# Proof that a limit is less than or equal to zero

## 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

$f'(0)\leq0$

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

$lim _{x\rightarrow0} \frac{f(x)-f(0)}{x-0}\leq0$

Any help would be very appreciated. Thanks.

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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?

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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!