Hi,
I am having a hard time trying to prove this.
If,
lim f(x) = L and f(x)>0, then L>=0.
x->a
(I am supposed to prove it by contradiction)
If $\displaystyle L<0$ then there exists $\displaystyle \epsilon > 0$ so that $\displaystyle L+\epsilon < 0$. This means there exists $\displaystyle \delta > 0$ so that for all $\displaystyle 0<|x-a|<\delta$ we have $\displaystyle f(x) < L + \epsilon < 0$ and so $\displaystyle f(x) < 0$ on $\displaystyle (a-\delta,a+\delta)\setminus \{ a\}$.