# Proving $g(c)$ isn't equal to $g(x)$ in delta-neighborhood

I was solving Stephen Abbot exercises and I came across this one I'm struggling with:

Assume $g$ is differentiable at point $c \in (a,b)$. If $g'(c) \neq 0$ show that there exists $\delta$ neighborhood around $c \in (a,b)$ for which $g(x)\neq g(c), \forall x$ in that neighborhood .

Although I think I understand intuition behind this, I still don't now how to approach this problem. I think I have to prove it by contradiction, assuming that for all $\delta$ neighborhoods around $c$, $g(x) = g(c)$ for at least one $x$, but I don't know where to go starting from here. I think, at some point I'll have to end up stating that in that case $g'(c)$ has to be equal to zero, which would be the desired contradiction.

I also tried to build sequences in $(a,b)$ that converge to $c$, but apparently that didn't help at all.

• You have all the right ideas to solve the problem written in your question. I think you can do it. – Lorenzo Najt May 6 '18 at 19:24

Your approach works indeed: suppose for sake of contradiction that for all $\delta$-neighbourhoods of $c$, there is an $x \ne c$ in said neighbourhood such that $g(x) = g(c)$. In particular, we can find for all integers $n$ big enough an $x_n \in (c - 1/n, c + 1/n)$ such that $x_n \ne c$ and $g(x_n) = g(c)$. Here, big enough means such that $(c - 1/n, c + 1/n) \subseteq (a, b)$. Now, what can you say about the sequence $(x_n)_n$?
• It seems that this sequence converges to c. So it is enough to state that $g'(c) = \lim_{x \to \infty }\frac{g(x_{n})-g(c)}{x_{n}-c}$ = 0 ? – Arci May 6 '18 at 20:15
You know that$$\lim_{x\to c}\frac{g(x)-g(c)}{x-c}=g'(c)\neq0.$$Take $\delta>0$ such that$$|x-c|<\delta\implies\left|\frac{g(x)-g(c)}{x-c}-g'(c)\right|<\bigl|g'(c)\bigr|.$$Then$$|x-c|<\delta\implies\frac{g(x)-g(c)}{x-c}\neq0\implies g(x)\neq g(c).$$