The question is as follows :
Let f : (a,b) $\to \mathbb{R}$ and c $\epsilon$ (a,b) be such that $\lim_{x\to c} f(x) > \alpha$. Prove that there exists some $\delta > 0 $ such that $f(c+h) >\alpha $ for all $|h| < \delta $
I have taken a rather lengthy approach of splitting it into 9 different cases as follows, but I am happy about my method.
Case 1 : monotonically increasing
Case 2 : monotonically decreasing
Case 3 : constant function
Case 4 : x is a point of minima
Case 5 : x is a point of maxima
Case 6 : constant function to the left of x = c and increasing to the right of x = c
Case 7 : constant function to the right of x = c and increasing to the left of x = c
Case 8 : constant function to the left of x = c and decreasing to the right of x = c
Case 9 : constant function to the right of x = c and decreasing to the left of x = c
I have managed to prove all cases but case 5 with ease.
I am unsure of how to proceed with a formal proof of case 5.
For all other cases I have used the basic definitions of increasing/decreasing/constant.
Can someone guide me how to proceed with proof of case 5 ?
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.