Suppose $f(x) \to L$, show that for some $h >0, f(x) > 0$ using $\epsilon, \delta$ 
Suppose $f(x) \to L$, as $x \to \alpha$. if $L>0$ show that for some $h >0, f(x) > 0$, and what happens if $L<0$?

my take:
for any $\epsilon >0$ there exists $\delta >0$ so that $|f(x)-L| < \epsilon$ provided $0 < |x-\alpha| < \delta$. Since this is true for all $\epsilon >0$, it is true for $\epsilon = L$ and say $\delta = h$, then we can say $|f(x)-L| < L$ which leads to $0 < f(x) < 2L$ provided with $0 < |x-\alpha| < h$ then $-h+\alpha < x < h+\alpha$ and we are done.
Did I do this correctly?? Also, if $L<0$ do we say ..for some $h<0$ and do the same thing as $L>0$?

Your proof is correct - you have found an $h$.
Here's how to start attacking the rest of the problem.
"[Find out] what happens when $L < 0$" asks you to rewrite the end of the sentence to make a true statement for that case. A picture and your intuition should tell you the assertion is false. The opposite of "for every $h > 0$" starts "for some $h > 0$" ...
Finish that sentence correctly and you should be able to write the proof you need.
