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I have an exercise that reads:

Suppose $f:\mathbb{R}\rightarrow\mathbb{R}$ satisfies $0\leq f(x)\leq x^2$ for all $x\in \mathbb{R}$. What is $f(0)$? Show that $f$ is differentiable at $0$ and find $f'(0)$.

Here is my proof...

Part (a)

Since $0\leq f(x) \leq x^2$, we have $$\lim\limits_{x\rightarrow 0^+}0 \leq \lim\limits_{x\rightarrow 0^+}f(x) \leq \lim\limits_{x\rightarrow 0^+}x^2$$ $$0\leq \lim\limits_{x\rightarrow 0^+} f(x) \leq 0$$ and we have that $$\lim\limits_{x\rightarrow 0^-}0 \leq \lim\limits_{x\rightarrow 0^-}f(x) \leq \lim\limits_{x\rightarrow 0^-}x^2$$ $$0\leq \lim\limits_{x\rightarrow 0^-} f(x) \leq 0$$ we can conclude that $$\lim\limits_{x\rightarrow 0^+}0 \leq \lim\limits_{x\rightarrow 0^+}f(x) \leq \lim\limits_{x\rightarrow 0^+}x^2$$ $$\lim\limits_{x\rightarrow 0}f(x)=f(0)=0$$ by the Squeeze Theorem.

Part (b)

Since $0\leq f(x) \leq x^2$, we have $$\frac{0-f(0)}{x-0}\leq \frac{f(x)-f(0)}{x-0} \leq \frac{x^2-f(0)}{x-0}$$ $$0\leq \frac{f(x)-f(0)}{x-0} \leq x$$ further, we have that $$\lim\limits_{x\rightarrow 0^+}0\leq \lim\limits_{x\rightarrow 0^+}\frac{f(x)-f(0)}{x-0} \leq \lim\limits_{x\rightarrow 0^+}x$$ $$0\leq \lim\limits_{x\rightarrow 0^+}\frac{f(x)-f(0)}{x-0} \leq 0$$ and we have that $$\lim\limits_{x\rightarrow 0^-}0\leq \lim\limits_{x\rightarrow 0^-}\frac{f(x)-f(0)}{x-0} \leq \lim\limits_{x\rightarrow 0^-}x$$ $$0\leq \lim\limits_{x\rightarrow 0^-}\frac{f(x)-f(0)}{x-0} \leq 0$$ we can conclude that $$\lim\limits_{x\rightarrow 0}\frac{f(x)-f(0)}{x-0}=f'(0)=0$$ by the Squeeze Theorem.

Is my proof for this correct?

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For part (a) you don't need to take a limit, just plug $x=0$ into the inequality $0 \leq f(x) \leq x^2$. – Antonio Vargas Dec 13 '12 at 6:50
Yes...the obvious always seems to evade me. Is part (b) correct then? – kaiserphellos Dec 13 '12 at 6:51
It seems to be correct. – Joe Z. Dec 13 '12 at 6:52
In (b) the inequalities should be reversed for negative $x$. Argument still works. – André Nicolas Dec 13 '12 at 6:55
Is it necessary to include both the left and right-hand limits in the proof of part (b)? That is, if we know that $\lim\limits_{x\rightarrow 0}x=0$ is it necessary to break the proof into left and right-hand limits? – kaiserphellos Dec 13 '12 at 6:55

The first part follows from evaluation: By assumption $0 \leq f(0) \leq 0$, so $f(0) = 0$.

I think the second part is more straightforward if you start from the definition of derivative.

You have $|\frac{f(x)}{x}|\le |x|$, hence $\lim_{x\to 0} \frac{f(x)}{x} = 0$. Since $f'(0) = \lim_{x\to 0} \frac{f(x)-f(0)}{x}$, and $f(0) = 0$, we have $f'(0) = 0$.

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