If $f(x) \leq M$ for all $x \in\mathbb{R}$ and $\lim_{x\to c} f(x) = L$, show that $L \leq M$.

Question

The problem makes sense intuitively, but what's a good, logical proof for it? Can you explain how to use Epsilon Delta?

(This was on stack overflow originally, but someone told me to post it here)

Answer 1

If $L>M$, then choose $\varepsilon>0$ small enough that $L-\varepsilon>M$.

Since $\lim_{x\to c}f(x)=L$, there exists $\delta>0$ such that if $|x-c|<\delta$ then $|f(x)-L|<\varepsilon$. It follows that $$ f(x)=L+f(x)-L>L-\varepsilon>M $$ if $|x-c|<\delta$, contrary to the hypothesis that $f(x)\leq M$ for all $x$.

Answer 2

Assume towards a contradiction $L > M$. $f(x)\rightarrow L$, therefore $\forall \epsilon>0$, $\exists \delta$ s.t. $|x-c|<\delta\implies |f(x)-L|<\epsilon$.

Specifically, choose $\epsilon=(L-M)/2$. Therefore $\exists \delta$ s.t. $|x-c|<\delta\implies |f(x)-L|<(M-L)/2\implies f(x)>M$. Which is a contradiction. Therefore we must have $L\leq M$.

Answer 3

Suppose that $L>M$. Since $f(x) \to L$ as $x\to c$, it follows that the values of $f$ can be as near to $L$ as we please. And if we get too near to $L$ we eventually exceed $M$ because $L>M$ and this contradicts $f(x) \leq M$ for all $x$. Epsilon delta stuff is used to convert this simple and obvious argument presented above into a high brow technical argument without adding any extra rigor. One should avoid Epsilon delta for simpler problems unless specifically asked in exam.