The problem is "You have a function $f:(a,b) \rightarrow \mathbb{R}$. $\exists u \in$ $\mathbb{R}$ such that $f(x) < u \forall x \in (a,b)$. Prove that if the limit of f(x) as x approaches b exists, then the limit must be $\leq$ $u$."
Proof: We proceed by contradiction. Assume that the limit exists and call it $L$.
Also assume that $L > u$.
Then we know that $f(x) < u < L$. Here is where I don't understand what my friend is doing: "Since $f(x) < L,$ we can pick an $N \in \mathbb{N}$ such that $\forall n \in \mathbb{N}$ with $n > N$, $f(n) > f(x)$, which is a contradiction."
How is that a contradiction? The definition of convergence states that a function f(x) converges to L if for all epsilon greater than zero, there is an N such that for all n > N, |f(x) - L| < e.
It seems to me that this supposed contradiction could be rearranged as $f(x) - f(n) < 0$, which seems allowable.
Where am I going wrong? Thanks