Given $f(x)$ is differentiable at $x=0$ and $f(0)=0$ and $f'(0) > 0$
Prove that there exists an $\varepsilon > 0$ such that $f(x) > 0$ for all $0<x<\varepsilon$
My proof:
$$ f'(0) = \lim_{x \to 0} \frac{f(x) -f(0)}{x-0} = \lim_{x \to 0} \frac{f(x) - 0}{x} = \lim_{x \to 0} \frac{f(x)}{x} = f'(0) > 0$$
Thus:
$$ \frac{f(x)}{x} > 0 $$
So, assuming there is no such $\varepsilon>0$, it means our function is always negative ($f(x)<0$) otherwise we would have such $\varepsilon$ right?
But, if $x>0$ and $f(x)<0$ (because it is negative for all $x$'s) it means that $f(x)/x < 0$ contradicting the given information we got from the limit above...
I am really unsure of this proof, why does it seems very wrong? intuitively, I don't know where the red line is being drawn when talking about proving by contradiction..
Thank you!
Yes, you correctly sense that the negation of "$f(x)$ is positive in some right-sided neighborhood of $0$" is not "$f(x)$ is negative in some right-sided neighborhood of $0$."
The correct negation is "every right-sided neighborhood of $0$ contains a point $a$ at which $f(x)$ is nonpositive." This negation would imply that there exists a strictly decreasing sequence of such points, $a_{n} > 0$, converging to $0$: $$ {f(a_{n}) \over a_{n}} \leq 0 \mbox{ for all natural numbers }n. \quad \mbox{(*)} $$ Knowing that $f$ is differentiable at zero, the limit $$ \lim_{n \rightarrow +\infty} {f(a_{n}) \over a_{n}} $$ exists and equals $f'(0)$, which is positive. But, by inequality (*), this limit must be nonpositive, which is the contradiction you were after.