Show that this function is differentiable

[Oats]

Homework Statement

Exercise 4.2.4.

[/B]

Let $I \subseteq \mathbb{R}$ be an open interval, let $c \in I$, and let $f:I \rightarrow \mathbb{R}$. Suppose that $|f(x)| \leq (x - c)^2$ for all $x \in I$. Prove that $f$ is differentiable at $c$ and $f'(c) = 0$.

2. The attempt at a solution
I'm not really sure where to start. We just want to show that $\lim_{x \to c} \frac{f(x) - f(c)}{x - c} = 0$. I see that $\lim_{x \to c} (x - c)^2 = 0$. I feel that this may be a simple trick of inequalities, but I am having a complete brain fart at the moment. Can anyone provide any direction? Thanks in advance for any response.

 

[S.G. Janssens]

What is $f(c)$ equal to? Can you then find an upper bound for $\frac{|f(x) - f(c)|}{|x - c|}$?

 

[Oats]

What is $f(c)$ equal to? Can you then find an upper bound for $\frac{|f(x) - f(x)|}{|x - c|}$?

Aha, yes! I didn't even consider finding what $f(c)$ equals. Having $|f(x)| \leq (x - c)^2$ for all $x \in I$ forces $f(c) = 0$. Now, to find $f'(c)$ we must now find $$\lim_{x \to c} \frac{f(x)}{x - c}$$, which we can show to be $0$, if we can alternatively show that $$\lim_{x \to c}|\frac{f(x)}{x-c}| = \lim_{x \to c}\frac{|f(x)|}{|x-c|} = 0$$. Now, $|f(x)| \leq (x - c)^2$ for all $x \in I$, also implies that for $x \in I-\{c\}$, $\frac{|f(x)|}{|x-c|} \leq |x - c|$. Since the latter's limit is 0, and both are point-wise positive, we have that $$\lim_{x \to c}|\frac{f(x)}{x-c}| = 0$$, and we are done.