Showing that $\lim_{x\to c} f(x) > \alpha \Rightarrow f(c+h) > \alpha$ for $|h| < \delta $ where $\delta > 0 $
Solution 1:
By limit definition $\forall \epsilon>0, \exists \delta>0, |x-c|<\delta \implies |f(x)-L|<\epsilon$. By hypothesis, we are given $L>\alpha$.
$L-\epsilon<f(x)<L+\epsilon$, rewriting the limit definition as a compound inequality.
So:
$L-\epsilon-\alpha<f(x)-\alpha<L+\epsilon-\alpha$.
Select $\epsilon<L-\alpha$. Then we guarantee $f(x)-\alpha>0$ for some $\delta$ and $c-\delta<x<c+\delta$ by the earlier statements. This inequality still holds for $c-h<x<c+h, |h|<\delta$ since it overlaps.