Proof that a limit is less than or equal to zero

[Username007]

Homework Statement

Proof that, If f is a function such that

(1) f is differentiable at (open) the interval D,

(2) D includes 0 and f(0)=0, and

(3) for all x in D other than 0, f(x) and x have opposite signs

Then

f'(0)\leq0

Homework Equations

None.

The Attempt at a Solution

I managed to prove that for all x in D other than 0

\frac{f(x)-f(0)}{x-0}\leq0

I don't know how to get from there to the fact that

lim _{x\rightarrow0} \frac{f(x)-f(0)}{x-0}\leq0Any help would be very appreciated. Thanks.

 

[Axiomer]

If for some function g(x) we have lim _{x\rightarrow0}g(x)=L>0, then can you argue that g(x) must be positive in some (-δ,δ)\{0} by considering a certain ε>0?

 

[Username007]

Okay, I got it. If the limit equals k and k>0, then

\forallε>0\:\existsδ>0 (|x|<δ \rightarrow \left|\frac{f(x)-f(0)}{x-0}-k\right|<ε)

implies (for ε=k/2) that

\existsδ>0 (|x|<δ \rightarrow \frac{k}{2}<\frac{f(x)-f(0)}{x-0}<\frac{3k}{2})

But \frac{k}{2}<\frac{f(x)-f(0)}{x-0}<\frac{3k}{2} cannot be true at (0,δ), because it would contradict statement (3).

Thanks!