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Let $a>1$, let $x\in\mathbb{Q}$, and define $f(x)=x^a$. I am interesting in computing $f'(0)$ if it exists. I claim that $f'(0)=0$.

Attempt: Let $\epsilon > 0$. Suppose $0 < \lvert x-0 \rvert$ = $\lvert x \rvert$ < $\delta$. Consider $$\left\lvert \frac{f(x)-f(0)}{x-0}-L \right\rvert = \left\lvert \frac{x^a-0^a}{x-0}-0 \right\rvert = \left\lvert \frac{x^a}{x} \right\rvert = \left\lvert x^{a-1} \right\rvert$$ My initial thought was to let $\delta$ := min{1,$\epsilon$} so that $$\left\lvert x^{a-1} \right\rvert \leq \lvert x \rvert < \delta \leq \epsilon$$ however I think this fails if $1<a<2$. Is there a way to define $\delta$ that avoids having to worry about the value of $a$ or should I break the problem into cases?

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  • $\begingroup$ @ 5xum: Because $\displaystyle\left| x^{1.5-1} \right| \not\le \left| x \right|$ if $0 < x < 1$. ($a=1.5$ here.) $\endgroup$ – Christopher Carl Heckman Sep 3 '15 at 7:24
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    $\begingroup$ $\delta=\epsilon^{1/(a-1)}$. $\endgroup$ – Did Sep 3 '15 at 8:22
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Since x belong to the set of rational numbers, you can prove rigorously that f(x) is not continuous at x = 0, and hence f'(0) does not exist.

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