I am studying from the book "Introduction to Real Analysis" by Bartle and Sherbert. In the proof of the Interior Extremum Theorem, to prove that $f'(c) = 0$, they rule out the other two possibilities by deriving contradictions whenever $f'(c) > 0$ and $f'(c) < 0$. To do so, they use the fact (theorem 4.2.9) that if the limit of a function $f$ is strictly positive at a cluster point, say $c$, then there exists a $\delta$-neighbourhood of $c$ such that $f(x) > 0$ in that neighbourhood. Assuming for a contradiction that,
$f'(c) > 0$ which means : $\lim_{x \rightarrow c} \dfrac{f(x) - f(c)}{x - c} > 0$
Since this limit is strictly positive, there must exist a $V_{\delta}(c)$ such that the difference quotient is strictly positive. Then they make the statement that if $x \in V_{\delta}(c)$ and $x > c$... we get a contradiction. At this point why do we disregard the possibility that $x < c$? In order for a quotient to be positive both numerator and denominator must have the same sign. Since $f(x) < f(c)$ because $f(c)$ is a local maximum, the denominator must also be less than $0$,thus forcing $x < c$.
Doesn't the contradiction in fact show that we may not assume $x>c$? A similar case holds when $f'(c) < 0$ where, we need opposite signs forcing $x > c$. Having forced this condition on $x$, should we not proceed further (I don't know how) to somehow show that even with this restriction on $x$, we still get some contradiction, and thus rule out $f'(c) > 0$? What mistake am I making, or what fact am I overlooking?
Please help. Thank you.
