I can't find the contradiction in this proof

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

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You do not seem to understand the definition of convergence (the one you gave does not make sense, what is the connection between $x$ and $n$ ?). More precisely, you seem to mix the definition of convergence for a sequence, and for a function. – beauby Nov 23 '12 at 0:41

2 Answers

Here's the proof:

Suppose that $\lim_{x \to b} f(x) = L > u$. Then $L - u = \epsilon > 0$. Find a $\delta > 0$ so that if $0 < |x - b| < \delta$, then $|f(x) - L| < \epsilon$. So, $L - u > L - f(x)$. Rearranging, $f(x) > u$, which is a contradiction.

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It is not a contradiction, and the proposed proof does not work.

Suppose, as you did, that $L\gt u$. Let $\epsilon=\frac{L-u}{2}$. There are $x$ such that $|L-f(x)|\lt \epsilon$. This condition forces $f(x)$ to be bigger than $L-\epsilon$, which is bigger than $u$. (Draw an informal sketch as a guide. On a line, put down $u$, $L$, and $L-\epsilon$.)

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