R -> R be a function such that [tex] \lim_{z\to 0^+} zf(z) \gt 0 [/tex] Prove that there is no function g(x) such that g'(x) = f(x) for all x in R.

## Homework Equations

Supposed to use the mean value theorem. If f(x) is continuous on [a,b] and differentiable on (a,b) then [tex] \exists c \in (a,b), f'(c) = \frac{f(b) - f(a)}{b-a}[/tex]

## The Attempt at a Solution

Assume there exists a function g(x) such that g'(x) = f(x) for all x in R. By the mean value theorem:

[tex] \exists c \in (0,x), f(c) = g'(c) = \frac{g(x) - g(0)}{x - 0}[/tex]

This means [tex] \lim_{x\to 0^+} \left(\frac{g(x) - g(0)}{x} - f(c) \right)= 0

\\

\lim_{x\to 0^+} \left(g(x) - g(0) - xf(c)\right) = 0

\\

\lim_{x\to 0^+} \left(g(x) - g(0)\right) = \lim_{x\to 0^+} xf(c)

[/tex]Since g(x) is differentiable everywhere (and hence continuous), the left side is equal to 0.

Now the right side is where I'm not completely comfortable. If it was the limit of x(fx) then it's exactly what I'm given in the question. I'm thinking since x is going to 0 from the right then c also has to be going to 0 and so the limit on the right is greater than 0, which is a contradiction.

Can anyone tell me if I'm using faulty reasoning? I'm not really looking for the correct answer if I'm wrong. I'd just like to know where the flaw in my logic is.

Thanks